Chapter 19
The Second Law of Thermodynamics
Conceptual Problems
1
•
Modern automobile gasoline engines have efficiencies of about 25%.
About what percentage of the heat of combustion is not used for work but
released as heat? (a) 25%, (b) 50%, (c) 75%, (d) 100%, (e) You cannot tell from
the data given.
Determine the Concept The efficiency of a heat engine is the ratio of the work
done per cycle W to the heat absorbed from the high-temperature reservoir Qh.
The percentage of the heat of combustion (heat absorbed from the hightemperature reservoir) is the ratio of Qc to Qh. We can use the relationship
between W, Qh, and Qc ( W = Qh − Qc ) to find Qc/ Qh.
Use the definition of efficiency and
the relationship between W, Qh, and
Qc to obtain:
ε=
Q
W Qh − Qc
=
= 1− c
Qh
Qh
Qh
Solving for Qc/ Qh yields:
Qc
= 1− ε
Qh
Substitute for ε to obtain:
Qc
= 1 − 0.25 = 0.75
Qh
and
(c )
is correct.
2
•
If a heat engine does 100 kJ of work per cycle while releasing 400 kJ
of heat, what is its efficiency? (a) 20%, (b) 25%, (c) 80%, (d) 400%, (e) You
cannot tell from the data given.
Determine the Concept The efficiency of a heat engine is the ratio of the work
done per cycle W to the heat absorbed from the high-temperature reservoir Qh. We
can use the relationship between W, Qh, and Qc ( W = Qh − Qc ) to express the
efficiency of the heat engine in terms of Qc and W.
Use the definition of efficiency and
the relationship between W, Qh, and
Qc to obtain:
ε=
1867
W
W
=
=
Qh W + Qc
1
Q
1+ c
W
1868 Chapter 19
Substitute for Qc and W to obtain:
1
= 0.2
400 kJ
1+
100 kJ
and (a ) is correct.
ε=
3
•
If the heat absorbed by a heat engine is 600 kJ per cycle, and it
releases 480 kJ of heat each cycle, what is its efficiency? (a) 20%, (b) 80%,
(c) 100%, (d) You cannot tell from the data given.
Determine the Concept The efficiency of a heat engine is the ratio of the work
done per cycle W to the heat absorbed from the high-temperature reservoir Qh. We
can use the relationship between W, Qh, and Qc ( W = Qh − Qc ) to express the
efficiency of the heat engine in terms of Qc and Qh.
Use the definition of efficiency and
the relationship between W, Qh, and
Qc to obtain:
Substitute for Qc and Qh to obtain:
4
•
ε=
Q
W Qh − Qc
=
= 1− c
Qh
Qh
Qh
480 kJ
= 0.2
600 kJ
and (a ) is correct.
ε = 1−
Explain what distinguishes a refrigerator from a ″heat pump.″
Determine the Concept The job of a refrigerator is to move heat from its cold
interior to the warmer kitchen environment. This process moves heat in a
direction that is opposite its ″natural″ direction of flow, analogous to the use of a
water pump to pump water out of a boat. The term heat pump is used to describe
devices, such as air conditioners, that are used to cool living and working spaces
in the summer and warm them in the winter.
An air conditioner’s COP is mathematically identical to that
Q
of a refrigerator, that is, COPAC = COPref = c . However a heat pump’s COP is
W
Q
defined differently, as COPhp = h . Explain clearly why the two COPs are
W
defined differently. Hint: Think of the end use of the three different devices.
5
•
[SSM]
Determine the Concept The COP is defined so as to be a measure of the
effectiveness of the device. For a refrigerator or air conditioner, the important
quantity is the heat drawn from the already colder interior, Qc. For a heat pump,
the ideas is to focus on the heat drawn into the warm interior of the house, Qh.
The Second Law of Thermodynamics 1869
6
•
Explain why you cannot cool your kitchen by leaving your refrigerator
door open on a hot day. (Why does turning on a room air conditioner cool down
the room, but opening a refrigerator door does not?)
Determine the Concept As described by the second law of thermodynamics,
more heat must be transmitted to the outside world than is removed by a
refrigerator or air conditioner. The heating coils on a refrigerator are inside the
room and so the refrigerator actually heats the room in which it is located. The
heating coils on an air conditioner are outside one’s living space, so the waste
heat is vented to the outside.
7
•
Why do steam-power-plant designers try to increase the temperature
of the steam as much as possible?
Determine the Concept Increasing the temperature of the steam increases the
Carnot efficiency, and generally increases the efficiency of any heat engine.
8
•
To increase the efficiency of a Carnot engine, you should
(a) decrease the temperature of the hot reservoir, (b) increase the temperature of
the cold reservoir, (c) increase the temperature of the hot reservoir, (d) change the
ratio of maximum volume to minimum volume.
Determine the Concept Because the efficiency of a Carnot cycle engine is given
T
by ε C = 1 − c , you should increase the temperature of the hot reservoir. (c ) is
Th
correct.
9
•• [SSM] Explain why the following statement is true: To increase the
efficiency of a Carnot engine, you should make the difference between the two
operating temperatures as large as possible; but to increase the efficiency of a
Carnot cycle refrigerator, you should make the difference between the two
operating temperatures as small as possible.
Determine the Concept A Carnot-cycle refrigerator is more efficient when the
temperatures are close together because it is easier to extract heat from an already
cold interior if the temperature of the exterior is close to the temperature of the
interior of the refrigerator. A Carnot-cycle heat engine is more efficient when the
temperature difference is large because then more work is done by the engine for
each unit of heat absorbed from the hot reservoir.
10 ••
A Carnot engine operates between a cold temperature reservoir of
o
27 C and a high temperature reservoir of 127°C. Its efficiency is (a) 21%,
(b) 25%, (c) 75%, (d) 79%.
1870 Chapter 19
Determine the Concept The efficiency of a Carnot cycle engine is given by
T
ε C = 1 − c where Tc and Th (in kelvins) are the temperatures of the cold and hot
Th
reservoirs, respectively.
Substituting numerical values for Tc
and Th yields:
300 K
= 0.25
400 K
is correct.
εC = 1−
(b )
11 •• The Carnot engine in Problem 10 is run in reverse as a refrigerator. Its
COP is (a) 0.33, (b) 1.3, (c) 3.0 (d) 4.7.
Determine the Concept The coefficient of performance of a Carnot cycle engine
Q
run in reverse as refrigerator is given by COPref = c . We can use the relationship
W
between W, Qc, and Qh to eliminate W from this expression and then use the
Q
T
relationship, applicable only to a device operating in a Carnot cycle, c = c to
Qh Th
express the refrigerator’s COP in terms of Tc and Th.
The coefficient of performance of a
refrigerator is given by:
Qc
W
or, because W = Qh − Qc ,
COPref =
COPref =
Qc
Qh − Qc
Dividing the numerator and
denominator of this fraction by Qc
yields:
COPref =
For a device operating in a Carnot
cycle:
Qc Tc
=
Qh Th
Substitute in the expression for
COPref to obtain:
COPref, C =
1
Qh
−1
Qc
1
Th
−1
Tc
The Second Law of Thermodynamics 1871
Substitute numerical values and
evaluate COPref, C:
1
= 3.0
400 K
−1
300 K
(c ) is correct.
COPref, C =
12 ••
On a humid day, water vapor condenses on a cold surface. During
condensation, the entropy of the water (a) increases, (b) remains constant,
(c) decreases, (d) may decrease or remain unchanged. Explain your answer.
Determine the Concept When water vapor condenses, its entropy decreases (the
liquid state is a more ordered state than is the vapor state) and the entropy of the
universe increases. (a ) is correct.
13 ••
An ideal gas is taken reversibly from an initial state Pi, Vi, Ti to the
final state Pf, Vf, Tf. Two possible paths are (A) an isothermal expansion followed
by an adiabatic compression and (B) an adiabatic compression followed by an
isothermal expansion. For these two paths, (a) ΔEint A > ΔEint B, (b) ΔSA > ΔSB,
(c) ΔSA < ΔSB, (d) None of the above.
Determine the Concept The two paths
are shown on the PV diagram to the
right. We can use the concept of a state
function to choose from among the
alternatives given as possible answers
to the problem.
P
B
B
Ti
f
Pf
Pi
i
A
Tf
A
Vi
Vf
V
(a) Because Eint is a state function and the initial and final states are the same for
the two paths, ΔEint, A = ΔEint, B .
(b) and (c) S, like Eint, is a state function and its change when the system moves
from one state to another depends only on the system’s initial and final states. It
is not dependent on the process by which the change occurs. Thus ΔS A = ΔS B .
(d) (d ) is correct.
Figure 19-12 shows a thermodynamic cycle for an ideal gas on an ST
14 ••
diagram. Identify this cycle and sketch it on a PV diagram.
1872 Chapter 19
Determine the Concept The processes
A→B and C→D are adiabatic and the
processes B→C and D→A are
isothermal. Therefore, the cycle is the
Carnot cycle shown in the adjacent PV
diagram.
P
B
C
A
D
V
15 ••
Figure 19-13 shows a thermodynamic cycle for an ideal gas on an SV
diagram. Identify the type of engine represented by this diagram.
Determine the Concept Note that A→B is an adiabatic expansion, B→C is a
constant-volume process in which the entropy decreases, C→D is an adiabatic
compression and D→A is a constant-volume process that returns the gas to its
original state. The cycle is that of the Otto engine (see Figure 19-3). The points A,
B, C, and D in Figure 19-13 correspond to points c, d, a, and b, respectively, in
Figure 19-3.
16 ••
Sketch an ST diagram of the Otto cycle. (The Otto cycle is discussed
in Section 19-1.)
Determine the Concept The Otto cycle consists of four quasi-static steps. Refer
to Figure 19-3. There a→b is an adiabatic compression, b→c is a constant volume
heating, c→d is an adiabatic expansion and d→a is a constant-volume cooling.
So, from a to b, S is constant and T increases, from b to c, heat is added to the
system and both S and T increase, from c→d S is constant while T decreases, and
from d to a both S and T decrease.
To determine how S depends on T
along b→c and d→a, consider the
entropy change of the gas from point
b to an arbitrary point on the path
b→c where the entropy and
temperature of the gas are S and T,
respectively:
Q
T
where, because heat is entering the
system, Q is positive.
Because Won = 0 for this constantvolume process:
ΔEint = Qin = Q = C V ΔT = C V (T − Tb )
Substituting for Q yields:
ΔS =
ΔS =
C V (T − Tb )
⎛ T ⎞
= C V ⎜1 − b ⎟
T
⎝ T ⎠
The Second Law of Thermodynamics 1873
On path b→c the entropy is given by:
⎛ T ⎞
S = S b + ΔS = S b + C V ⎜1 − b ⎟
⎝ T ⎠
The first and second derivatives,
dS dT and d 2 S dT 2 , give the slope
and concavity of the path. Calculate
these derivatives assuming CV is
constant. (For an ideal gas CV is a
positive constant.):
T
dS
= C V b2
dT
T
2
T
d S
= −2C V b3
2
dT
T
These results tell us that, along path b→c, the slope of the path is positive and the
slope decreases as T increases. The concavity of the path is negative for all T.
Following the same procedure on
path d→a gives:
⎛ T ⎞
S = S d + C V ⎜1 − d ⎟
⎝ T ⎠
T
dS
= C V d2
dT
T
2
T
d S
= −2C V d3
2
dT
T
These results tell us that, along path d→a, the slope of the path is positive and the
slope decreases as T increases. The concavity of the path is negative for all T.
An ST diagram for the Otto cycle is
shown to the right.
S
d
a
c
b
T
17 ••
[SSM]
Sketch an SV diagram of the Carnot cycle for an ideal gas.
Determine the Concept Referring to Figure 19-8, process 1→2 is an isothermal
expansion. In this process heat is added to the system and the entropy and volume
increase. Process 2→3 is adiabatic, so S is constant as V increases. Process 3→4
is an isothermal compression in which S decreases and V also decreases. Finally,
process 4→1 is adiabatic, that is, isentropic, and S is constant while V decreases.
1874 Chapter 19
During the isothermal expansion (from
point 1 to point 2) the work done by
the gas equals the heat added to the
gas. The change in entropy of the gas
from point 1 (where the temperature is
T1) to an arbitrary point on the curve is
given by:
ΔS =
For an isothermal expansion, the
work done by the gas, and thus the
heat added to the gas, are given by:
⎛V ⎞
Q = W = nRT1 ln⎜⎜ ⎟⎟
⎝ V1 ⎠
Substituting for Q yields:
⎛V ⎞
ΔS = nR ln⎜⎜ ⎟⎟
⎝ V1 ⎠
Since S = S1 + ΔS , we have:
⎛V ⎞
S = S1 + nR ln⎜⎜ ⎟⎟
⎝ V1 ⎠
S
The graph of S as a function of V for an
isothermal expansion shown to the right
was plotted using a spreadsheet
program. This graph establishes the
curvature of the 1→2 and 3→4 paths
for the SV graph.
Q
T1
V
An SV graph for the Carnot cycle
(see Figure 19-8) is shown to the
right.
S
2
1
3
4
V
18 •• Sketch an SV diagram of the Otto cycle. (The Otto cycle is discussed in
Section 19-1.)
The Second Law of Thermodynamics 1875
Determine the Concept The Otto cycle is shown in Figure 19-3. Process a→b
takes place adiabatically and so both Q = 0 and ΔS = 0 along this path. Process
b→c takes place at constant volume. Qin, however, is positive and so, while
ΔV = 0 along this path, Q > 0 and, therefore ΔS > 0. Process c→d also takes place
adiabatically and so, again, both Q = 0 and ΔS = 0 along this path. Finally, process
d→a is a constant-volume process, this time with heat leaving the system and
ΔS < 0. A sketch of the SV diagram for the Otto cycle follows:
S
c
d
b
a
V
19 ••
Figure 19-14 shows a thermodynamic cycle for an ideal gas on an SP
diagram. Make a sketch of this cycle on a PV diagram.
Determine the Concept Process A→B
is at constant entropy; that is, it is an
adiabatic process in which the pressure
increases. Process B→C is one in
which P is constant and S decreases;
heat is exhausted from the system and
the volume decreases. Process C→D is
an adiabatic compression. Process
D→A returns the system to its original
state at constant pressure. The cycle is
shown in the adjacent PV diagram.
P
C
B
D
A
V
20 ••
One afternoon, the mother of one of your friends walks into his room
and finds a mess. She asks your friend how the room came to be in such a state,
and your friend replies, ″Well, it is the natural destiny of any closed system to
degenerate toward greater and greater levels of entropy. That’s all, Mom.″ Her
reply is a sharp ″Nevertheless, you’d better clean your room!″ Your friend retorts,
″But that can’t happen. It would violate the second law of thermodynamics.″
Critique your friend’s response. Is his mother correct to ground him for not
cleaning his room, or is cleaning the room really impossible?
Determine the Concept The son is out of line, here, but besides that, he’s also
wrong. While it is true that systems tend to degenerate to greater levels of
1876 Chapter 19
disorder, it is not true that order cannot be brought forth from disorder. What is
required is an agent doing work – for example, your friend – on the system in
order to reduce the level of chaos and bring about order. His cleanup efforts will
be rewarded with an orderly system after a sufficient time for him to complete the
task. It is true that order will not come about from the disordered chaos of his
room – unless he applies some elbow grease.
Estimation and Approximation
21 •
Estimate the change in COP of your electric food freezer when it is
removed from your kitchen to its new location in your basement, which is 8°C
cooler than your kitchen.
Picture the Problem We can use the definition of the coefficient of performance
to express the ratio of the coefficient of performance in your basement to the
coefficient of performance in the kitchen. If we further assume that the freezer
operates in a Carnot cycle, then we can use the proportion Qh Qc = Th Tc to
express the ratio of the coefficients of performance in terms of the temperatures in
the kitchen, basement, and freezer.
The ratio of the coefficients of
performance in the basement and
kitchen is given by:
COPbasement
COPkit
Qc,basement
W
= c,basement
Qc,kit
Wc,kit
COPbasement
COPkit
Qc,basement
Q
− Qc,basement
= h,basement
Qc,kit
Qh,kit − Qc,kit
Because W = Qh − Qc for a heat
engine or refrigerator:
Divide the numerators and
denominators by Qc,basement and Qc,kit
and simplify to obtain:
1
COPbasement
COPkit
Qh,basement
−1
Qc,basement
=
1
Qh,kit
−1
Qc,kit
=
Qh,kit
−1
Qc,kit
Qh,basement
−1
Qc,basement
The Second Law of Thermodynamics 1877
If we assume that the freezer unit
operates in a Carnot cycle, then
Qh Th
=
and our expression for the
Qc Tc
ratio of the COPs becomes:
Assuming that the temperature in
your kitchen is 20°C and that the
temperature of the interior of your
freezer is −5°C, substitute numerical
values and evaluate the ratio of the
coefficients of performance:
Th,kit
−1
Tc,kit
COPbasement
=
Th,basement
COPkit
−1
Tc,basement
293 K
−1
COPbasement 268 K
= 1.47
=
285 K
COPkit
−1
268 K
or an increase of 47% in the
performance of the freezer!
22 ••
Estimate the probability that all the molecules in your bedroom are
located in the (open) closet which accounts for about 10% of the total volume of
the room.
Picture the Problem The probability that all the molecules in your bedroom are
N
⎛V ⎞
located in the (open) closet is given by p = ⎜⎜ 2 ⎟⎟ where N is the number of air
⎝ V1 ⎠
molecules in your bedroom and V1 and V2 are the volumes of your bedroom and
closet, respectively. We can use the ideal-gas law to find the number of molecules
N. We’ll assume that the volume of your room is about 50 m3 and that the
temperature of the air is 20°C.
If the original volume of the air in
your bedroom is V1, the probability
p of finding the N molecules,
normally in your bedroom, confined
to your closet whose volume is V2 is
given by:
⎛V ⎞
p = ⎜⎜ 2 ⎟⎟
⎝ V1 ⎠
or, because V2 = 101 V1 ,
Use the ideal-gas law to express N:
N=
Substitute numerical values and
evaluate N:
(
101.325 kPa ) (50 m 3 )
N=
(1.381×10−23 J/K )(293 K )
N
⎛1⎞
p=⎜ ⎟
⎝ 10 ⎠
N
PV
kT
= 1.252 × 1027 molecules
(1)
1878 Chapter 19
Substitute for N in equation (1) and
evaluate p:
1.252×1027
⎛1⎞
p=⎜ ⎟
⎝ 10 ⎠
≈ 10−10
=
1
101.252×10
27
= 10 −1.252×10
27
27
23 •• [SSM] Estimate the maximum efficiency of an automobile engine
that has a compression ratio of 8.0:1.0. Assume the engine operates according to
the Otto cycle and assume γ = 1.4. (The Otto cycle is discussed in Section 19-1. )
Picture the Problem The maximum efficiency of an automobile engine is given
by the efficiency of a Carnot engine operating between the same two
temperatures. We can use the expression for the Carnot efficiency and the
equation relating V and T for a quasi-static adiabatic expansion to express the
Carnot efficiency of the engine in terms of its compression ratio.
Express the Carnot efficiency of an
engine operating between the
temperatures Tc and Th:
Relate the temperatures Tc and Th to
the volumes Vc and Vh for a quasistatic adiabatic compression from Vc
to Vh:
Substitute for
Tc
to obtain:
Th
Express the compression ratio r:
Substituting for r yields:
Substitute numerical values for r and
γ (1.4 for diatomic gases) and
evaluate εC:
εC = 1 −
γ −1
TcVc
Tc
Th
γ −1
= ThVh
⎛V ⎞
ε C = 1 − ⎜⎜ h ⎟⎟
⎝ Vc ⎠
r=
T V γ −1 ⎛ V ⎞
⇒ c = hγ −1 = ⎜⎜ h ⎟⎟
Th Vc
⎝ Vc ⎠
γ −1
γ −1
Vc
Vh
εC = 1 −
ε C = 1−
1
r
γ −1
1
≈ 56%
(8.0)1.4−1
24 •• You are working as an appliance salesperson during the summer. One
day, your physics professor comes into your store to buy a new refrigerator.
Wanting to buy the most efficient refrigerator possible, she asks you about the
efficiencies of the available models. She decides to return the next day to buy the
most efficient refrigerator. To make the sale, you need to provide her with the
following estimates: (a) the highest COP possible for a household refrigerator,
and (b) and the highest rate possible for the heat to be released by the refrigerator
if the refrigerator uses 600 W of electrical power.
The Second Law of Thermodynamics 1879
Picture the Problem If we assume that the temperature on the inside of the
refrigerator is 0°C (273 K) and the room temperature to be about 30°C (303 K),
then the refrigerator must be able to maintain a temperature difference of about
30 K. We can use the definition of the COP of a refrigerator and the relationship
between the temperatures of the hot and cold reservoir and Qh and Qc to find an
upper limit on the COP of a household refrigerator. In (b) we can solve the
definition of COP for Qc and differentiate the resulting equation with respect to
time to estimate the rate at which heat is being drawn from the refrigerator
compartment.
(a) Using its definition, express the
COP of a household refrigerator:
COP =
Apply conservation of energy to
the refrigerator to obtain:
W + Qc = Qh ⇒ W = Qh − Qc
Substitute for W and simplify to
obtain:
COP =
Assume, for the sake of finding the
upper limit on the COP, that the
refrigerator is a Carnot refrigerator
and relate the temperatures of the hot
and cold reservoirs to Qh and Qc:
Substitute for
Qh
to obtain:
Qc
Qc
W
(1)
Qc
1
=
Qh − Qc Qh
−1
Qc
Qh Th
=
Qc Tc
COPmax =
1
Th
−1
Tc
Substitute numerical values and
evaluate COPmax:
COPmax =
(b) Solve equation (1) for Qc:
Qc = W (COP )
Differentiate equation (2) with
respect to time to obtain:
dQc
dW
= (COP )
dt
dt
Substitute numerical values and
dQc
evaluate
:
dt
dQc
= (9.1)(600 J/s ) = 5.5 kW
dt
1
= 9.1
303 K
−1
273 K
(2)
1880 Chapter 19
25 •• [SSM] The average temperature of the surface of the Sun is about
5400 K, the average temperature of the surface of Earth is about 290 K. The solar
constant (the intensity of sunlight reaching Earth’s atmosphere) is about
1.37 kW/m2. (a) Estimate the total power of the sunlight hitting Earth. (b)
Estimate the net rate at which Earth’s entropy is increasing due to this solar
radiation.
Picture the Problem We can use the definition of intensity to find the total power
of sunlight hitting the earth and the definition of the change in entropy to find the
changes in the entropy of Earth and the Sun resulting from the radiation from the
Sun.
(a) Using its definition, express the
intensity of the Sun’s radiation on
Earth in terms of the power P
delivered to Earth and Earth’s cross
sectional area A:
I=
P
A
Solve for P and substitute for A to
obtain:
P = IA = IπR 2
where R is the radius of Earth.
Substitute numerical values and
evaluate P:
P = π 1.37 kW/m 2 6.37 × 10 6 m
(b) Express the rate at which Earth’s
entropy SEarth changes due to the
flow of solar radiation:
dS Earth
P
=
dt
TEarth
Substitute numerical values and
dS Earth
evaluate
:
dt
dS Earth 1.746 ×1017 W
=
dt
290 K
(
)(
)
2
= 1.746 × 1017 W = 1.75 × 1017 W
= 6.02 ×1014 J/K ⋅ s
26 ••
A 1.0-L box contains N molecules of an ideal gas, and the positions of
the molecules are observed 100 times per second. Calculate the average time it
should take before we observe all N molecules in the left half of the box if N is
equal to (a) 10, (b) 100, (c) 1000, and (d) 1.0 mole. (e) The best vacuums that
have been created to date have pressures of about 10–12 torr. If a vacuum chamber
has the same volume as the box, how long will a physicist have to wait before all
of the gas molecules in the vacuum chamber occupy only the left half of it?
Compare that to the expected lifetime of the universe, which is about 1010 years.
Picture the Problem If you had one molecule in a box, it would have a 50%
chance of being on one side or the other. We don’t care which side the molecules
are on as long as they all are on one side, so with one molecule you have a 100%
chance of it being on one side or the other. With two molecules, there are four
The Second Law of Thermodynamics 1881
possible combinations (both on one side, both on the other, one on one side and
one on the other, and the reverse), so there is a 25% (1 in 4) chance of them both
being on a particular side, or a 50% chance of them both being on either side.
Extending this logic, the probability of N molecules all being on one side of the
box is P = 2/2N, which means that, if the molecules shuffle 100 times a second,
the time it would take them to cover all the combinations and all get on one side
2N
or the other is t =
. In (e) we can apply the ideal gas law to find the number
2(100 )
of molecules in 1.0 L of air at a pressure of 10−12 torr and an assumed temperature
of 300 K.
(a) Evaluate t for N = 10 molecules:
(b) Evaluate t for N = 100 molecules:
210
t=
= 5.12 s ≈ 5 s
2(100 s −1 )
t=
2100
2 100 s −1
(
)
= 6.34 × 1027 s ×
1y
3.156 × 107 s
≈ 2 × 1020 y
(c) Evaluate t for N = 1000 molecules:
t=
21000
2 100 s −1
(
)
To evaluate 21000 let 10 x = 21000 and
take the logarithm of both sides of
the equation to obtain:
(1000)ln 2 = x ln10 ⇒ x = 301
Substitute to obtain:
10301
t=
2 100 s −1
(
)
= 0.5 × 10 299 s ×
1y
3.156 × 107 s
≈ 2 × 10 291 y
(d) Evaluate t for
N = 1.0 mol =6.022 ×1023 molecules:
23
To evaluate 26.022×10 let
23
10 x = 26.022×10 and take the logarithm
of both sides of the equation to
obtain:
23
26.022×10
t=
2 100 s −1
(
)
(6.022 ×10 )ln 2 = x ln10 ⇒ x ≈ 10
23
23
1882 Chapter 19
Substituting for x yields:
23
1010
1y
⎛
⎞
t≈
−1 ⎜
7 ⎟
2 100 s ⎝ 3.156 ×10 s ⎠
(
)
23
≈ 1010 y
(e) Solve the ideal gas law for the
number of molecules N in the gas:
Assuming the gas to be at room
temperature (300 K), substitute
numerical values and evaluate N:
Evaluate t for N = 3.22×107
molecules:
7
To evaluate 23.22×10 let
7
10 x = 23.22×10 and take the logarithm
of both sides of the equation to
obtain:
Substituting for x yields:
N=
PV
kT
(10
N=
torr )(133.32 Pa/torr )(1.0 L )
(1.381×10−23 J/K )(300 K )
−12
= 3.22 × 107 molecules
7
23.22×10
t=
2 100 s −1
(
)
(3.22 × 10 )ln 2 = x ln10 ⇒ x ≈ 10
7
7
7
1010
1y
×
t=
−1
2 100 s
3.156 × 107 s
(
)
7
≈ 1010 y
Express the ratio of this waiting time
to the lifetime of the universe tuniverse:
t
tuniverse
or
7
7
1010 y
= 10 ≈ 1010
10 y
7
t ≈ 1010 tuniverse
Heat Engines and Refrigerators
[SSM] A heat engine with 20.0% efficiency does 0.100 kJ of work
27 •
during each cycle. (a) How much heat is absorbed from the hot reservoir during
each cycle? (b) How much heat is released to the cold reservoir during each
cycle?
Picture the Problem (a) The efficiency of the engine is defined to be
ε = W Qh where W is the work done per cycle and Qh is the heat absorbed from
the hot reservoir during each cycle. (b) Because, from conservation of energy,
Qh = W + Qc , we can express the efficiency of the engine in terms of the heat Qc
The Second Law of Thermodynamics 1883
released to the cold reservoir during each cycle.
100 J
= 500 J
0.200
(a) Qh absorbed from the hot reservoir
during each cycle is given by:
Qh =
(b) Use Qh = W + Qc to obtain:
Qc = Qh − W = 500 J − 100 J = 400 J
W
ε
=
28 •
A heat engine absorbs 0.400 kJ of heat from the hot reservoir and does
0.120 kJ of work during each cycle. (a) What is its efficiency? (b) How much heat
is released to the cold reservoir during each cycle?
Picture the Problem (a) The efficiency of the engine is defined to be
ε = W Qh where W is the work done per cycle and Qh is the heat absorbed from
the hot reservoir during each cycle. (b) We can apply conservation of energy to
the engine to obtain Qh = W + Qc and solve this equation for the heat Qc released
to the cold reservoir during each cycle.
W 120 J
=
= 30%
Qh 400 J
(a) The efficiency of the heat engine
is given by:
ε=
(b) Apply conservation of energy to
the engine to obtain:
Qh = W + Qc ⇒ Qc = Qh − W
Substitute numerical values and
evaluate Qc:
Qc = 400 J − 120 J = 280 J
29 •
A heat engine absorbs 100 J of heat from the hot reservoir and releases
60 J of heat to the cold reservoir during each cycle. (a) What is its efficiency? (b)
If each cycle takes 0.50 s, find the power output of this engine.
Picture the Problem We can use its definition to find the efficiency of the engine
and the definition of power to find its power output.
Q
W Qh − Qc
=
= 1− c
Qh
Qh
Qh
(a) The efficiency of the heat engine
is given by:
ε=
Substitute numerical values and
evaluate ε:
ε = 1−
60 J
= 40%
100 J
1884 Chapter 19
(b) The power output P of this
engine is the rate at which it does
work:
P=
Substitute numerical values and
evaluate P:
⎛ 100 J ⎞
⎟⎟ = 80 W
P = (0.40 )⎜⎜
⎝ 0.500 s ⎠
dQh
dW d
= ε Qh = ε
dt
dt
dt
30 •
A refrigerator absorbs 5.0 kJ of heat from a cold reservoir and releases
8.0 kJ to a hot reservoir. (a) Find the coefficient of performance of the
refrigerator. (b) The refrigerator is reversible. If it is run backward as a heat
engine between the same two reservoirs, what is its efficiency?
Picture the Problem We can apply their definitions to find the COP of the
refrigerator and the efficiency of the heat engine.
(a) The COP of a refrigerator is
defined to be:
COP =
Apply conservation of energy to
relate the work done per cycle to
Qh and Qc:
W = Qh − Qc
Substitute for W to obtain:
COP =
Qc
Qh − Qc
Substitute numerical values and
evaluate COP:
COP =
5.0 kJ
= 1.7
8.0 kJ − 5.0 kJ
(b) The efficiency of a heat pump
is defined to be:
ε=
W
Qh
Apply conservation of energy to the
heat pump to obtain:
ε=
Qh − Qc
Q
= 1− c
Qh
Qh
Substitute numerical values and
evaluate ε :
ε = 1−
Qc
W
5.0 kJ
= 38%
8.0 kJ
31 ••
[SSM] The working substance of an engine is 1.00 mol of a
monatomic ideal gas. The cycle begins at P1 = 1.00 atm and V1 = 24.6 L. The gas
is heated at constant volume to P2 = 2.00 atm. It then expands at constant pressure
until its volume is 49.2 L. The gas is then cooled at constant volume until its
The Second Law of Thermodynamics 1885
pressure is again 1.00 atm. It is then compressed at constant pressure to its
original state. All the steps are quasi-static and reversible. (a) Show this cycle on
a PV diagram. For each step of the cycle, find the work done by the gas, the heat
absorbed by the gas, and the change in the internal energy of the gas. (b) Find the
efficiency of the cycle.
Picture the Problem To find the heat added during each step we need to find the
temperatures in states 1, 2, 3, and 4. We can then find the work done on the gas
during each process from the area under each straight-line segment and the heat
that enters the system from Q = CV ΔT and Q = CP ΔT . We can use the 1st law of
thermodynamics to find the change in internal energy for each step of the cycle.
Finally, we can find the efficiency of the cycle from the work done each cycle
and the heat that enters the system each cycle.
(a) The cycle is shown to the right:
Apply the ideal-gas law to state 1 to find T1:
T1 =
P1V1
=
nR
(1.00 atm )(24.6 L )
(1.00 mol)⎛⎜ 8.206 ×10 − 2 L ⋅ atm ⎞⎟
mol ⋅ K ⎠
⎝
The pressure doubles while the
volume remains constant between
states 1 and 2. Hence:
T2 = 2T1 = 600 K
The volume doubles while the
pressure remains constant between
states 2 and 3. Hence:
T3 = 2T2 = 1200 K
The pressure is halved while the
volume remains constant
between states 3 and 4. Hence:
T4 = 12 T3 = 600 K
= 300 K
1886 Chapter 19
For path 1→2:
W12 = PΔV12 = 0
and
J ⎞
⎛
Q12 = C V ΔT12 = 32 RΔT12 = 32 ⎜ 8.314
⎟ (600 K − 300 K ) = 3.74 kJ
mol ⋅ K ⎠
⎝
The change in the internal energy of
the system as it goes from state 1 to
state 2 is given by the 1st law of
thermodynamics:
ΔEint = Qin + Won
Because W12 = 0 :
ΔEint,12 = Q12 = 3.74 kJ
For path 2→3:
⎛ 101.325 J ⎞
Won = −W23 = − PΔV23 = −(2.00 atm )(49.2 L − 24.6 L )⎜
⎟ = − 4.99 kJ
⎝ L ⋅ atm ⎠
J ⎞
⎛
Q23 = C P ΔT23 = 52 RΔT23 = 52 ⎜ 8.314
⎟ (1200 K − 600 K ) = 12.5 kJ
mol ⋅ K ⎠
⎝
Apply ΔEint = Qin + Won to obtain:
ΔEint, 23 = 12.5 kJ − 4.99 kJ = 7.5 kJ
For path 3→4:
W34 = PΔV34 = 0
and
J ⎞
⎛
Q34 = ΔEint, 34 = C V ΔT34 = 32 RΔT34 = 32 ⎜ 8.314
⎟ (600 K − 1200 K ) = − 7.48 kJ
mol ⋅ K ⎠
⎝
Apply ΔEint = Qin + Won to obtain:
ΔEint, 34 = −7.48 kJ + 0 = − 7.48 kJ
For path 4→1:
⎛ 101.325 J ⎞
Won = −W41 = − PΔV41 = −(1.00 atm )(24.6 L − 49.2 L )⎜
⎟ = 2.49 kJ
⎝ L ⋅ atm ⎠
and
J ⎞
⎛
Q41 = C P ΔT41 = 52 RΔT41 = 52 ⎜ 8.314
⎟ (300 K − 600 K ) = − 6.24 kJ
mol ⋅ K ⎠
⎝
The Second Law of Thermodynamics 1887
Apply ΔEint = Qin + Won to obtain:
ΔEint, 41 = −6.24 kJ + 2.49 kJ = − 3.75 kJ
For easy reference, the results of the preceding calculations are summarized in the
following table:
Process Won , kJ Qin , kJ ΔEint (= Qin + Won ) , kJ
1→2
2→3
3→4
4→1
0
−4.99
0
2.49
3.74
12.5
−7.48
−6.24
3.74
7.5
−7.48
−3.75
(b) The efficiency of the cycle is
given by:
ε=
Substitute numerical values and
evaluate ε:
ε=
Wby
Qin
=
− W23 + (− W41 )
Q12 + Q23
4.99 kJ − 2.49 kJ
≈ 15%
3.74 kJ + 12.5 kJ
Remarks: Note that the work done per cycle is the area bounded by the
rectangular path. Note also that, as expected because the system returns to its
initial state, the sum of the changes in the internal energy for the cycle is zero.
32 •• The working substance of an engine is 1.00 mol of a diatomic ideal
gas. The engine operates in a cycle consisting of three steps: (1) an adiabatic
expansion from an initial volume of 10.0 L to a pressure of 1.00 atm and a
volume of 20.0 L, (2) a compression at constant pressure to its original volume of
10.0 L, and (3) heating at constant volume to its original pressure. Find the
efficiency of this cycle.
P (atm)
Picture the Problem The three steps
in the process are shown on the PV
diagram. We can find the efficiency of
the cycle by finding the work done by
the gas and the heat that enters the
system per cycle.
1
2.639
2
1
2
3
0
0
The pressures and volumes at the end
points of the adiabatic expansion are
related according to:
20.0
10.0
γ
⎛V ⎞
P1V1 = P2V2 ⇒ P1 = ⎜⎜ 2 ⎟⎟ P2
⎝ V1 ⎠
γ
γ
V(L)
1888 Chapter 19
1.4
Substitute numerical values and
evaluate P1:
⎛ 20.0 L ⎞
P1 = ⎜
⎟
⎝ 10.0 L ⎠
Express the efficiency of the cycle:
ε=
No heat enters or leaves the system
during the adiabatic expansion:
Q12 = 0
Find the heat entering or leaving
the system during the isobaric
compression:
Q23 = C V ΔT23 = 72 RΔT23 = 72 PΔV23
Find the heat entering or leaving
the system during the constantvolume process:
Q31 = C V ΔT31 = 52 RΔT31 = 52 ΔPV31
Apply the 1st law of thermodynamics
to the cycle ( ΔEint, cycle = 0 ) to obtain:
Won = ΔEint − Qin = −Qin
(1.00 atm) = 2.639 atm
W
Qh
=
7
2
(1)
(1.00 atm )(10.0 L − 20.0 L )
= −35.0 atm ⋅ L
=
5
2
(2.639 atm − 1.00 atm )(10.0 L )
= 41.0 atm ⋅ L
= Q12 + Q23 + Q31
= 0 − 35.0 atm ⋅ L + 41.0 atm ⋅ L
= 6.0 atm ⋅ L
Substitute numerical values in
equation (1) and evaluate ε :
ε=
6.0 atm ⋅ L
= 15%
41atm ⋅ L
33 ••
An engine using 1.00 mol of an ideal gas initially at a volume of
24.6 L and a temperature of 400 K performs a cycle consisting of four steps: (1)
an isothermal expansion at 400 K to twice its initial volume, (2) cooling at
constant volume to a temperature of 300 K (3) an isothermal compression to its
original volume, and (4) heating at constant volume to its original temperature of
400 K. Assume that Cv = 21.0 J/K. Sketch the cycle on a PV diagram and find its
efficiency.
Picture the Problem We can find the efficiency of the cycle by finding the work
done by the gas and the heat that enters the system per cycle.
The Second Law of Thermodynamics 1889
2
P (atm)
The PV diagram of the cycle is
shown to the right. A, B, C, and D
identify the four states of the gas and
the numerals 1, 2, 3, and 4 represent
the four steps through which the gas
is taken.
A
1.5
4
1
1
D
3
0.5
0
C
0
10
20
30 40
V (L)
B
2
50
Express the efficiency of the cycle:
ε=
W1 + W2 + W3 + W4
W
=
Qh Qh,1 + Qh, 2 + Qh,3 + Qh, 4
Because steps 2 and 4 are constantvolume processes, W2 = W4 = 0:
ε=
W1 + 0 + W3 + 0
W
=
Qh Qh,1 + Qh, 2 + Qh,3 + Qh, 4
Because the internal energy of the
gas increases in step 4 while no work
is done, and because the internal
energy does not change during step 1
while work is done by the gas, heat
enters the system only during these
processes:
ε=
W + W3
W
= 1
Qh Qh,1 + Qh, 4
The work done during the isothermal
expansion (1) is given by:
⎛V
W1 = nRT ln⎜⎜ B
⎝ VA
The work done during the isothermal
compression (3) is given by:
⎛V ⎞
W3 = nRTc ln⎜⎜ D ⎟⎟
⎝ VC ⎠
Because there is no change in the
internal energy of the system during
step 1, the heat that enters the system
during this isothermal expansion is
given by:
⎛V
Q1 = W1 = nRTh ln⎜⎜ B
⎝ VA
The heat that enters the system
during the constant-volume step 4 is
given by:
(1)
⎞
⎟⎟
⎠
⎞
⎟⎟
⎠
Q4 = C V ΔT = C V (Th − Tc )
400 K
300 K
60
1890 Chapter 19
Substituting in equation (1) yields:
Noting the
⎛V ⎞
⎛V ⎞
nRTh ln⎜⎜ B ⎟⎟ + nRTc ln⎜⎜ D ⎟⎟
⎝ VA ⎠
⎝ VC ⎠
ε=
⎛V ⎞
nRTh ln⎜⎜ B ⎟⎟ + C V (Th − Tc )
⎝ VA ⎠
VB
V
1
= 2 and D = , substitute and simplify to obtain:
VA
VC 2
⎛1⎞
Th ln (2) + Tc ln⎜ ⎟
Th − Tc
⎝ 2 ⎠ = Th ln (2) − Tc ln (2) =
ε=
CV
C
C
(Th − Tc )
Th ln (2) + V (Th − Tc ) Th ln (2) + V (Th − Tc ) Th +
nR
nR
nR ln (2)
Substitute numerical values and evaluate ε:
ε=
400 K − 300 K
= 13.1%
J
21.0
K
(400 K − 300 K )
400 K +
J ⎞
⎛
(1.00 mol)⎜ 8.314
⎟ ln (2)
mol ⋅ K ⎠
⎝
34 ••
Figure 19-15 shows the cycle followed by 1.00 mol of an ideal
monatomic gas initially at a volume of 25.0 L. All the processes are quasi-static.
Determine (a) the temperature of each numbered state of the cycle, (b) the heat
transfer for each part of the cycle, and (c) the efficiency of the cycle.
Picture the Problem We can use the ideal-gas law to find the temperatures of
each state of the gas and the heat capacities at constant volume and constant
pressure to find the heat flow for the constant-volume and isobaric processes.
Because the change in internal energy is zero for the isothermal process, we can
use the expression for the work done on or by a gas during an isothermal process
to find the heat flow during such a process. Finally, we can find the efficiency of
the cycle from its definition.
(a) Use the ideal-gas law to find the
temperature at point 1:
T1 =
P1V1
=
nR
(100 kPa )(25.0 L )
(1.00 mol)⎛⎜ 8.314
⎝
= 301 K
J ⎞
⎟
mol ⋅ K ⎠
The Second Law of Thermodynamics 1891
Use the ideal-gas law to find the
temperatures at points 2 and 3:
P2V2
nR
(
200 kPa )(25.0 L )
=
(1.00 mol)⎛⎜ 8.314 J ⎞⎟
mol ⋅ K ⎠
⎝
T2 = T3 =
= 601 K
(b) Find the heat entering the system for the constant-volume process from 1 → 2:
J ⎞
⎛
Q12 = C V ΔT12 = 32 RΔT12 = 32 ⎜ 8.314
⎟ (601 K − 301 K ) = 3.74 kJ
mol ⋅ K ⎠
⎝
Find the heat entering or leaving the system for the isothermal process from
2 → 3:
⎛V ⎞
⎛ 5 0 .0 L ⎞
J ⎞
⎛
⎟⎟ = 3.46 kJ
Q23 = nRT2 ln⎜⎜ 3 ⎟⎟ = (1.00 mol)⎜ 8.314
⎟ (601 K )ln⎜⎜
mol ⋅ K ⎠
⎝
⎝ 25.0 L ⎠
⎝ V2 ⎠
Find the heat leaving the system during the isobaric compression from 3 → 1:
J ⎞
⎛
Q31 = C P ΔT31 = 52 RΔT31 = 52 ⎜ 8.314
⎟ (301 K − 601 K ) = − 6.24 kJ
mol ⋅ K ⎠
⎝
(c) Express the efficiency of the
cycle:
ε=
Apply the 1st law of thermodynamics
to the cycle:
W = ∑ Q = Q12 + Q23 + Q31
W
W
=
Qin Q12 + Q23
(1)
= 3.74 kJ + 3.46 kJ − 6.24 kJ
= 0.96 kJ
because, for the cycle, ΔEint = 0 .
Substitute numerical values in equation
(1) and evaluate ε :
ε=
0.96 kJ
= 13%
3.74 kJ + 3.46 kJ
35 ••
An ideal diatomic gas follows the cycle shown in Figure 19-16. The
temperature of state 1 is 200 K. Determine (a) the temperatures of the other three
numbered states of the cycle and (b) the efficiency of the cycle.
Picture the Problem We can use the ideal-gas law to find the temperatures of
each state of the gas. We can find the efficiency of the cycle from its definition;
1892 Chapter 19
using the area enclosed by the cycle to find the work done per cycle and the heat
entering the system between states 1 and 2 and 2 and 3 to determine Qin.
(a) Use the ideal-gas law for a fixed
amount of gas to find the
temperature in state 2 to the
temperature in state 1:
PV
P
P1V1 P2V2
=
⇒ T2 = T1 2 2 = T1 2
T1
T2
P1V1
P1
Substitute numerical values and
evaluate T2:
T2 = (200 K )
Apply the ideal-gas law for a fixed
amount of gas to states 2 and 3 to
obtain:
T3 = T2
Substitute numerical values and
evaluate T3:
T3 = (600 K )
Apply the ideal-gas law for a fixed
amount of gas to states 3 and 4 to
obtain:
T4 = T3
Substitute numerical values and
evaluate T4:
T4 = (1800 K )
(b) The efficiency of the cycle is:
ε=
Use the area of the rectangle to
find the work done each cycle:
W = ΔPΔV
Apply the ideal-gas law to state 1
to find the product of n and R:
nR =
(3.0 atm ) =
(1.0 atm )
600 K
P3V3
V
= T2 3
P2V2
V2
(300 L ) =
(100 L )
1800 K
P4V4
P
= T3 4
P3V3
P3
(1.0 atm ) =
(3.0 atm )
600 K
W
Qin
(1)
= (300 L − 100 L )(3.0 atm − 1.0 atm )
= 400 atm ⋅ L
P1V1 (1.0 atm )(100 L )
=
T1
200 K
= 0.50 L ⋅ atm/K
Noting that heat enters the system
between states 1 and 2 and states 2
and 3, express Qin:
Qin = Q12 + Q23 = C V ΔT12 + C P ΔT23
= 52 nRΔT12 + 72 nRΔT23
= ( 52 ΔT12 + 72 ΔT23 )nR
The Second Law of Thermodynamics 1893
Substitute numerical values and evaluate Qin:
L ⋅ atm ⎞
⎛
Qin = [ 52 (600 K − 200 K ) + 72 (1800 K − 600 K )]⎜ 0.50
⎟ = 2600 atm ⋅ L
K ⎠
⎝
Substitute numerical values in
equation (1) and evaluate ε :
ε=
400 atm ⋅ L
= 15%
2600 atm ⋅ L
36 ••• Recently, an old design for a heat engine, known as the Stirling engine
has been promoted as a means of producing power from solar energy. The cycle
of a Stirling engine is as follows: (1) isothermal compression of the working gas
(2) heating of the gas at constant volume, (3) an isothermal expansion of the gas,
and (4) cooling of the gas at constant volume. (a) Sketch PV and ST diagrams for
the Stirling cycle. (b) Find the entropy change of the gas for each step of the cycle
and show that the sum of these entropy changes is equal to zero.
Picture the Problem (a) The PV and ST cycles are shown below. (b) We can
show that the entropy change during one Stirling cycle is zero by adding up the
entropy changes for the four processes.
P
S
2
ΔV = 0
(2)
3
(3)
Th
Th
Tc
1
(4)
(1)
Tc
4
V
ΔV = 0
T
(b) The change in entropy for one
Stirling cycle is the sum of the
entropy changes during the cycle:
ΔS cycle = ΔS12 + ΔS 23 + ΔS 34 + ΔS 41 (1)
Express the entropy change for the
isothermal process from state 1 to
state 2:
⎛V ⎞
ΔS12 = nR ln⎜⎜ 2 ⎟⎟
⎝ V1 ⎠
1894 Chapter 19
Similarly, the entropy change for the
isothermal process from state 3 to
state 4 is:
The change in entropy for a constantvolume process is given by:
⎛V ⎞
ΔS 34 = nR ln⎜⎜ 4 ⎟⎟
⎝ V3 ⎠
or, because V2 = V3 and V1 = V4,
⎛V ⎞
⎛V ⎞
ΔS 34 = nR ln⎜⎜ 1 ⎟⎟ = − nR ln⎜⎜ 2 ⎟⎟
⎝ V2 ⎠
⎝ V1 ⎠
T
ΔS isochoric
dQ f nC V dT
=∫
=∫
T
T
Ti
⎛T
= nC V ln⎜⎜ f
⎝ Ti
⎞
⎟⎟
⎠
For the constant-volume process
from state 2 to state 3:
⎛T
ΔS 23 = C V ln⎜⎜ c
⎝ Th
⎞
⎟⎟
⎠
For the constant-volume process
from state 4 to state 1:
⎛T
ΔS 41 = C V ln⎜⎜ h
⎝ Tc
⎞
⎛T
⎟⎟ = −C V ln⎜⎜ c
⎝ Th
⎠
⎞
⎟⎟
⎠
Substituting in equation (1) yields:
⎛V ⎞
⎛T
ΔS cycle = −nR ln⎜⎜ 2 ⎟⎟ + C V ln⎜⎜ c
⎝ V1 ⎠
⎝ Th
⎞
⎛V ⎞
⎛T
⎟⎟ − nR ln⎜⎜ 2 ⎟⎟ − C V ln⎜⎜ c
⎠
⎝ V1 ⎠
⎝ Th
⎞
⎟⎟ = 0
⎠
37 ••
″As far as we know, Nature has never evolved a heat engine″—Steven
Vogel, Life’s Devices, Princeton University Press (1988). (a) Calculate the
efficiency of a heat engine operating between body temperature (98.6ºF) and a
typical outdoor temperature (70ºF), and compare this to the human body’s
efficiency for converting chemical energy into work (approximately 20%). Does
this efficiency comparison contradict the second law of thermodynamics?
(b) From the result of Part (a), and a general knowledge of the conditions under
which most warm-blooded organisms exist, give a reason why no warm-blooded
organisms have evolved heat engines to increase their internal energies.
Picture the Problem We can use the efficiency of a Carnot engine operating
between reservoirs at body temperature and typical outdoor temperatures to find
an upper limit on the efficiency of an engine operating between these
temperatures.
(a) Express the maximum efficiency
of an engine operating between body
temperature and 70°F:
εC = 1 −
Tc
Th
The Second Law of Thermodynamics 1895
Use T =
5
9
(tF − 32) + 273 to obtain:
Substitute numerical values and
evaluate ε C :
Tbody = 310 K and Troom = 294 K
ε C = 1−
294 K
= 5.16%
310 K
The fact that this efficiency is considerably less than the actual efficiency of a
human body does not contradict the second law of thermodynamics. The
application of the second law to chemical reactions such as the ones that supply
the body with energy have not been discussed in the text but we can note that we
don’t get our energy from heat swapping between our body and the environment.
Rather, we eat food to get the energy that we need.
(b) Most warm-blooded animals survive under roughly the same conditions as
humans. To make a heat engine work with appreciable efficiency, internal body
temperatures would have to be maintained at an unreasonably high level.
38 ••• The diesel cycle shown in Figure 19-17 approximates the behavior of a
diesel engine. Process ab is an adiabatic compression, process bc is an expansion
at constant pressure, process cd is an adiabatic expansion, and process da is
cooling at constant volume. Find the efficiency of this cycle in terms of the
volumes Va, Vb and Vc.
Picture the Problem The working fluid will be modeled as an ideal gas and the
process will be modeled as quasistatic. To find the efficiency of the diesel cycle
we can find the heat that enters the system and the heat that leaves the system and
use the expression that gives the efficiency in terms of these quantities. Note that
no heat enters or leaves the system during the adiabatic processes ab and cd. Heat
enters the system during the isobaric process bc and leaves the system during the
isovolumetric process da.
Express the efficiency of the cycle in
terms of Qc and Qh:
Express Q for the isobaric warming
process bc:
Because CV is independent of T,
Qda (the constant-volume cooling
process) is given by:
Substitute for Qh and Qc and
simplify using γ = C P C V to obtain:
ε=
Q
W Qh − Qc
=
= 1− c
Qh
Qh
Qh
Qbc = Qh = C P (Tc − Tb )
Qda = Qc = C V (Td − Ta )
ε = 1−
(T − Ta )
C V (Td − Ta )
= 1− d
C P (Tc − Tb )
γ (Tc − Tb )
1896 Chapter 19
Using an equation for a quasistatic
adiabatic process, relate the
temperatures Ta and Tb to the
volumes Va and Vb:
γ −1
TaVa
Proceeding similarly, relate the
temperatures Tc and Td to the
volumes Vc and Vd:
γ −1
= TbVb
Vbγ −1
⇒ Ta = Tb γ −1 (1)
Va
Vcγ −1
(2)
Vdγ −1
TcVcγ −1 = TdVdγ −1 ⇒ Td = Tc
Use equations (1) and (2) to eliminate
Ta and Td:
⎛ Vcγ −1
Vbγ −1 ⎞
⎜⎜ Tc γ −1 − Tb γ −1 ⎟⎟
Vd
Va ⎠
ε = 1− ⎝
γ (Tc − Tb )
Because Va = Vd:
⎛⎛ V
⎜⎜ c
⎜ ⎜⎝ Va
ε = 1− ⎝
⎞
⎟⎟
⎠
γ −1
T ⎛V
− b ⎜⎜ b
Tc ⎝ Va
⎛
γ ⎜⎜1 −
⎝
Noting that Pb = Pc, apply the idealgas law to relate Tb and Tc:
⎞
⎟⎟
⎠
γ −1
⎞
⎟
⎟
⎠
Tb ⎞
⎟
Tc ⎟⎠
Tb Vb
=
Tc Vc
Substitute for the ratio of Tb to Tc and simplify to obtain:
⎛ Vc
⎜⎜
V
ε = 1− ⎝ a
⎞
⎟⎟
⎠
γ −1
⎛ Vb
⎜⎜
⎝ Va
⎛ V ⎞
γ ⎜⎜1 − b ⎟⎟
⎝ Vc ⎠
γ
V
− b
Vc
⎞
⎟⎟
⎠
γ −1
⎛ Vc
Vc
⎜⎜
V
Va
⋅
= 1− ⎝ a
Vc
Va
γ −1
⎛ Vb
⎜⎜
⎝ Va
⎛V V ⎞
γ ⎜⎜ c − b ⎟⎟
⎝ Va Va ⎠
⎞
⎟⎟
⎠
V
− b
Vc
⎞
⎟⎟
⎠
γ −1
γ
⎛ Vc ⎞ ⎛ Vb ⎞
⎜⎜ ⎟⎟ − ⎜⎜ ⎟⎟
V
V
V γ − Vbγ
= 1 − ⎝ a ⎠ ⎝ a ⎠ = 1 − γ −c1
γVa (Vc − Vb )
⎛V V ⎞
γ ⎜⎜ c − b ⎟⎟
⎝ Va Va ⎠
Second Law of Thermodynamics
39 ••
[SSM] A refrigerator absorbs 500 J of heat from a cold reservoir and
releases 800 J to a hot reservoir. Assume that the heat-engine statement of the
second law of thermodynamics is false, and show how a perfect engine working
The Second Law of Thermodynamics 1897
with this refrigerator can violate the refrigerator statement of the second law of
thermodynamics.
Determine the Concept The following diagram shows an ordinary refrigerator
that uses 300 J of work to remove 500 J of heat from a cold reservoir and releases
800 J of heat to a hot reservoir (see (a) in the diagram). Suppose the heat-engine
statement of the second law is false. Then a ″perfect″ heat engine could remove
energy from the hot reservoir and convert it completely into work with 100
percent efficiency. We could use this perfect heat engine to remove 300 J of
energy from the hot reservoir and do 300 J of work on the ordinary refrigerator
(see (b) in the diagram). Then, the combination of the perfect heat engine and the
ordinary refrigerator would be a perfect refrigerator; transferring 500 J of heat
from the cold reservoir to the hot reservoir without requiring any work (see (c) in
the diagram).This violates the refrigerator statement of the second law.
Hot reservoir at temperature Th
⇓
⇓
Ordinary
refrigerator
300 J
300 J
Perfect
heat
engine
⇓
500 J
Perfect
refrigerator
⇓
⇓
⇓
500 J
⇓
300 J
800 J
500 J
Cold reservoir at temperature Tc
(a)
(b)
(c )
If two curves that represent quasi-static adiabatic processes could
40 ••
intersect on a PV diagram, a cycle could be completed by an isothermal path
between the two adiabatic curves shown in Figure 19-18. Show that such a cycle
violates the second law of thermodynamics.
Determine the Concept The work done by the system is the area enclosed by the
cycle, where we assume that we start with the isothermal expansion. It is only in
this expansion that heat is extracted from a reservoir. There is no heat transfer in
the adiabatic expansion or compression. Thus, we would completely convert heat
to mechanical energy, without exhausting any heat to a cold reservoir, in violation
of the second law of thermodynamics.
Carnot Cycles
41 •
[SSM] A Carnot engine works between two heat reservoirs at
temperatures Th = 300 K and Tc = 200 K. (a) What is its efficiency? (b) If it
1898 Chapter 19
absorbs 100 J of heat from the hot reservoir during each cycle, how much work
does it do each cycle? (c) How much heat does it release during each cycle?
(d) What is the COP of this engine when it works as a refrigerator between the
same two reservoirs?
Picture the Problem We can find the efficiency of the Carnot engine using
ε = 1 − Tc / Th and the work done per cycle from ε = W / Qh . We can apply
conservation of energy to find the heat rejected each cycle from the heat absorbed
and the work done each cycle. We can find the COP of the engine working as a
refrigerator from its definition.
Tc
200 K
= 1−
= 33.3%
Th
300 K
(a) The efficiency of the Carnot
engine depends on the temperatures
of the hot and cold reservoirs:
ε C = 1−
(b) Using the definition of efficiency,
relate the work done each cycle to the
heat absorbed from the hot reservoir:
W = ε C Qh = (0.333)(100 J ) = 33.3 J
(c) Apply conservation of energy to
relate the heat given off each cycle
to the heat absorbed and the work
done:
Qc = Qh − W = 100 J − 33.3 J = 66.7 J
(d) Using its definition, express and
evaluate the refrigerator’s coefficient
of performance:
COP =
= 67 J
Qc 66.7 J
=
= 2 .0
W 33.3 J
An engine absorbs 250 J of heat per cycle from a reservoir at 300 K
42 •
and releases 200 J of heat per cycle to a reservoir at 200 K. (a) What is its
efficiency? (b) How much additional work per cycle could be done if the engine
were reversible?
Picture the Problem We can find the efficiency of the engine from its definition
and the additional work done if the engine were reversible from W = ε CQh , where
εC is the Carnot efficiency.
(a) Express the efficiency of the
engine in terms of the heat absorbed
from the high-temperature reservoir
and the heat exhausted to the lowtemperature reservoir:
ε=
Q
W Qh − Qc
= 1− c
=
Qh
Qh
Qh
= 1−
200 J
= 20.0%
250 J
The Second Law of Thermodynamics 1899
(b) Express the additional work done
if the engine is reversible:
ΔW = WCarnot − WPart (a )
Relate the work done by a reversible
engine to its Carnot efficiency:
⎛ T
W = ε C Qh = ⎜⎜1 − c
⎝ Th
Substitute numerical values and
evaluate W:
⎛ 200 K ⎞
⎟⎟ (250 J ) = 83.3 J
W = ⎜⎜1 −
⎝ 300 K ⎠
Substitute numerical values in
equation (1) and evaluate ΔW:
ΔW = 83.3 J − 50 J = 33 J
(1)
⎞
⎟⎟Qh
⎠
43 ••
A reversible engine working between two reservoirs at temperatures Th
and Tc has an efficiency of 30%. Working as a heat engine, it releases 140 J per
cycle of heat to the cold reservoir. A second engine working between the same
two reservoirs also releases 140 J per cycle to the cold reservoir. Show that if the
second engine has an efficiency greater than 30%, the two engines working
together would violate the heat-engine statement of the second law.
Determine the Concept Let the first engine be run as a refrigerator. Then it will
remove 140 J from the cold reservoir, deliver 200 J to the hot reservoir, and
require 60 J of energy to operate. Now take the second engine and run it between
the same reservoirs, and let it eject 140 J into the cold reservoir, thus replacing the
heat removed by the refrigerator. If ε2, the efficiency of this engine, is greater than
30%, then Qh2, the heat removed from the hot reservoir by this engine, is
140 J/(1 − ε2) > 200 J, and the work done by this engine is W = ε2Qh2 > 60 J. The
end result of all this is that the second engine can run the refrigerator, replacing
the heat taken from the cold reservoir, and do additional mechanical work. The
two systems working together then convert heat into mechanical energy without
rejecting any heat to a cold reservoir, in violation of the second law.
44 •• A reversible engine working between two reservoirs at temperatures Th
and Tc has an efficiency of 20%. Working as a heat engine, it does 100 J of work
per cycle. A second engine working between the same two reservoirs also does
100 J of work per cycle. Show that if the efficiency of the second engine is greater
than 20%, the two engines working together would violate the refrigerator
statement of the second law.
Determine the Concept If the reversible engine is run as a refrigerator, it will
require 100 J of mechanical energy to take 400 J of heat from the cold reservoir
and deliver 500 J to the hot reservoir. Now let the second engine, with ε2 > 0.2,
operate between the same two heat reservoirs and use it to drive the refrigerator.
1900 Chapter 19
Because ε2 > 0.2, this engine will remove less than 500 J from the hot reservoir in
the process of doing 100 J of work. The net result is then that no net work is done
by the two systems working together, but a finite amount of heat is transferred
from the cold reservoir to the hot reservoir, in violation of the refrigerator
statement of the second law.
45 ••
A Carnot engine works between two heat reservoirs as a refrigerator.
During each cycle, 100 J of heat are absorbed and 150 J are released to the hot
reservoir. (a) What is the efficiency of the Carnot engine when it works as a heat
engine between the same two reservoirs? (b) Show that no other engine working
as a refrigerator between the same two reservoirs can have a COP greater than
2.00.
Picture the Problem We can use the definition of efficiency to find the efficiency
of the Carnot engine operating between the two reservoirs.
(a) The efficiency of the Carnot
engine is given by:
εC =
W
50 J
=
= 33%
Qh 150 J
(b) If the COP > 2, then 50 J of work will remove more than 100 J of heat from
the cold reservoir and put more than 150 J of heat into the hot reservoir. So
running engine described in Part (a) to operate the refrigerator with a COP > 2
will result in the transfer of heat from the cold to the hot reservoir without doing
any net mechanical work in violation of the second law.
46 ••
A Carnot engine works between two heat reservoirs at temperatures
Th = 300 K and Tc = 77.0 K. (a) What is its efficiency? (b) If it absorbs 100 J of
heat from the hot reservoir during each cycle, how much work does it do?
(c) How much heat does it release to the low-temperature reservoir during each
cycle? (d) What is the coefficient of performance of this engine when it works as a
refrigerator between these two reservoirs?
Picture the Problem We can use the definitions of the efficiency of a Carnot
engine and the coefficient of performance of a refrigerator to find these quantities.
The work done each cycle by the Carnot engine is given by W = ε CQh and we can
use the conservation of energy to find the heat rejected to the low-temperature
reservoir.
(a) The efficiency of a Carnot engine
depends on the temperatures of the
hot and cold reservoirs:
εC = 1−
77.0 K
Tc
= 1−
= 74.3%
300 K
Th
The Second Law of Thermodynamics 1901
(b) Express the work done each
cycle in terms of the efficiency of
the engine and the heat absorbed
from the high-temperature reservoir:
W = ε C Qh = (0.743)(100 J ) = 74.3 J
(c) Apply conservation of energy to
obtain:
Qc = Qh − W = 100 J − 74.3 J = 26 J
(d) Using its definition, express and
evaluate the refrigerator’s coefficient
of performance:
COP =
Qc
26 J
=
= 0.35
W 74.3 J
47 ••
[SSM] In the cycle shown 1 in Figure 19-19, 1.00 mol of an ideal
diatomic gas is initially at a pressure of 1.00 atm and a temperature of 0.0ºC. The
gas is heated at constant volume to T2 = 150ºC and is then expanded adiabatically
until its pressure is again 1.00 atm. It is then compressed at constant pressure back
to its original state. Find (a) the temperature after the adiabatic expansion,
(b) the heat absorbed or released by the system during each step, (c) the efficiency
of this cycle, and (d) the efficiency of a Carnot cycle operating between the
temperature extremes of this cycle.
Picture the Problem We can use the ideal-gas law for a fixed amount of gas and
the equations of state for an adiabatic process to find the temperatures, volumes,
and pressures at the end points of each process in the given cycle. We can use
Q = C V ΔT and Q = C P ΔT to find the heat entering and leaving during the
constant-volume and isobaric processes and the first law of thermodynamics to
find the work done each cycle. Once we’ve calculated these quantities, we can
use its definition to find the efficiency of the cycle and the definition of the
Carnot efficiency to find the efficiency of a Carnot engine operating between the
extreme temperatures.
(a) Apply the ideal-gas law for a
fixed amount of gas to relate the
temperature at point 3 to the
temperature at point 1:
Apply the ideal-gas law for a fixed
amount of gas to relate the pressure
at point 2 to the temperatures at
points 1 and 2 and the pressure at 1:
P1V1 P3V3
=
T1
T3
or, because P1 = P3,
V
T3 = T1 3
V1
PV T
P1V1 P2V2
=
⇒ P2 = 1 1 2
T1
T2
V2T1
(1)
1902 Chapter 19
T2
423 K
= (1.00 atm )
= 1.55 atm
T1
273 K
Because V1 = V2:
P2 = P1
Apply an equation for an adiabatic
process to relate the pressures and
volumes at points 2 and 3:
⎛ P ⎞γ
P1V1γ = P3V3γ ⇒ V3 = V1 ⎜⎜ 1 ⎟⎟
⎝ P3 ⎠
Noting that V1 = 22.4 L, evaluate V3:
⎛ 1.55 atm ⎞ 1.4
⎟⎟ = 30.6 L
V3 = (22.4 L )⎜⎜
1
atm
⎝
⎠
Substitute numerical values in
equation (1) and evaluate T3 and t3:
T3 = (273 K )
1
1
30.6 L
= 373 K
22.4 L
and
t 3 = T3 − 273 = 100°C
(b) Process 1→2 takes place at
constant volume (note that γ = 1.4
corresponds to a diatomic gas and
that CP – CV = R):
Q12 = CV ΔT12 = 52 RΔT12
J ⎞
⎛
= 52 ⎜ 8.314
⎟ (423 K − 273 K )
mol ⋅ K ⎠
⎝
= 3.12 kJ
Process 2→3 takes place adiabatically:
Q23 = 0
Process 3→1 is isobaric (note that
CP = CV + R):
Q31 = C P ΔT31 = 72 RΔT12
J ⎞
⎛
= 72 ⎜ 8.314
⎟ (273 K − 373 K )
mol ⋅ K ⎠
⎝
= − 2.91 kJ
(c) The efficiency of the cycle is
given by:
ε=
Apply the first law of thermodynamics
to the cycle:
ΔEint = Qin + Won
or, because ΔEint, cycle = 0 (the system
W
Qin
(2)
begins and ends in the same state) and
Won = −Wby the gas = Qin .
Evaluating W yields:
W = ∑ Q = Q12 + Q23 + Q31
= 3.12 kJ + 0 − 2.91 kJ = 0.21 kJ
The Second Law of Thermodynamics 1903
0.21 kJ
= 6.7%
3.12 kJ
Substitute numerical values in
equation (2) and evaluate ε :
ε=
(d) Express and evaluate the
efficiency of a Carnot cycle
operating between 423 K and
273 K:
ε C = 1−
273 K
Tc
= 1−
= 35.5%
423 K
Th
48 ••
You are part of a team that is completing a mechanical-engineering
project. Your team built a steam engine that takes in superheated steam at 270ºC
and discharges condensed steam from its cylinder at 50.0ºC. Your team has
measured its efficiency to be 30.0%. (a) How does this efficiency compare with
the maximum possible efficiency for your engine? (b) If the useful power output
of the engine is known to be 200 kW, how much heat does the engine release to
its surroundings in 1.00 h?
Picture the Problem We can find the maximum efficiency of the steam engine
by calculating the Carnot efficiency of an engine operating between the given
temperatures. We can apply the definition of efficiency to find the heat
discharged to the engine’s surroundings in 1.00 h.
(a) The efficiency of the steam
engine as a percentage of the
maximum possible efficiency is
given by:
ε steam engine 0.300
=
ε max
ε max
The efficiency of a Carnot engine
operating between temperatures Fc
and Th is:
ε max = 1 −
Substituting for εmax yields:
ε steam engine 0.300
=
= 74.05%
0.4052
ε max
Tc
323 K
= 1−
= 40.52%
Th
543 K
or
ε steam engine = 0.740ε max
(b) Relate the heat Qc discharged to
the engine’s surroundings to Qh and
the efficiency of the engine:
ε=
W Qh − Qc
=
⇒ Qc = (1 − ε )Qh
Qh
Qh
1904 Chapter 19
PΔt
Using its definition, relate the
efficiency of the engine to the heat
intake of the engine and the work it
does each cycle:
Qh =
Substitute for Qh in the expression
for Qc and simplify to obtain:
Qc = (1 − ε )
W
ε
=
ε
PΔt
ε
⎛1 ⎞
= ⎜ − 1⎟ PΔt
⎝ε ⎠
Substitute numerical values and evaluate Qc (1.00 h ) :
kJ ⎞
⎛ 1
⎞⎛
Qc (1.00 h ) = ⎜
− 1⎟ ⎜ 200 ⎟ (3600 s ) = 1.68 GJ
s ⎠
⎝ 0.300 ⎠ ⎝
*Heat Pumps
49 •
[SSM] As an engineer, you are designing a heat pump that is capable
of delivering heat at the rate of 20 kW to a house. The house is located where, in
January, the average outside temperature is –10ºC. The temperature of the air in
the air handler inside the house is to be 40ºC. (a) What is maximum possible
COP for a heat pump operating between these temperatures? (b) What must be the
minimum power of the electric motor driving the heat pump? (c) In reality, the
COP of the heat pump will be only 60 percent of the ideal value. What is the
minimum power of the engine when the COP is 60 percent of the ideal value?
Picture the Problem We can use the definition of the COPHP and the Carnot
efficiency of an engine to express the maximum efficiency of the refrigerator in
terms of the reservoir temperatures. We can apply the definition of power to find
the minimum power needed to run the heat pump.
(a) Express the COPHP in terms of Th
and Tc:
COPHP =
=
Substitute numerical values and
evaluate COPHP:
COPHP =
Qh
Qh
=
W Qh − Qc
Th
1
1
=
=
Q
T
Th − Tc
1− c 1− c
Qh
Th
313 K
= 6.26
313 K − 263 K
= 6.3
The Second Law of Thermodynamics 1905
(b) The COPHP is also given by:
COPHP =
Substitute numerical values and
evaluate Pmotor:
Pmotor =
(c) The minimum power of the
engine is given by:
Pmin
Pout
Pout
⇒ Pmotor =
Pmotor
COPHP
20 kW
= 3.2 kW
6.26
dQc
= dt =
ε HP
dQc
dt
ε (COPHP, max )
where εHP is the efficiency of the heat
pump.
Substitute numerical values and
evaluate Pmin:
Pmin =
20 kW
= 5.3 kW
(0.60)(6.26)
50 •
A refrigerator is rated at 370 W. (a) What is the maximum amount of
heat it can absorb from the food compartment in 1.00 min if the foodcompartment temperature of the refrigerator is 0.0ºC and it releases heat into a
room at 20.0ºC? (b) If the COP of the refrigerator is 70% of that of a reversible
refrigerator, how much heat can it absorb from the food compartment in 1.00 min
under these conditions?
Picture the Problem We can use the definition of the COP to relate the heat
removed from the refrigerator to its power rating and operating time. By
expressing the COP in terms of Tc and Th we can write the amount of heat
removed from the refrigerator as a function of Tc, Th, P, and Δt.
(a) Express the amount of heat the
refrigerator can remove in a given
period of time as a function of its
COP:
Express the COP in terms of Th and
Tc and simplify to obtain:
Qc = (COP )W
= (COP )PΔt
COP =
=
=
Qc
Q
Q −W
= c = h
ε Qh
W ε Qh
1− ε
ε
=
Tc
Th − Tc
1
ε
−1 =
1
−1
Tc
1−
Th
1906 Chapter 19
Substituting for COP yields:
⎛ Tc ⎞
⎟⎟ PΔt
Qc = ⎜⎜
⎝ Th − Tc ⎠
Substitute numerical values and evaluate Qc:
273 K
60 s ⎞
⎛
⎞
⎛
Qc = ⎜
⎟ (370 W )⎜1.00 min ×
⎟ = 303 kJ = 0.30 MJ
min ⎠
⎝ 293 K − 273 K ⎠
⎝
(b) If the COP is 70% of the
efficiency of an ideal pump:
Qc' = (0.70 )(303 kJ ) = 0.21 MJ
51 •
A refrigerator is rated at 370 W. (a) What is the maximum amount of
heat it can absorb for the food compartment in 1.00 min if the temperature in the
compartment is 0.0ºC and it releases heat into a room at 35ºC? (b) If the COP of
the refrigerator is 70% of that of a reversible pump, how much heat can it absorb
from the food compartment in 1.00 min? Is the COP for the refrigerator greater
when the temperature of the room is 35ºC or 20ºC? Explain.
Picture the Problem We can use the definition of the COP to relate the heat
removed from the refrigerator to its power rating and operating time. By
expressing the COP in terms of Tc and Th we can write the amount of heat
removed from the refrigerator as a function of Tc, Th, P, and Δt.
(a) Express the amount of heat the
refrigerator can remove in a given
period of time as a function of its
COP:
Express the COP in terms of Th
and Tc and simplify to obtain:
Qc = (COP )W
= (COP )PΔt
COP =
=
=
Substituting for COP yields:
Qc
Q
Q −W
= c = h
εQh
W εQh
1− ε
ε
=
1
ε
−1
Tc
1
−1 =
T
Th − Tc
1− c
Th
⎛ Tc ⎞
⎟⎟ PΔt
Qc = ⎜⎜
T
T
−
⎝ h c⎠
The Second Law of Thermodynamics 1907
Substitute numerical values and evaluate Qc:
273 K
60 s ⎞
⎞
⎛
⎛
Qc = ⎜
⎟ = 173 kJ = 0.17 MJ
⎟ (370 W )⎜1.00 min ×
min ⎠
⎝ 308 K − 273 K ⎠
⎝
(b) If the COP is 70% of the
efficiency of an ideal pump:
Qc' = (0.70 )(173 kJ ) = 0.12 MJ
Because the temperature difference increases when the room is warmer, the COP
decreases.
52 ••• You are installing a heat pump, whose COP is half the COP of a
reversible heat pump. You will use the pump on chilly winter nights to increase
the air temperature in your bedroom. Your bedroom’s dimensions are 5.00 m ×
3.50 m × 2.50 m. The air temperature should increase from 63°F to 68°F. The
outside temperature is 35°F , and the temperature at the air handler in the room is
112°F. If the pump’s electric power consumption is 750 W, how long will you
have to wait in order for the room’s air to warm (take the specific heat of air to be
1.005 kJ/(kg·°C)? Assume you have good window draperies and good wall
insulation so that you can neglect the release of heat through windows, walls,
ceilings and floors. Also assume that the heat capacity of the floor, ceiling, walls
and furniture are negligible.
Picture the Problem We can use the definition of the coefficient of performance
of a heat pump and the relationship between the work done per cycle and the
pump’s power consumption to find your waiting time.
The coefficient of performance of the
heat pump is defined as:
We’re given that the coefficient of
performance of the heat pump is half
the coefficient of performance of an
ideal heat pump:
Substituting for COPHP yields:
Qh
Q
Qh
= h ⇒ Δt =
(COPHP )P
W PΔt
where Qh is the heat required to raise
the temperature of your bedroom, P is
the power consumption of the heat
pump, and Δt is the time required to
warm the bedroom.
COPHP =
Qh 1
= COPmax
W 2
⎛ Th ⎞
⎟⎟
= 12 ⎜⎜
T
T
−
⎝ h c⎠
COPHP =
Δt =
2Qh
⎛ Th ⎞
⎜⎜
⎟⎟ P
⎝ Th − Tc ⎠
1908 Chapter 19
The heat required to warm the room
is related to the volume of the room,
the density of air, and the desired
increase in temperature:
Substitute for Qh to obtain:
Qh = mcΔT = ρVcΔT
where ρ is the density of air and c is its
specific heat capacity.
Δt =
2 ρVcΔT
⎛ Th ⎞
⎜⎜
⎟⎟ P
⎝ Th − Tc ⎠
Substitute numerical values and evaluate Δt:
⎛
kg ⎞
J ⎞⎛
5 C° ⎞
⎛
⎟⎟ ⎜ 5 F° ×
2⎜1.293 3 ⎟ (5.00 m × 3.50 m × 2.50 m )⎜⎜1005
⎟
m ⎠
kg ⋅ C° ⎠ ⎝
9 F° ⎠
⎝
⎝
Δt =
= 56 s
317 K
⎛
⎞
⎜
⎟ (750 W )
⎝ 317 K − 275 K ⎠
Entropy Changes
53 •
[SSM] You inadvertently leave a pan of water boiling away on the
hot stove. You return just in time to see the last drop converted into steam. The
pan originally held 1.00 L of boiling water. What is the change in entropy of the
water associated with its change of state from liquid to gas?
Picture the Problem Because the water absorbed heat in the vaporization process
Qabsorbed
by H 2 O
. See Table 18-2
its change in entropy is positive and given by ΔS H 2 O =
T
for the latent heat of vaporization of water.
The change in entropy of the water is
given by:
The heat absorbed by the water as it
vaporizes is the product of its mass
and latent heat of vaporization:
Substituting for Qabsorbed yields:
by H 2 O
ΔS H 2 O =
Qabsorbed
by H 2 O
T
Qabsorbed = mLv = ρVLv
by H 2 O
ΔS H 2 O =
ρVLv
T
The Second Law of Thermodynamics 1909
Substitute numerical values and
evaluate ΔS H 2O :
ΔS H 2O
⎛
kg ⎞
kJ ⎞
⎛
⎜1.00 ⎟ (1.00 L )⎜⎜ 2257 ⎟⎟
L⎠
kg ⎠
⎝
⎝
=
373 K
kJ
= 6.05
K
54 •
What is the change in entropy of 1.00 mol of liquid water at 0.0ºC that
freezes to ice at 0.0°C?
Picture the Problem We can use the definition of entropy change to find the
change in entropy of the liquid water as it freezes. Because heat is removed from
liquid water when it freezes, the change in entropy of the liquid water is negative.
See Appendix C for the molar mass of water and Table 18-2 for the latent heat of
fusion of water.
The change in entropy of the water is
given by::
The heat removed from the water as
it freezes is the product of its mass
and latent heat of fusion:
ΔS H 2 O =
Qremoved
from H 2 O
T
Qremoved = −mLf
from H 2 O
or, because m = nM H 2 O ,
Qremoved = − nM H 2 O Lf
from H 2 O
Substitute numerical values and evaluate ΔS H 2O :
ΔS H2O
J⎞
g ⎞⎛
⎛
− (1.00 mol)⎜18.015
⎟ ⎜⎜ 333.5 ⎟⎟
g⎠
mol ⎠ ⎝
J
⎝
= − 22.0
=
K
273 K
Consider the freezing of 50.0 g of water once it is placed in the freezer
55 •
compartment of a refrigerator. Assume the walls of the freezer are maintained at
–10ºC. The water, initially liquid at 0.0ºC, is frozen into ice and cooled to
–10ºC. Show that even though the entropy of the water decreases, the net entropy
of the universe increases.
Picture the Problem The change in the entropy of the universe resulting from the
freezing of this water and the cooling of the ice formed is the sum of the entropy
changes of the water-ice and the freezer. Note that, while the entropy of the water
decreases, the entropy of the freezer increases.
1910 Chapter 19
The change in entropy of the
universe resulting from this freezing
and cooling process is given by:
ΔS u = ΔS water + ΔS freezer
(1)
Express ΔS water :
ΔS water = ΔS freezing + ΔS cooling
(2)
Express ΔS freezing :
ΔS freezing =
− Qfreezing
(3)
Tfreezing
where the minus sign is a consequence
of the fact that heat is leaving the water
as it freezes.
Relate Qfreezing to the latent heat of
Qfreezing = mLf
fusion and the mass of the water:
− mLf
Tfreezing
Substitute in equation (3) to obtain:
ΔS freezing =
Express ΔS cooling :
⎛T
ΔS cooling = mC p ln⎜⎜ f
⎝ Ti
Substitute in equation (2) to obtain:
Noting that the freezer gains heat
(at 263 K) from the freezing
water and cooling ice, express
ΔS freezer :
ΔS water =
⎞
⎟⎟
⎠
⎛T
− mLf
+ mCp ln⎜⎜ f
Tfreezing
⎝ Ti
ΔS freezer =
=
ΔQice ΔQcooling ice
+
Tfreezer
Tfreezer
mC p ΔT
mLf
+
Tfreezer
Tfreezer
Substitute for ΔS water and ΔS freezer in equation (1):
ΔS u =
⎛T
− mLf
+ mCp ln⎜⎜ f
Tfreezing
⎝ Ti
mCp ΔT
⎞ mLf
⎟⎟ +
+
Tfreezer
⎠ Tfreezer
⎡ − Lf
⎛T
= m⎢
+ C p ln⎜⎜ f
⎝ Ti
⎣⎢ Tfreezing
⎞ Lf + C p ΔT ⎤
⎟⎟ +
⎥
Tfreezer ⎦⎥
⎠
⎞
⎟⎟
⎠
The Second Law of Thermodynamics 1911
Substitute numerical values and evaluate ΔSu:
⎡
3 J
⎢ 333.5 ×10 kg ⎛
J ⎞ ⎛ 263 K ⎞
⎟⎟
⎟⎟ ln⎜⎜
+ ⎜⎜ 2100
ΔS u = (0.0500 kg ) ⎢−
⋅
273
K
kg
K
273 K
⎢
⎝
⎠ ⎝
⎠
⎢⎣
⎤
J ⎞
J ⎛
⎟⎟ (273 K − 263 K ) ⎥
+ ⎜⎜ 2100
333.5 ×10 3
kg ⋅ K ⎠
kg ⎝
⎥ = 2.40 J/K
+
⎥
263 K
⎥
⎦
and, because ΔSu > 0, the entropy of the universe increases.
In this problem, 2.00 mol of an ideal gas at 400 K expand quasi56 •
statically and isothermally from an initial volume of 40.0 L to a final volume of
80.0 L. (a) What is the entropy change of the gas? (b) What is the entropy change
of the universe for this process?
Picture the Problem We can use the definition of entropy change and the 1st law
of thermodynamics to express ΔS for the ideal gas as a function of its initial and
final volumes.
(a) The entropy change of the gas is
given by:
ΔS gas =
Apply the first law of thermodynamics
to the isothermal process to express Q
in terms of Won:
Q = ΔEint − Won
Q
T
(1)
or, because ΔEint = 0 for an isothermal
expansion of a gas,
Q = −Won
⎞
⎛V
⎟⎟ ⇒ Q = −nRT ln⎜⎜ i
⎠
⎝ Vf
The work done on the gas is given
by:
⎛V
Won = nRT ln⎜⎜ i
⎝ Vf
Substitute for Q in equation (1) to
obtain:
⎛V
ΔS gas = −nR ln⎜⎜ i
⎝ Vf
⎞
⎟⎟
⎠
Substitute numerical values and evaluate ΔS:
J ⎞ ⎛ 40.0 L ⎞
J
⎛
⎟⎟ = 11.5
ΔS gas = −(2.00 mol)⎜ 8.314
⎟ ln⎜⎜
mol ⋅ K ⎠ ⎝ 80.0 L ⎠
K
⎝
⎞
⎟⎟
⎠
1912 Chapter 19
(b) Because the process is reversible:
ΔS u = 0
Remarks: The entropy change of the environment of the gas is −11.5 J/K.
57 ••
[SSM] A system completes a cycle consisting of six quasi-static
steps, during which the total work done by the system is 100 J. During step 1 the
system absorbs 300 J of heat from a reservoir at 300 K, during step 3 the system
absorbs 200 J of heat from a reservoir at 400 K, and during step 5 it absorbs heat
from a reservoir at temperature T3. (During steps 2, 4 and 6 the system undergoes
adiabatic processes in which the temperature of the system changes from one
reservoir’s temperature to that of the next.) (a) What is the entropy change of the
system for the complete cycle? (b) If the cycle is reversible, what is the
temperature T3?
Picture the Problem We can use the fact that the system returns to its original
state to find the entropy change for the complete cycle. Because the entropy
change for the complete cycle is the sum of the entropy changes for each process,
we can find the temperature T3 from the entropy changes during the 1st two
processes and the heat released during the third.
(a) Because S is a state function of
the system, and because the system’s
final state is identical to its initial
state:
ΔSsystem
(b) Relate the entropy changes for
each of the three heat reservoirs and
the system for one complete cycle of
the system:
ΔS1 + ΔS 2 + ΔS 3 + ΔS system = 0
= 0
1 complete cycle
or
Q1 Q2 Q3
+
+
+0 = 0
T1 T2 T3
Substitute numerical values. Heat is
rejected by the two high-temperature
reservoirs and absorbed by the cold
reservoir:
− 300 J − 200 J 400 J
+
+
=0
300 K 400 K
T3
Solving for T3 yields:
T3 = 267 K
58 ••
In this problem, 2.00 mol of an ideal gas initially has a temperature of
400 K and a volume of 40.0 L. The gas undergoes a free adiabatic expansion to
twice its initial volume. What is (a) the entropy change of the gas and (b) the
entropy change of the universe?
The Second Law of Thermodynamics 1913
Picture the Problem The initial and final temperatures are the same for a free
expansion of an ideal gas. Thus, the entropy change ΔS for a free expansion from
Vi to Vf is the same as ΔS for an isothermal process from Vi to Vf. We can use the
definition of entropy change and the 1st law of thermodynamics to express ΔS for
the ideal gas as a function of its initial and final volumes.
(a) The entropy change of the gas is
given by:
ΔS gas =
Apply the first law of thermodynamics
to the isothermal process to express Q:
Q = ΔEint − Won
Q
T
(1)
or, because ΔEint = 0 for a free
expansion of a gas,
Q = −Won
⎞
⎛V
⎟⎟ ⇒ Q = −nRT ln⎜⎜ i
⎠
⎝ Vf
The work done on the gas is given
by:
⎛V
Won = nRT ln⎜⎜ i
⎝ Vf
Substitute for Q in equation (1) to
obtain:
⎛V
ΔS gas = −nR ln⎜⎜ i
⎝ Vf
⎞
⎟⎟
⎠
Substitute numerical values and evaluate ΔS:
J ⎞ ⎛ 40.0 L ⎞
J
⎛
⎟⎟ = 11.5
ΔS gas = −(2.00 mol)⎜ 8.314
⎟ ln⎜⎜
mol ⋅ K ⎠ ⎝ 80.0 L ⎠
K
⎝
(b) The change in entropy of the
universe is the sum of the entropy
changes of the gas and the
surroundings:
ΔS u = ΔS gas + ΔS surroundings
For the change in entropy of the
surroundings we use the fact that,
during the free expansion, the
surroundings are unaffected:
ΔS surroundings =
The change in entropy of the universe
is the change in entropy of the gas:
ΔS u = 11.5
Qrev 0
= =0
T
T
J
K
59 ••
A 200-kg block of ice at 0.0ºC is placed in a large lake. The
temperature of the lake is just slightly higher than 0.0ºC, and the ice melts very
⎞
⎟⎟
⎠
1914 Chapter 19
slowly. (a) What is the entropy change of the ice? (b) What is the entropy change
of the lake? (c) What is the entropy change of the universe (the ice plus the lake)?
Picture the Problem Because the ice gains heat as it melts, its entropy change is
positive and can be calculated from its definition. Because the temperature of the
lake is just slightly greater than 0°C and the mass of water is so much greater than
that of the block of ice, the absolute value of the entropy change of the lake will
be approximately equal to the entropy change of the ice as it melts.
(a) The entropy change of the ice is
given by:
Substitute numerical values and
evaluate ΔSice :
(b) Relate the entropy change of the
lake to the entropy change of the ice:
(c) The entropy change of the
universe due to this melting process
is the sum of the entropy changes of
the ice and the lake:
ΔS ice =
ΔS ice =
mLf
T
(200 kg )⎛⎜⎜ 333.5 kJ ⎞⎟⎟
⎝
273 K
kg ⎠
ΔSlake ≈ −ΔSice = − 244
= 244
kJ
K
kJ
K
ΔS u = ΔSice + ΔSlake
Because the temperature of the lake is slightly greater than that of the ice, the
magnitude of the entropy change of the lake is less than 244 kJ/K and the entropy
change of the universe is greater than zero. The melting of the ice is an
irreversible process and ΔS u > 0 .
60 ••
A 100-g piece of ice at 0.0ºC is placed in an insulated calorimeter with
negligible heat capacity containing 100 g of water at 100ºC. (a) What is the final
temperature of the water once thermal equilibrium is established? (b) Find the
entropy change of the universe for this process.
Picture the Problem We can use conservation of energy to find the equilibrium
temperature of the water and apply the equations for the entropy change during a
melting process and for constant-pressure processes to find the entropy change of
the universe (the entropy change of the piece of ice plus the entropy change of the
water in the insulated container).
The Second Law of Thermodynamics 1915
∑Q
(a) Apply conservation of energy
to obtain:
i
=0
i
or
Qmelting + Qwarming − Qcooling = 0
ice
water
water
Substitute to relate the masses of the ice and water to their temperatures,
specific heats, and the final temperature of the water:
(100 g )⎛⎜⎜ 333.5 kJ ⎞⎟⎟ + (100 g )⎛⎜⎜ 4.18
kg ⎠
⎝
⎝
kJ ⎞
⎟t
kg ⋅ C° ⎟⎠
⎛
kJ ⎞
⎟ (100°C − t ) = 0
− (100 g )⎜⎜ 4.18
kg ⋅ C° ⎟⎠
⎝
Solving for t yields:
t = 10.1°C
(b) The entropy change of the
universe is the sum of the entropy
changes of the ice and the water:
ΔS u = ΔSice + ΔS water
Using the expression for the entropy
change for a constant-pressure
process, express the entropy change
of the melting ice and warming icewater:
ΔSice = ΔS melting ice + ΔS warming water
=
⎛T
mLf
+ mcP ln⎜⎜ f
Tf
⎝ Ti
⎞
⎟⎟
⎠
Substitute numerical values to obtain:
ΔSice =
(0.100 kg )⎛⎜⎜ 333.5 kJ ⎞⎟⎟
⎝
273 K
kg ⎠
⎛
J
kJ ⎞ ⎛ 283 K ⎞
⎟⎟ = 137
⎟⎟ ln⎜⎜
+ (0.100 kg )⎜⎜ 4.18
K
kg ⋅ K ⎠ ⎝ 273 K ⎠
⎝
Find the entropy change of the cooling water:
⎛
kJ ⎞ ⎛ 283 K ⎞
J
⎟⎟ = −115
⎟⎟ ln⎜⎜
ΔS water = (0.100 kg )⎜⎜ 4.18
kg ⋅ K ⎠ ⎝ 373 K ⎠
K
⎝
Substitute for ΔSice and ΔSwater and
evaluate the entropy change of the
universe:
ΔS u = 137
J
J
J
− 115 = 22
K
K
K
1916 Chapter 19
Remarks: The result that ΔSu > 0 tells us that this process is irreversible.
61 ••
[SSM] A 1.00-kg block of copper at 100ºC is placed in an insulated
calorimeter of negligible heat capacity containing 4.00 L of liquid water at 0.0ºC.
Find the entropy change of (a) the copper block, (b) the water, and (c) the
universe.
Picture the Problem We can use conservation of energy to find the equilibrium
temperature of the water and apply the equations for the entropy change during a
constant pressure process to find the entropy changes of the copper block, the
water, and the universe.
(a) Use the equation for the entropy
change during a constant-pressure
process to express the entropy
change of the copper block:
⎛T
ΔS Cu = mCu cCu ln⎜⎜ f
⎝ Ti
Apply conservation of energy to
obtain:
∑Q
i
⎞
⎟⎟
⎠
(1)
=0
i
or
Qcopper block + Qwarming water = 0
Substitute to relate the masses of the block and water to their temperatures,
specific heats, and the final temperature Tf of the water:
(1.00 kg )⎛⎜⎜ 0.386
⎝
kJ ⎞
⎟ (Tf − 373 K )
kg ⋅ K ⎟⎠
kg ⎞ ⎛
kJ ⎞
⎛
⎟ (Tf − 273 K ) = 0
+ (4.00 L )⎜1.00 ⎟ ⎜⎜ 4.18
L ⎠⎝
kg ⋅ K ⎟⎠
⎝
Solve for Tf to obtain:
Tf = 275.26 K
Substitute numerical values in equation (1) and evaluate ΔS Cu :
⎛
kJ ⎞ ⎛ 275.26 K ⎞
J
⎟⎟ = − 117
⎟⎟ln⎜⎜
ΔS Cu = (1.00 kg )⎜⎜ 0.386
kg ⋅ K ⎠ ⎝ 373 K ⎠
K
⎝
(b) The entropy change of the
water is given by:
⎛T
ΔS water = mwater c water ln⎜⎜ f
⎝ Ti
⎞
⎟⎟
⎠
The Second Law of Thermodynamics 1917
Substitute numerical values and evaluate ΔS water :
⎛
J
kJ ⎞ ⎛ 275.26 K ⎞
⎟⎟ = 138
⎟⎟ ln⎜⎜
ΔS water = (4.00 kg )⎜⎜ 4.18
kg ⋅ K ⎠ ⎝ 273 K ⎠
K
⎝
(c) Substitute for ΔS Cu and ΔS water
and evaluate the entropy change of
the universe:
ΔS u = ΔSCu + ΔS water = −117
= 20
J
J
+ 138
K
K
J
K
Remarks: The result that ΔSu > 0 tells us that this process is irreversible.
If a 2.00-kg piece of lead at 100ºC is dropped into a lake at 10ºC, find
62 ••
the entropy change of the universe.
Picture the Problem Because the mass of the water in the lake is so much
greater than the mass of the piece of lead, the temperature of the lake will
increase only slightly and we can reasonably assume that its final temperature is
10°C. We can apply the equation for the entropy change during a constant
pressure process to find the entropy changes of the piece of lead, the water in the
lake, and the universe.
Express the entropy change of the
universe in terms of the entropy
changes of the lead and the water in
the lake:
ΔS u = ΔS Pb + ΔS w
(1)
Using the equation for the entropy change during a constant-pressure
process, express and evaluate the entropy change of the lead:
⎛T
ΔS Pb = mPb cPb ln⎜⎜ f
⎝ Ti
⎞
⎛
kJ ⎞ ⎛ 283 K ⎞
J
⎟⎟ = (2.00 kg )⎜⎜ 0.128
⎟⎟ = −70.69
⎟⎟ ln⎜⎜
kg ⋅ K ⎠ ⎝ 373 K ⎠
K
⎝
⎠
The entropy change of the water in
the lake is given by:
ΔS w =
Qw QPb mPb cPb ΔTPb
=
=
Tw
Tw
Tw
1918 Chapter 19
Substitute numerical values and
evaluate ΔSw:
ΔS w =
kJ ⎞
⎟ (90 K )
kg ⋅ K ⎟⎠
⎝
283 K
(2.00 kg )⎛⎜⎜ 0.128
= 81.41 J/K
Substitute numerical values in
equation (1) and evaluate ΔSu:
ΔS u = −70.69
J
J
J
+ 81.41 = 11
K
K
K
Entropy and ″Lost″ Work
63 ••
[SSM] A a reservoir at 300 K absorbs 500 J of heat from a second
reservoir at 400 K. (a) What is the change in entropy of the universe, and (b) how
much work is lost during the process?
Picture the Problem We can find the entropy change of the universe from the
entropy changes of the high- and low-temperature reservoirs. The maximum
amount of the 500 J of heat that could be converted into work can be found from
the maximum efficiency of an engine operating between the two reservoirs.
(a) The entropy change of the
universe is the sum of the
entropy changes of the two
reservoirs:
ΔS u = ΔS h + ΔSc = −
Substitute numerical values and
evaluate ΔSu:
⎛ 1
1 ⎞
⎟⎟
−
ΔS u = (− 500 J ) ⎜⎜
⎝ 400 K 300 K ⎠
Q Q
+
Th Tc
⎛1 1⎞
= −Q⎜⎜ − ⎟⎟
⎝ Th Tc ⎠
= 0.42 J/K
(b) Relate the heat that could have
been converted into work to the
maximum efficiency of an engine
operating between the two
reservoirs:
W = ε max Qh
The maximum efficiency of an
engine operating between the two
reservoir temperatures is the
efficiency of a Carnot device
operating between the reservoir
temperatures:
ε max = ε C = 1 −
Tc
Th
The Second Law of Thermodynamics 1919
Substitute for εmax to obtain:
⎛ T ⎞
W = ⎜⎜1 − c ⎟⎟Qh
⎝ Th ⎠
Substitute numerical values and
evaluate W:
⎛ 300 K ⎞
⎟⎟ (500 J ) = 125 J
W = ⎜⎜1 −
⎝ 400 K ⎠
64 ••
In this problem, 1.00 mol of an ideal gas at 300 K undergoes a free
adiabatic expansion from V1 = 12.3 L to V2 = 24.6 L. It is then compressed
isothermally and reversibly back to its original state. (a) What is the entropy
change of the universe for the complete cycle? (b) How much work is lost in this
cycle? (c) Show that the work lost is TΔSu.
Picture the Problem Although no heat is lost by the gas in the adiabatic free
expansion, the process is irreversible and the entropy of the gas increases. In the
isothermal reversible process that returns the gas to its original state, the gas
releases heat to the surroundings. However, because the process is reversible, the
entropy change of the universe is zero. Consequently, the net entropy change is
the negative of that of the gas in the isothermal compression.
(a) Relate the entropy change of the
universe to the entropy changes of
the gas during 1 complete cycle:
ΔS u = ΔS gas during
free expansion
+ ΔS gas during
isothermal compresion
or, because ΔSgas during
=0,
ΔS u = ΔS gas during
=
free expansion
isothermal compresion
−Q
T
The work done by the gas during its
isothermal compression is given by:
⎛V
Wby = −Won = −Q = −nRT ln⎜⎜ f
⎝ Vi
Substituting for ΔS gas in the
⎛V
ΔS u = − nR ln⎜⎜ f
⎝ Vi
expression for ΔS u yields:
⎞
⎟⎟
⎠
⎞
⎟⎟
⎠
(1)
Substitute numerical values and evaluate ΔSu:
J ⎞ ⎛ 12.3 L ⎞
J
J
⎛
⎟⎟ = 5.763 = 5.76
ΔS u = −(1.00 mol)⎜ 8.314
⎟ln⎜⎜
mol ⋅ K ⎠ ⎝ 24.6 L ⎠
K
K
⎝
(b) Use Equation 19-22 to find the
amount of energy that becomes
unavailable for doing work during
this process:
J⎞
⎛
Wlost = TΔS u = (300 K )⎜ 5.763 ⎟
K⎠
⎝
= 1.73 kJ
1920 Chapter 19
(c) Use equation (1) to express the
product of T and ΔSu:
⎛
⎛V
TΔS u = T ⎜⎜ − nR ln⎜⎜ f
⎝ Vi
⎝
= Wby gas during its
⎞⎞
⎛V
⎟⎟ ⎟ = − nRT ln⎜⎜ f
⎟
⎠⎠
⎝ Vi
⎞
⎟⎟
⎠
isothermal compression
General Problems
65 •
A heat engine with an output of 200 W has an efficiency of 30%. It
operates at 10.0 cycles/s. (a) How much work is done by the engine during each
cycle? (b) How much heat is absorbed from the hot reservoir and how much is
released to the cold reservoir during each cycle?
Picture the Problem We can use the definition of power to find the work done
each cycle and the definition of efficiency to find the heat that is absorbed each
cycle. Application of the first law of thermodynamics will yield the heat given off
each cycle.
(a) Use the definition of power to
relate the work done in each cycle to
the frequency of each cycle:
Substitute numerical values and
evaluate Wcycle:
(b) Express the heat absorbed in each
cycle in terms of the work done and
the efficiency of the engine:
Apply the 1st law of thermodynamics
to find the heat given off in each
cycle:
Wcycle = PΔt =
P
f
where f is the frequency of the engine.
Wcycle =
200 W
= 20.0 J
10.0 s −1
Qh,cycle =
Wcycle
ε
=
20.0 J
= 67 J
0.30
Qc,cycle = Qh,cycle − W = 67 J − 20 J
= 47 J
During each cycle, a heat engine operating between two heat
66 •
reservoirs absorbs 150 J from the reservoir at 100ºC and releases 125 J to the
reservoir at 20ºC. (a) What is the efficiency of this engine? (b) What is the ratio
of its efficiency to that of a Carnot engine working between the same reservoirs?
(This ratio is called the second law efficiency.)
Picture the Problem We can use their definitions to find the efficiency of the
engine and that of a Carnot engine operating between the same reservoirs.
The Second Law of Thermodynamics 1921
Q
W Qh − Qc
=
= 1− c
Qh
Qh
Qh
(a) The efficiency of the engine
is given by:
ε=
Substitute numerical values and
evaluate ε:
ε = 1−
(b) Find the efficiency of a Carnot
engine operating between the same
reservoirs:
εC = 1−
Express the ratio of the two
efficiencies:
ε 16.67%
=
= 0.777
ε C 21.45%
125 J
= 16.67% = 16.7%
150 J
Tc
293 K
= 1−
= 21.45%
Th
373 K
67 •
[SSM] An engine absorbs 200 kJ of heat per cycle from a reservoir
at 500 K and releases heat to a reservoir at 200 K. Its efficiency is 85 percent of
that of a Carnot engine working between the same reservoirs. (a) What is the
efficiency of this engine? (b) How much work is done in each cycle? (c) How
much heat is released to the low-temperature reservoir during each cycle?
Picture the Problem We can use the definition of efficiency to find the work
done by the engine during each cycle and the first law of thermodynamics to find
the heat released to the low-temperature reservoir during each cycle.
(a) Express the efficiency of the
engine in terms of the efficiency of a
Carnot engine working between the
same reservoirs:
Substitute numerical values and
evaluate ε :
(b) Use the definition of efficiency to
find the work done in each cycle:
(c) Apply the first law of
thermodynamics to the cycle to obtain:
⎛
ε = 0.85ε C = 0.85⎜⎜1 −
⎝
⎛
ε = 0.85⎜⎜1 −
⎝
Tc ⎞
⎟
Th ⎟⎠
200 K ⎞
⎟ = 0.510 = 51%
500 K ⎟⎠
W = ε Qh = (0.510)(200 kJ ) = 102 kJ
= 0.10 MJ
Qc,cycle = Qh,cycle − W = 200 kJ − 102 kJ
= 98 kJ
Estimate the change in entropy of the universe associated with an
68 •
Olympic diver diving into the water from the 10-m platform.
1922 Chapter 19
Picture the Problem Assume that the mass of the diver is 75 kg and that the
temperature of the water in the pool is 25°C. The energy added to the water in the
pool is the change in the gravitational potential energy of the diver during the
dive.
The change in entropy of the
universe associated with a dive is
given by:
Qadded to water
Twater
where Qadded to water is the energy entering
ΔS u = ΔS water =
the water as a result of the kinetic
energy of the diver as he enters the
water.
The energy added to the water is the
change in the gravitational potential
energy of the diver:
Substitute numerical values and
evaluate ΔS u :
ΔS u =
mgh
Twater
(
75 kg ) (9.81 m/s 2 )(10 m )
ΔS u =
(25 + 273)K
≈ 25
J
K
To maintain the temperature inside a house at 20ºC, the electric power
69 •
consumption of the electric baseboard heaters is 30.0 kW on a day when the
outside temperature is –7ºC. At what rate does this house contribute to the
increase in the entropy of the universe?
Picture the Problem The change in entropy of the universe is the change in
entropy of the house plus the change in entropy of the environment. We can find
the change in entropy of the house by exploiting the given information that the
temperature inside the house is maintained at a constant temperature. We can find
the change in entropy of the surrounding by dividing the heat added by the
temperature.
Entropy is a state function, and the
state of the house does not change.
Therefore the entropy of the house
does not change:
Heat is absorbed by the surroundings
at the same rate R that energy is
delivered to the house:
ΔS u = ΔS house + ΔSsurroundings
or, because ΔS house = 0 ,
ΔS u = ΔSsurroundings
ΔS surroundings =
Q
Tsurroundings
=
RΔt
Tsurroundings
The Second Law of Thermodynamics 1923
Substitute for ΔSsurroundings yields:
Substitute numerical values and
evaluate ΔSu/Δt:
ΔS u =
RΔt
Tsurroundings
⇒
ΔS u
R
=
Δt
Tsurroundings
ΔS u 30.0 kW
W
=
= 113
K
Δt
266 K
70 •• Calvin Cliffs Nuclear Power Plant, located on the Hobbes River,
generates 1.00 GW of power. In this plant, liquid sodium circulates between the
reactor core and a heat exchanger located in the superheated steam that drives the
turbine. Heat is absorbed by the liquid sodium in the core, and released by the
liquid sodium (and into the superheated steam) in the heat exchanger. The
temperature of the superheated steam is 500 K. Heat is released into the river, and
the water in the river flows by at a temperature of 25ºC. (a) What is the highest
efficiency that this plant can have? (b) How much heat is released into the river
every second? (c) How much heat must be released by the core to supply
1.00 GW of electrical power? (d) Assume that new environmental laws have been
passed to preserve the unique wildlife of the river. Because of these laws, the
plant is not allowed to heat the river by more than 0.50ºC. What is the minimum
flow rate that the water in the Hobbes River must have?
Picture the Problem We can use the expression for the Carnot efficiency of the
plant to find the highest efficiency this plant can have. We can then use this
efficiency to find the power that must be supplied to the plant to generate
1.00 GW of power and, from this value, the power that is wasted. The rate at
which heat is being released to the river is related to the requisite flow rate of the
river by dQ dt = cΔTρ dV dt .
(a) The Carnot efficiency of a plant
operating between temperatures Tc
and Th is given by:
ε max = ε C = 1 −
Substitute numerical values and
evaluate εC:
ε max = 1 −
298 K
= 0.404
500 K
Psupplied =
Poutput
(c) The power that must be supplied,
at 40.4% efficiency, to produce an
output of 1.00 GW is given by:
(b) Relate the wasted power to the
power generated and the power
supplied:
ε max
Tc
Th
=
1.00 GW
0.404
= 2.48 GW
Pwasted = Psupplied − Pgenerated
1924 Chapter 19
Substitute numerical values and
evaluate Pwasted :
Pwasted = 2.48 GW − 1.00 GW
(d) Express the rate at which heat is
being dumped into the river:
dQ
dm
d
= cΔT
= cΔT (ρV )
dt
dt
dt
dV
= cΔTρ
dt
Solve for the flow rate dV/dt of the
river:
dV dQ dt
=
dt
cΔTρ
Substitute numerical values (see
Table 19-1 for the specific heat of
water) and evaluate dV/dt:
J
dV
s
=
dt ⎛
kg ⎞
J ⎞
⎛
⎜⎜ 4180 ⎟⎟ (0.50 K )⎜10 3 3 ⎟
kg ⎠
m ⎠
⎝
⎝
= 1.48 GW
1.48 × 10 9
= 7.1× 10 5 L/s
71 •• An inventor comes to you to explain his new invention. It is a novel
heat engine using water vapor as the working substance. He claims that the water
vapor absorbs heat at 100°C, does work at the rate of 125 W, and releases heat to
the air at the rate of only 25.0 W, when the air temperature is 25°C. (a) Explain
to him why he cannot be correct. (b) After careful analysis of the data in his
prospectus folder, you decide he has made an error in the measurement of his
exhausted-heat value. What is the minimum rate of exhausting heat that would
make you consider believing him?
Picture the Problem We can use the inventor’s data to calculate the thermal
efficiency of his steam engine and then compare this value to the efficiency of a
Carnot engine operating between the same temperatures.
(a) The Carnot efficiency of an
engine operating between these
temperatures is:
The thermal efficiency of the
inventor’s device, in terms of the rate
at which it expels heat to the air and
does work is:
ε C = 1−
Tc
298 K
= 1−
= 20.1%
373 K
Th
dW
dW
dt
ε = dt =
dQh
dW dQc
+
dt
dt
dt
125 W
=
= 83.3%
125 W + 25.0 W
The Second Law of Thermodynamics 1925
You should explain to him that, because the efficiency he claims for his invention
is greater than the efficiency of a Carnot engine operating between the same two
temperatures, his data is not consistent with what is known about the
thermodynamics of engines. He must have made a mistake in his analysis of his
data−or he is a con man looking for people to swindle.
(b) The maximum efficiency of a steam engine that has ever been achieved is
about 50% of the Carnot efficiency of an engine operating between the same
temperatures.
Setting the efficiency of his steam
engine equal to half the Carnot
efficiency of the engine yields:
dW
dW
dt
1
ε = dt =
2 C
dQh
dW dQc
+
dt
dt
dt
Solve for dQc/dt to obtain:
⎞ dW
dQc ⎛ 2
= ⎜⎜ − 1⎟⎟
dt
⎝ ε C ⎠ dt
Assuming that the inventor has
measured the work done per cycle
by his invention correctly:
dQc ⎛ 2
⎞
=⎜
− 1⎟ (125 W ) ≈ 1100 W
dt
⎝ 0.201 ⎠
a value totally inconsistent with the
inventor’s claims for his engine.
Ignoring his claim that 25.0 W of
work are done per cycle, let’s
assume that his device does take in
150 W of energy each cycle and find
how much work would do with an
efficiency half that of a Carnot
engine:
dW
dW 1 dQh
1
ε = dt ⇒
= 2 εC
2 C
dQh
dt
dt
dt
Substituting numerical values yields:
dW
=
dt
dQc dQh dW
=
−
,a
dt
dt
dt
reasonable value for dQc/dt is:
dQc
= 150 W − 15 W = 135 W
dt
Because
1
2
(0.201)(150 W ) ≈ 15 W
The cycle represented in Figure 19-12 (next to Problem 19-14) is for
72 ••
1.00 mol of an ideal monatomic gas. The temperatures at points A and B are 300
and 750 K, respectively. What is the efficiency of the cyclic process ABCDA?
1926 Chapter 19
Picture the Problem Because the cycle represented in Figure 19-12 is a Carnot
cycle, its efficiency is that of a Carnot engine operating between the temperatures
of its isotherms.
The Carnot efficiency of the cycle is
given by:
εC = 1−
Tc
Th
Substitute numerical values and
evaluate εC:
ε C = 1−
300 K
= 60.0%
750 K
73 ••
[SSM] (a) Which of these two processes is more wasteful? (1) A
block moving with 500 J of kinetic energy being slowed to rest by sliding
(kinetic) friction when the temperature of the environment is 300 K, or (2) A
reservoir at 400 K releasing 1.00 kJ of heat to a reservoir at 300 K? Explain your
choice. Hint: How much of the 1.00 kJ of heat could be converted into work by an
ideal cyclic process? (b) What is the change in entropy of the universe for each
process?
Picture the Problem All 500 J of mechanical energy are lost, i.e., transformed
into heat in process (1). For process (2), we can find the heat that would be
converted to work by a Carnot engine operating between the given temperatures
and subtract that amount of work from 1.00 kJ to find the energy that is lost. In
Part (b) we can use its definition to find the change in entropy for each process.
(a) For process (2):
W2,max = Wrecovered = ε CQin
The efficiency of a Carnot engine
operating between temperatures
Th and Tc is given by:
ε C = 1−
Tc
Th
and hence
⎛ T
Wrecovered = ⎜⎜1 − c
⎝ Th
Substitute for ε C to obtain:
⎞
⎟⎟Qin
⎠
⎛ 300 K ⎞
Wrecovered = ⎜1 −
⎟(1.00 kJ ) = 250 J
⎝ 400 K ⎠
or
750 J are lost.
Process (2) is more wasteful of available work.
(b) Find the change in entropy of the
universe for process (1):
ΔS1 =
ΔQ 500 J
=
= 1.67 J/K
T
300 K
The Second Law of Thermodynamics 1927
Express the change in entropy of
the universe for process (2):
ΔS 2 = ΔS h + ΔSc = −
ΔQ ΔQ
+
Th
Tc
⎛1 1⎞
= ΔQ ⎜⎜ − ⎟⎟
⎝ Tc Th ⎠
Substitute numerical values and
evaluate ΔS2:
⎛ 1
1 ⎞
⎟⎟
ΔS 2 = (1.00 kJ )⎜⎜
−
⎝ 300 K 400 K ⎠
= 0.833 J/K
Helium, a monatomic gas, is initially at a pressure of 16 atm, a volume
74 ••
of 1.0 L, and a temperature of 600 K. It is quasi-statically expanded at constant
temperature until its volume is 4.0 L and is then quasi-statically compressed at
constant pressure until its volume and temperature are such that a quasi-static
adiabatic compression will return the gas to its original state. (a) Sketch this
cycle on a PV diagram. (b) Find the volume and temperature after the
compression at constant pressure. (c) Find the work done during each step of the
cycle. (d) Find the efficiency of the cycle.
Picture the Problem Denote the three states of the gas as 1, 2, and 3 with 1 being
the initial state. We can use the ideal-gas law and the equation of state for an
adiabatic process to find the temperatures, volumes, and pressures at points 1, 2,
and 3. To find the work done during each cycle, we can use the equations for the
work done during isothermal, isobaric, and adiabatic processes. Finally, we find
the efficiency of the cycle from the work done each cycle and the heat that enters
the system during the isothermal expansion.
(a) The PV diagram of the cycle is
shown to the right.
(b) Apply the ideal-gas law to the
isothermal expansion 1→2 to find
P2:
P2 = P1
⎛ 1.0 L ⎞
V1
⎟⎟ = 4.0 atm
= (16 atm )⎜⎜
4.0
L
V2
⎝
⎠
1928 Chapter 19
1γ
Apply an equation for an adiabatic
process to relate the pressures and
volumes at 1 and 3:
⎛P⎞
P1V1 = P3V3 ⇒ V3 = V1 ⎜⎜ 1 ⎟⎟
⎝ P3 ⎠
Substitute numerical values and
evaluate V3:
⎛ 16 atm ⎞
⎟⎟
V3 = (1.0 L )⎜⎜
⎝ 4.0 atm ⎠
γ
γ
1 1.67
= 2.294 L
= 2.3 L
Apply an equation for an adiabatic
process (γ =1.67) to relate the
temperatures and volumes at 1 and 3:
T3V3
Substitute numerical values and
evaluate T3:
⎛ 1.0 L ⎞
⎟⎟
T3 = (600 K )⎜⎜
⎝ 2.294 L ⎠
γ −1
γ −1
= T1V1
⎛V ⎞
⇒ T3 = T1 ⎜⎜ 1 ⎟⎟
⎝ V3 ⎠
γ −1
1.67 −1
= 344 K
= 3.4 ×10 2 K
(c) Express the work done each
cycle:
W = W12 + W23 + W31
For the process 1→2:
⎛V ⎞
⎛V ⎞
W12 = nRT1 ln⎜⎜ 2 ⎟⎟ = P1V1 ln⎜⎜ 2 ⎟⎟
⎝ V1 ⎠
⎝ V1 ⎠
⎛ 4.0 L ⎞
⎟⎟
= (16 atm )(1.0 L )ln⎜⎜
⎝ 1.0 L ⎠
(1)
= 22.18 atm ⋅ L
For the process 2→3:
W23 = P2 ΔV23
= (4.0 atm )(2.294 L − 4.00 L )
= −6.824 atm ⋅ L
For the process 3→1:
W31 = −C V ΔT31 = − 32 nR(T1 − T3 ) = − 32 (P1V1 − P3V3 )
= − 32 [(16 atm )(1.0 L ) − (4.0 atm )(2.294 L )]
= −10.24 atm ⋅ L
Substitute numerical values in
equation (1) and evaluate W:
W = 22.18 atm ⋅ L − 6.824 atm ⋅ L
− 10.24 atm ⋅ L
= 5.116 atm ⋅ L = 5 atm ⋅ L
The Second Law of Thermodynamics 1929
(d) Use its definition to express
the efficiency of the cycle:
ε=
W
W
W
=
=
Qin Q12 W12
Substitute numerical values and
evaluate ε:
ε=
5.116 atm ⋅ L
≈ 20%
22.18 atm ⋅ L
75 ••
[SSM] A heat engine that does the work of blowing up a balloon at a
pressure of 1.00 atm absorbs 4.00 kJ from a reservoir at 120ºC. The volume of the
balloon increases by 4.00 L, and heat is released to a reservoir at a temperature Tc,
where Tc < 120ºC. If the efficiency of the heat engine is 50% of the efficiency of a
Carnot engine working between the same two reservoirs, find the temperature Tc.
Picture the Problem We can express the temperature of the cold reservoir as a
function of the Carnot efficiency of an ideal engine and, given that the efficiency
of the heat engine is half that of a Carnot engine, relate Tc to the work done by and
the heat input to the real heat engine.
Using its definition, relate the
efficiency of a Carnot engine
working between the same reservoirs
to the temperature of the cold
reservoir:
εC = 1−
Relate the efficiency of the heat
engine to that of a Carnot engine
working between the same
temperatures:
ε=
Substitute for ε C to obtain:
⎛ 2W
Tc = Th ⎜⎜1 −
⎝ Qin
The work done by the gas in
expanding the balloon is:
Tc
⇒ Tc = Th (1 − ε C )
Th
2W
W
= 12 ε C ⇒ ε C =
Qin
Qin
⎞
⎟⎟
⎠
W = PΔV = (1.00 atm )(4.00 L )
= 4.00 atm ⋅ L
1930 Chapter 19
Substitute numerical values and evaluate Tc:
⎛
101.325 J ⎞ ⎞
⎛
⎜ 2 ⎜ 4.00 atm ⋅ L ×
⎟⎟
atm ⋅ L ⎠ ⎟
⎝
⎜
Tc = (393 K ) 1 −
= 313 K
⎜
⎟
4.00 kJ
⎜
⎟
⎝
⎠
78 ••
Show that the coefficient of performance of a Carnot engine run as a
refrigerator is related to the efficiency of a Carnot engine operating between the
same two temperatures by ε C × COPC = Tc Th .
Picture the Problem We can use the definitions of the COP and εC to show that
their relationship is ε C × COPC = TC Th .
Using the definition of the COP,
relate the heat removed from the cold
reservoir to the work done each
cycle:
COP =
Apply energy conservation to relate
Qc, Qh, and W:
Qc = Qh − W
Substitute for Qc to obtain:
COP =
Divide the numerator and
denominator by Qh and simplify to
obtain:
W
Q −W
Qh
=
COP = h
W
W
Qh
Because ε C =
T
W
= 1− c :
Qh
Th
Qc
W
Qh − W
W
1−
⎛ T
1 − ⎜⎜1 − c
Th
1− ε C
COPc =
= ⎝
εC
εC
⎞ Tc
⎟⎟
⎠ = Th
εC
and
ε C × COPc =
Tc
Th
1
77 •• A freezer has a temperature Tc = –23ºC. The air in the kitchen has
a temperature Th = 27ºC. The freezer is not perfectly insulated and some heat
The Second Law of Thermodynamics 1931
leaks through the walls of the freezer at a rate of 50 W. Find the power of the
motor that is needed to maintain the temperature in the freezer.
Picture the Problem We can use the definition of the COP to express the work
the motor must do to maintain the temperature of the freezer in terms of the rate at
which heat flows into the freezer. Differentiation of this expression with respect
to time will yield an expression for the power of the motor that is needed to
maintain the temperature in the freezer.
Using the definition of the COP,
relate the heat that must be removed
from the freezer to the work done by
the motor:
COP =
Differentiate this expression with
respect to time to express the power
of the motor:
P=
Express the maximum COP of the
motor:
COPmax =
Substitute for COPmax to obtain:
P=
Substitute numerical values and
evaluate P:
⎛ 50 K ⎞
⎟⎟ = 10 W
P = (50 W )⎜⎜
⎝ 250 K ⎠
Qc
Q
⇒W = c
W
COP
dW dQc dt
=
COP
dt
Tc
ΔT
dQc ΔT
dt Tc
78 ••
In a heat engine, 2.00 mol of a diatomic gas are taken through the
cycle ABCA as shown in Figure 19-20. (The PV diagram is not drawn to scale.)
At A the pressure and temperature are 5.00 atm and 600 K. The volume at B is
twice the volume at A. The segment BC is an adiabatic expansion and the
segment CA is an isothermal compression. (a) What is the volume of the gas at
A? (b) What are the volume and temperature of the gas at B? (c) What is the
temperature of the gas at C? (d) What is the volume of the gas at C? (e) How
much work is done by the gas in each of the three segments of the cycle? (f) How
much heat is absorbed or released by the gas in each segment of this cycle?
Picture the Problem We can use the ideal-gas law to find the unknown
temperatures, pressures, and volumes at points A, B, and C and then find the work
done by the gas and the efficiency of the cycle by using the expressions for the
work done on or by the gas and the heat that enters the system for the constantpressure, adiabatic, and isothermal processes of the cycle.
1932 Chapter 19
(a) Apply the ideal-gas law to find
the volume of the gas at A:
VA =
nRTA
PA
(2.00 mol)⎛⎜ 8.314
J ⎞
⎟ (600 K )
mol ⋅ K ⎠
⎝
=
101.325 kPa
5.00 atm ×
atm
1
L
= 1.969 ×10 − 2 m 3 × −3 3 = 19.69 L
10 m
= 19.7 L
(b) We’re given that VB = 2VA .
Hence:
VB = 2(19.69 L ) = 39.38 L = 39.4 L
Apply the ideal-gas law to this
constant-pressure process to
obtain:
TB = TA
(c) Because the process C→A is
isothermal:
TC = TA = 600 K
VB
2V
= (600 K ) A
VA
VA
= 1200 K
1
(d) Apply an equation for an
adiabatic process (γ = 1.4) to find the
volume of the gas at C:
TBVB
Substitute numerical values and
evaluate VC:
⎛ 1200 K ⎞ 1.4−1
⎟⎟ = 222.77 L
VC = (39.38 L )⎜⎜
600
K
⎠
⎝
γ −1
γ −1
= TCVC
⎛ TB ⎞ γ −1
⇒ VC = VB ⎜⎜ ⎟⎟
⎝ TC ⎠
1
= 223 L
(e) The work done by the gas
during the constant-pressure
process AB is given by:
Substitute numerical values and
evaluate WAB:
WAB = PA (VB − VA ) = PA (2VA − VA )
= PAVA
WAB = (5.00 atm )(19.69 L )
= 98.45 atm ⋅ L ×
101.325 J
atm ⋅ L
= 9.9754 ×10 3 J = 9.98 kJ
The Second Law of Thermodynamics 1933
Apply the first law of thermodynamics
to express the work done on the gas
during the adiabatic expansion BC:
WBC = ΔEint, BC − Qin, BC = ΔEint, BC − 0
= ΔEint, BC = −ncV ΔTBC
= − 52 nRΔTBC
Substitute numerical values and evaluate WBC:
J ⎞
⎛
4
WBC = − 52 (2.00 mol)⎜ 8.314
⎟ (600 K − 1200 K ) = 2.494 ×10 J
mol ⋅ K ⎠
⎝
= 24.9 kJ
The work done by the gas during the isothermal compression CA is:
⎛V
WCA = nRTC ln⎜⎜ A
⎝ VC
⎞
⎛ 19.69 L ⎞
J ⎞
⎛
⎟⎟ = (2.00 mol)⎜ 8.314
⎟⎟
⎟ (600 K ) ln⎜⎜
mol ⋅ K ⎠
⎝
⎝ 222.77 L ⎠
⎠
= −24.20 kJ = − 24.2 kJ
(f) The heat absorbed during the constant-pressure expansion AB is:
QAB = ncP ΔTA − B = 72 nRΔTA − B =
7
2
(2.00 mol)⎛⎜ 8.314
⎝
J ⎞
⎟ (1200 K − 600 K )
mol ⋅ K ⎠
= 34.92 kJ = 34.9 kJ
The heat absorbed during the
adiabatic expansion BC is:
QBC = 0
Use the first law of thermodynamics
to find the heat absorbed during the
isothermal compression CA:
QCA = WCA + ΔEint, CA = WCA
= − 24.2 kJ
because ΔEint, CA = 0 for an isothermal
process.
79 ••
[SSM] In a heat engine, 2.00 mol of a diatomic gas are carried
through the cycle ABCDA shown in Figure 19-21. (The PV diagram is not drawn
to scale.) The segment AB represents an isothermal expansion, the segment BC an
adiabatic expansion. The pressure and temperature at A are 5.00 atm and 600 K.
The volume at B is twice the volume at A. The pressure at D is 1.00 atm.
(a) What is the pressure at B? (b) What is the temperature at C? (c) Find the total
work done by the gas in one cycle.
1934 Chapter 19
Picture the Problem We can use the ideal-gas law to find the unknown
temperatures, pressures, and volumes at points B, C, and D. We can then find the
work done by the gas and the efficiency of the cycle by using the expressions for
the work done on or by the gas and the heat that enters the system for the various
thermodynamic processes of the cycle.
(a) Apply the ideal-gas law for a
fixed amount of gas to the
isothermal process AB to find the
pressure at B:
PB = PA
VA
V
= (5.00 atm ) A
2VA
VB
= 2.50 atm ×
101.325 kPa
= 253.3 kPa
1atm
= 253 kPa
(b) Apply the ideal-gas law for a
fixed amount of gas to the
adiabatic process BC to express the
temperature at C:
TC = TB
Use the ideal-gas law to find the
volume of the gas at B:
VB =
PCVC
PBVB
(1)
nRTB
PB
(2.00 mol)⎛⎜ 8.314
J ⎞
⎟ (600 K )
mol ⋅ K ⎠
⎝
253.3 kPa
=
= 39.39 L
1γ
Use the equation of state for an
adiabatic process and γ = 1.4 to
find the volume occupied by the
gas at C:
⎛P ⎞
VC = VB ⎜⎜ B ⎟⎟
⎝ PC ⎠
= 75.78 L
Substitute numerical values in
equation (1) and evaluate TC:
TC = (600 K )
(1.00 atm)(75.78 L )
(2.50 atm)(39.39 L)
= 462 K
(c) The work done by the gas in
one cycle is given by:
1 1.4
⎛ 2.50 atm ⎞
⎟⎟
= (39.39 L )⎜⎜
⎝ 1.00 atm ⎠
W = WAB + WBC + WCD + WDA
The Second Law of Thermodynamics 1935
The work done during the isothermal expansion AB is:
⎛V
WAB = nRTA ln⎜⎜ B
⎝ VA
⎞
⎛ 2V
J ⎞
⎛
⎟⎟ = (2.00 mol)⎜ 8.314
⎟ (600 K ) ln⎜⎜ A
mol ⋅ K ⎠
⎝
⎠
⎝ VA
⎞
⎟⎟ = 6.915 kJ
⎠
The work done during the adiabatic expansion BC is:
J ⎞
⎛
WBC = −C V ΔTBC = − 52 nRΔTBC = − 52 (2.00 mol)⎜ 8.314
⎟ (462 K − 600 K )
mol ⋅ K ⎠
⎝
= 5.737 kJ
The work done during the isobaric compression CD is:
WCD = PC (VD − VC ) = (1.00 atm )(19.7 L − 75.78 L ) = −56.09 atm ⋅ L ×
101.325 J
atm ⋅ L
= −5.680 kJ
Express and evaluate the work
done during the constant-volume
process DA:
WDA = 0
Substitute numerical values and
evaluate W:
W = 6.915 kJ + 5.737 kJ − 5.680 kJ + 0
= 6.972 kJ = 6.97 kJ
80 ••
In a heat engine, 2.00 mol of a monatomic gas are taken through the
cycle ABCA as shown in Figure 19-20. (The PV diagram is not drawn to scale.)
At A the pressure and temperature are 5.00 atm and 600 K. The volume at B is
twice the volume at A. The segment BC is an adiabatic expansion and the
segment CA is an isothermal compression. (a) What is the volume of the gas at
A? (b) What are the volume and temperature of the gas at B? (c) What is the
temperature of the gas at C? (d) What is the volume of the gas at C? (e) How
much work is done by the gas in each of the three segments of the cycle? (f) How
much heat is absorbed by the gas in each segment of the cycle?
Picture the Problem We can use the ideal-gas law to find the unknown
temperatures, pressures, and volumes at points A, B, and C and then find the work
done by the gas and the efficiency of the cycle by using the expressions for the
work done on or by the gas and the heat that enters the system for the isobaric,
adiabatic, and isothermal processes of the cycle.
1936 Chapter 19
(a) Apply the ideal-gas law to find
the volume of the gas at A:
VA =
nRTA
PA
(2.00 mol)⎛⎜ 8.314
J ⎞
⎟ (600 K )
mol ⋅ K ⎠
⎝
101.325 kPa
5.00 atm ×
atm
=
= 19.69 L = 19.7 L
(b) We’re given that:
VB = 2VA = 2(19.69 L ) = 39.39 L
= 39.4 L
2V
VB
= (600 K ) A
VA
VA
Apply the ideal-gas law to this
isobaric process to find the
temperature at B:
TB = TA
(c) Because the process CA is
isothermal:
TC = TA = 600 K
= 1200 K
1
(d) Apply an equation for an
adiabatic process (γ = 5/3) to express
the volume of the gas at C:
TBVB
Substitute numerical values and
evaluate VC:
⎛ 1200 K ⎞ 2
⎟⎟
VC = (39.39 L )⎜⎜
⎝ 600 K ⎠
γ −1
γ −1
= TCVC
⎛ TB ⎞ γ −1
⇒ VC = VB ⎜⎜ ⎟⎟
⎝ TC ⎠
3
= 111.4 L = 111 L
(e) The work done by the gas
during the isobaric process AB is
given by:
Substitute numerical values and
evaluate WAB:
WAB = PA (VB − VA ) = PA (2VA − VA )
= PAVA
WAB = (5.00 atm )(19.69 L )
= 98.45 atm ⋅ L ×
101.325 J
atm ⋅ L
= 9.975 kJ = 9.98 kJ
The Second Law of Thermodynamics 1937
Apply the first law of thermodynamics
to express the work done by the gas
during the adiabatic expansion BC:
WBC = ΔEint, BC − Qin, BC
= ΔEint, BC − 0
= ΔEint, BC = −(ncV ΔTBC )
= − 32 nRΔTBC
Substitute numerical values and evaluate WBC:
J ⎞
⎛
WBC = − 32 (2.00 mol)⎜ 8.314
⎟ (600 K − 1200 K ) = 14.97 kJ
mol ⋅ K ⎠
⎝
= 15.0 kJ
The work done by the gas during the isothermal compression CA is:
⎛V
WCA = nRTC ln⎜⎜ A
⎝ VC
⎞
⎛ 19.69 L ⎞
J ⎞
⎟⎟ = (2.00 mol)⎛⎜ 8.314
⎟⎟
⎟ (600 K ) ln⎜⎜
mol ⋅ K ⎠
⎝
⎝ 111.4 L ⎠
⎠
= −17.29 kJ − 17.3 kJ
(f) The heat absorbed during the isobaric expansion AB is:
Qin, AB = ncP ΔTAB = 52 nRΔTAB =
5
2
(2.00 mol)⎛⎜ 8.314
⎝
J ⎞
⎟ (1200 K − 600 K )
mol ⋅ K ⎠
= 24.9 kJ
Express and evaluate the heat
absorbed during the adiabatic
expansion BC:
QBC = 0
Use the first law of thermodynamics
to express and evaluate the heat
absorbed during the isothermal
compression CA:
QCA = WCA + ΔEint, CA = WCA
= − 17.3 kJ
because ΔEint = 0 for an isothermal
process.
81 ••
In a heat engine, 2.00 mol of a monatomic gas are carried through the
cycle ABCDA shown in Figure 19-21. (The PV diagram is not drawn to scale.)
The segment AB represents an isothermal expansion, the segment BC an
adiabatic expansion. The pressure and temperature at A are 5.00 atm and 600 K.
The volume at B is twice the volume at A. The pressure at D is 1.00 atm.
1938 Chapter 19
(a) What is the pressure at B? (b) What is the temperature at C? (c) Find the total
work done by the gas in one cycle.
Picture the Problem We can use the ideal-gas law to find the unknown
temperatures, pressures, and volumes at points B, C, and D and then find the work
done by the gas and the efficiency of the cycle by using the expressions for the
work done on or by the gas and the heat that enters the system for the various
thermodynamic processes of the cycle.
(a) Apply the ideal-gas law for a
fixed amount of gas to the isothermal
process AB:
PB = PA
VA
V
= (5.00 atm ) A
2VA
VB
= 2.50 atm ×
101.325 kPa
1atm
= 253.3 kPa = 253 kPa
(b) Apply the ideal-gas law for a
fixed amount of gas to the adiabatic
process BC:
TC = TB
Use the ideal-gas law to find the
volume at B:
VB =
PCVC
PBVB
(1)
nRTB
PB
(2.00 mol)⎛⎜ 8.314
J ⎞
⎟ (600 K )
mol ⋅ K ⎠
⎝
253.3 kPa
=
= 39.39 L
1γ
Use the equation of state for an
adiabatic process and γ = 5/3 to
find the volume occupied by the
gas at C:
⎛P ⎞
VC = VB ⎜⎜ B ⎟⎟
⎝ PC ⎠
= 68.26 L
Substitute numerical values in
equation (1) and evaluate TC:
TC = (600 K )
⎛ 2.50 atm ⎞
⎟⎟
= (39.39 L )⎜⎜
⎝ 1.00 atm ⎠
(1.00 atm )(68.26 L )
(2.50 atm )(39.39 L )
= 415.9 K = 416 K
(c) The work done by the gas in one
cycle is given by:
W = WAB + WBC + WCD + WDA
(2)
35
The Second Law of Thermodynamics 1939
The work done during the isothermal expansion AB is:
⎛V
WAB = nRTA ln⎜⎜ B
⎝ VA
⎞
⎛ 2V
J ⎞
⎛
⎟⎟ = (2.00 mol)⎜ 8.314
⎟ (600 K ) ln⎜⎜ A
mol ⋅ K ⎠
⎝
⎠
⎝ VA
⎞
⎟⎟ = 6.915 kJ
⎠
The work done during the adiabatic expansion BC is:
WBC = −C V ΔTBC = − 32 nRΔTBC
J ⎞
⎛
= − 32 (2.00 mol)⎜ 8.314
⎟ (415.9 K − 600 K )
mol ⋅ K ⎠
⎝
= 4.592 kJ
The work done during the isobaric compression CD is:
WCD = PC (VD − VC ) = (1.00 atm )(19.7 L − 68.26 L ) = −48.56 atm ⋅ L ×
101.325 J
atm ⋅ L
= −4.920 kJ
The work done during the constantvolume process DA is:
WDA = 0
Substitute numerical values in
equation (2) to obtain:
W = 6.915 kJ + 4.592 kJ − 4.920 kJ + 0
= 6.59 kJ
82 ••
Compare the efficiency of the Otto cycle to the efficiency of the
Carnot cycle operating between the same maximum and minimum temperatures.
(The Otto cycle is discussed in Section 19-1.)
Picture the Problem We can express the efficiency of the Otto cycle using the
result from Example 19-2. We can apply the relation TV γ −1 = constant to the
adiabatic processes of the Otto cycle to relate the end-point temperatures to the
volumes occupied by the gas at these points and eliminate the temperatures at c
and d. We can use the ideal-gas law to find the highest temperature of the gas
during its cycle and use this temperature to express the efficiency of a Carnot
engine. Finally, we can compare the efficiencies by examining their ratio.
1940 Chapter 19
The efficiency of the Otto engine is
given in Example 19-2:
ε Otto = 1 −
Td − Ta
Tc − Tb
(1)
where the subscripts refer to the various
points of the cycle as shown in Figure
19-3.
Apply the relation TV γ −1 = constant
to the adiabatic process a→b to
obtain:
⎛V ⎞
Tb = Ta ⎜⎜ a ⎟⎟
⎝ Vb ⎠
γ −1
Apply the relation TV γ −1 = constant
to the adiabatic process c→d to
obtain:
⎛V
Tc = Td ⎜⎜ d
⎝ Vc
γ −1
Subtract the first of these equations
from the second to obtain:
⎛V
Tc − Tb = Td ⎜⎜ d
⎝ Vc
In the Otto cycle, Va = Vd and
Vc = Vb. Substitute to obtain:
⎛V ⎞
Tc − Tb = Td ⎜⎜ a ⎟⎟
⎝ Vb ⎠
⎞
⎟⎟
⎠
γ −1
⎛V ⎞
− Ta ⎜⎜ a ⎟⎟
⎝ Vb ⎠
γ −1
⎛V ⎞
− Ta ⎜⎜ a ⎟⎟
⎝ Vb ⎠
⎞
⎟⎟
⎠
⎛V ⎞
= (Td − Ta )⎜⎜ a ⎟⎟
⎝ Vb ⎠
Substitute in equation (1) and
simplify to obtain:
ε Otto = 1 −
γ −1
γ −1
Td − Ta
⎛
(Td − Ta )⎜⎜ Va
⎝ Vb
γ −1
⎞
⎟⎟
⎠
γ −1
γ −1
⎛V ⎞
T
= 1 − ⎜⎜ b ⎟⎟ = 1 − a
Tb
⎝ Va ⎠
Note that, while Ta is the lowest
temperature of the cycle, Tb is not the
highest temperature.
Apply the ideal-gas law to c and
b to obtain an expression for the
cycle’s highest temperature Tc:
P
Pc Pb
⇒ Tc = Tb c > Tb
=
Pb
Tc Tb
The Second Law of Thermodynamics 1941
The efficiency of a Carnot engine
operating between the maximum and
minimum temperatures of the Otto
cycle is given by:
Express the ratio of the efficiency of
a Carnot engine to the efficiency of
an Otto engine operating between
the same temperatures:
ε Carnot = 1 −
ε Carnot
ε Otto
Ta
Tc
Ta
Tc
=
> 1 because Tc > Tb.
Ta
1−
Tb
1−
Hence, ε Carnot > ε Otto
83 ••• [SSM] A common practical cycle, often used in refrigeration, is the
Brayton cycle, which involves (1) an adiabatic compression, (2) an isobaric
(constant pressure) expansion,(3) an adiabatic expansion, and (4) an isobaric
compression back to the original state. Assume the system begins the adiabatic
compression at temperature T1, and transitions to temperatures T2, T3 and T4 after
each leg of the cycle. (a) Sketch this cycle on a PV diagram. (b) Show that the
(T − T )
efficiency of the overall cycle is given by ε = 1 − 4 1 . (c) Show that this
(T3 − T2 )
efficiency, can be written as ε = 1 − r (1− γ ) γ , where r is the pressure ratio Phigh/Plow
of the maximum and minimum pressures in the cycle.
Picture the Problem The efficiency of the cycle is the ratio of the work done to
the heat that flows into the engine. Because the adiabatic transitions in the cycle
do not have heat flow associated with them, all we must do is consider the heat
flow in and out of the engine during the isobaric transitions.
(a) The Brayton heat engine cycle is
shown to the right. The paths 1→2
and 3→4 are adiabatic. Heat Qh
enters the gas during the isobaric
transition from state 2 to state 3 and
heat Qc leaves the gas during the
isobaric transition from state 4 to
state 1.
Qh
P
2
⇓
3
1
⇓
4
Qc
(b) The efficiency of a heat engine is
given by:
ε=
Q − Qc
W
= h
Qin
Qin
V
(1)
1942 Chapter 19
During the constant-pressure
expansion from state 1 to state 2 heat
enters the system:
During the constant-pressure
compression from state 3 to state 4
heat enters the system:
Substituting in equation (1) and
simplifying yields:
Q23 = Qh = nC P ΔT = nC P (T3 − T2 )
Q41 = −Qc = − nC P ΔT = − nC P (T1 − T4 )
ε=
nC P (T3 − T2 ) − (− nC P (T1 − T4 ))
nC P (T3 − T2 )
(T3 − T2 ) + (T1 − T4 )
(T3 − T2 )
(T − T )
= 1− 4 1
(T3 − T2 )
=
(c) Given that, for an adiabatic
transition, TV γ −1 = constant , use the
ideal-gas law to eliminate V and
obtain:
Let the pressure for the transition
from state 1 to state 2 be Plow and the
pressure for the transition from state
3 to state 4 be Phigh. Then for the
adiabatic transition from state 1 to
state 2:
Tγ
= constant
P γ −1
γ
γ
T1
T
= γ2−1
γ −1
Plow
Phigh
⎛P
⇒ T1 = ⎜ low
⎜P
⎝ high
γ −1
γ
T2
γ −1
γ
Similarly, for the adiabatic transition
from state 3 to state 4:
⎛P
T4 = ⎜ low
⎜P
⎝ high
Subtract T1 from T4 and simplify to
obtain:
⎛P
T4 − T1 = ⎜ low
⎜P
⎝ high
⎞
⎟
⎟
⎠
⎛P
= ⎜ low
⎜P
⎝ high
⎞
⎟
⎟
⎠
⎞
⎟
⎟
⎠
⎞
⎟
⎟
⎠
T3
γ −1
γ
γ −1
γ
⎛P
T3 − ⎜ low
⎜P
⎝ high
(T3 − T2 )
⎞
⎟
⎟
⎠
γ −1
γ
T2
The Second Law of Thermodynamics 1943
Dividing both sides of the equation
by T3 − T2 yields:
Substitute in the result of Part (b) and
simplify to obtain:
T4 − T1 ⎛⎜ Plow
=
T3 − T2 ⎜⎝ Phigh
⎛P
ε = 1 − ⎜ low
⎜P
⎝ high
⎞
⎟
⎟
⎠
⎞
⎟
⎟
⎠
γ −1
γ
γ −1
γ
⎛ Phigh
= 1 − ⎜⎜
⎝ Plow
⎞
⎟⎟
⎠
1−γ
γ
1−γ
= 1 − (r ) γ
where r =
Phigh
Plow
84 ••• Suppose the Brayton cycle engine (see Problem 83) is run in reverse as
a refrigerator in your kitchen. In this case, the cycle begins at temperature T1 and
expands at constant pressure until its temperature T4. Then the gas is adiabatically
compressed, until its temperature is T3. And then it is compressed at constant
pressure until its temperature T2. Finally, it adiabatically expands until it returns
to its initial state at temperature T1. (a) Sketch this cycle on a PV diagram. (b)
(T4 − T1 ) .
Show that the coefficient of performance is COPB =
(T3 − T2 − T4 + T1 )
(c) Suppose your ″Brayton cycle refrigerator″ is run as follows. The cylinder
containing the refrigerant (a monatomic gas) has an initial volume and pressure of
60 mL and 1.0 atm. After the expansion at constant pressure, the volume and
temperature are 75 mL and –25°C. The pressure ratio r = Phigh/Plow for the cycle is
5.0. What is the coefficient of performance for your refrigerator? (d) To absorb
heat from the food compartment at the rate of 120 W, what is the rate at which
electrical energy must be supplied to the motor of this refrigerator? (e) Assuming
the refrigerator motor is actually running for only 4.0 h each day, how much does
it add to your monthly electric bill. Assume 15 cents per kWh of electric energy
and thirty days in a month.
Picture the Problem The efficiency of the Brayton refrigerator cycle is the ratio
of the heat that enters the system to the work done to operate the refrigerator.
Because the adiabatic transitions in the cycle do not have heat flow associated
with them, all we must do is consider the heat flow in and out of the refrigerator
during the isobaric transitions.
1944 Chapter 19
(a) The Brayton refrigerator cycle is
shown to the right. The paths 1→2
and 3→4 are adiabatic. Heat Qc
enters the gas during the constantpressure transition from state 1 to
state 4 and heat Qh leaves the gas
during the constant-pressure
transition from state 3 to state 2.
Qh
P
2
⇓
3
1
⇓
4
Qc
(b) The coefficient of performance of
the Brayton cycle refrigerator is
given by:
During the constant-pressure
compression from state 3 to state 2
heat leaves the system:
During the constant-pressure
expansion from state 1 to state 4 heat
enters the system:
Substituting in equation (1) and
simplifying yields:
COPB =
Qh
Qh
=
W Qh − Qc
V
(1)
Q32 = −Qh = −nC P ΔT = − nC P (T2 − T3 )
Q14 = Qc = nC P ΔT = nC P (T4 − T1 )
COPB =
=
=
nC P (T4 − T1 )
− nC P (T4 − T1 ) − nC P (T2 − T3 )
(T4 − T1 )
− (T4 − T1 ) − (T2 − T3 )
T4 − T1
T3 − T2 − T4 + T1
(c) The COPB requires the
temperatures corresponding to states
1, 2, 3, and 4. We’re given that the
temperature in state 4 is:
T4 = −25°C + 273 K = 248 K
For the constant-pressure transition
from state 1 to state 4, the quotient
T/V is constant:
⎛V ⎞
T1 T4
= ⇒ T1 = ⎜⎜ 1 ⎟⎟T4
V1 V4
⎝ V4 ⎠
Substitute numerical values and
evaluate T1:
⎛ 60 mL ⎞
T1 = ⎜
⎟ (248 K ) = 198 K
⎝ 75 mL ⎠
The Second Law of Thermodynamics 1945
Given that, for an adiabatic
transition, TV γ −1 = constant , use the
ideal-gas law to eliminate V and
obtain:
For the adiabatic transition from state
4 to state 3:
Substitute numerical values and
evaluate T3:
Similarly, for the adiabatic transition
from state 2 to state 1:
Tγ
= constant
P γ −1
⎛P ⎞
T3γ
T4γ
=
⇒ T3 = ⎜⎜ 3 ⎟⎟
γ −1
γ −1
P3
P4
⎝ P4 ⎠
T4
1.67 −1
T3 = (5) 1.67 (248 K ) = 473 K
γ −1
1.67 −1
⎛P ⎞ γ
T2 = ⎜⎜ 2 ⎟⎟ T1 = (5) 1.67 (198 K )
⎝ P1 ⎠
= 378 K
248 K − 198 K
473 K − 378 K − 248 K + 198 K
Substitute numerical values in the
expression derived in Part (a) and
evaluate COPB:
COPB =
(d) From the definition of COPB:
W=
The rate at which energy must be
supplied to this refrigerator is given
by:
1 dQc
dW
=
dt
COPB dt
Express the heat Qc that is drawn
from the cold reservoir:
γ −1
γ
= 1.1
Qc
COPB
or, if the frequency of the AC power
input is f,
fQc
dW
=
dt
COPB
Qc = nC P ΔT = nC P (T4 − T1 )
Substituting for Qc yields:
fnC P (T4 − T1 )
dW
=
dt
COPB
Use the ideal-gas law to express the
number of moles of the gas:
n=
P4V4
RT4
1946 Chapter 19
Because the gas is monatomic,
C P = 52 R . Substitute for n and CP to
obtain:
dW
=
dt
=
5
2
5
2
f
P4V4
R(T4 − T1 )
RT4
COPB
fP4V4 (T4 − T1 )
(COPB )T4
Substitute numerical values and evaluate dW/dt:
⎛
10 −3 m 3 ⎞
⎜
(60 s )(101.325 kPa )⎜ 75 mL × L ⎟⎟(248 K − 198 K )
dW
⎠
⎝
= 207 W
=
(1.11)(248 K )
dt
5
2
−1
= 0.21 kW
(e) The monthly cost of operation is given by
Monthly Cost = Cost Per Unit of Power × Power Consumption
= rate × daily consumption × number of days per month
Substitute numerical values and evaluate the monthly cost of operation:
Monthly Cost =
85
•••
$0.15
4.0 h
× 207 W ×
× 30 d ≈ $4
kWh
d
Using ΔS = Cv ln (T2 T1 ) − nR ln (V2 V1 ) (Equation 19-16) for the
entropy change of an ideal gas, show explicitly that the entropy change is zero for
a quasi-static adiabatic expansion from state (V1, T1) to state (V2, T2).
Picture the Problem We can use nR = CP − CV , γ = CP CV , and
TV γ −1 = a constant to show that the entropy change for a quasi-static adiabatic
expansion that proceeds from state (V1,T1) to state (V2,T2) is zero.
Express the entropy change for a
general process that proceeds from
state 1 to state 2:
⎛T ⎞
⎛V ⎞
ΔS = CV ln⎜⎜ 2 ⎟⎟ + nR ln⎜⎜ 2 ⎟⎟
⎝ T1 ⎠
⎝ V1 ⎠
For an adiabatic process:
T2 ⎛ V1 ⎞
=⎜ ⎟
T1 ⎜⎝ V2 ⎟⎠
γ −1
The Second Law of Thermodynamics 1947
Substitute for
T2
and simplify to obtain:
T1
γ −1
⎡
⎛ V1 ⎞ ⎤
⎢
CV ln⎜⎜ ⎟⎟ ⎥
γ −1
⎛ V1 ⎞
⎛ V2 ⎞
⎛ V2 ⎞ ⎢
⎝ V2 ⎠ ⎥
ΔS = CV ln⎜⎜ ⎟⎟ + nR ln⎜⎜ ⎟⎟ = ln⎜⎜ ⎟⎟ ⎢nR +
⎥
V
⎝ V2 ⎠
⎝ V1 ⎠
⎝ V1 ⎠ ⎢
ln 2
⎥
V1
⎢⎣
⎥⎦
⎡
⎛ ⎞⎤
(γ − 1)CV ln⎜⎜ V1 ⎟⎟ ⎥
⎢
⎛V ⎞
⎝ V2 ⎠ ⎥ = ln⎛⎜ V2 ⎞⎟ [nR − (γ − 1)C ]
= ln⎜⎜ 2 ⎟⎟ ⎢nR +
V
⎜V ⎟
V1
⎥
⎝ 1⎠
⎝ V1 ⎠ ⎢
ln
−
⎥
⎢
V2
⎥⎦
⎢⎣
Use the relationship between CP and
CV to obtain:
nR = CP − CV
Substituting for nR and γ and
simplifying yields:
⎞ ⎤
⎛C
⎛ V ⎞⎡
ΔS = ln⎜⎜ 2 ⎟⎟ ⎢CP − CV − ⎜⎜ p − 1⎟⎟CV ⎥
⎝ V1 ⎠ ⎣
⎠ ⎦
⎝ CV
= 0
86 ••• (a) Show that if the refrigerator statement of the second law of
thermodynamics were not true, then the entropy of the universe could decrease.
(b) Show that if the heat-engine statement of the second law were not true, then
the entropy of the universe could decrease. (c) A third statement of the second law
is that the entropy of the universe cannot decrease. Have you just proved that this
statement is equivalent to the refrigerator and heat-engine statements?
Picture the Problem
(a) Suppose the refrigerator statement of the second law is violated in the sense
that heat Qc is taken from the cold reservoir and an equal mount of heat is
transferred to the hot reservoir and W = 0. The entropy change of the universe is
then ΔSu = Qc/Th − Qc/Tc. Because Th > Tc, Su < 0, i.e., the entropy of the universe
would decrease.
(b) In this case, is heat Qh is taken from the hot reservoir and no heat is rejected to
the cold reservoir; that is, Qc = 0, then the entropy change of the universe is
ΔSu = −Qh/Th + 0, which is negative. Again, the entropy of the universe would
decrease.
1948 Chapter 19
(c) The heat-engine and refrigerator statements of the second law only state that
some heat must be rejected to a cold reservoir and some work must be done to
transfer heat from the cold to the hot reservoir, but these statements do not specify
the minimum amount of heat rejected or work that must be done. The statement
ΔSu ≥ 0 is more restrictive. The heat-engine and refrigerator statements in
conjunction with the Carnot efficiency are equivalent to ΔSu ≥ 0.
87 ••• Suppose that two heat engines are connected in series, such that the
heat released by the first engine is used as the heat absorbed by the second engine
as shown in Figure 19-22. The efficiencies of the engines are ε1 and ε2,
respectively. Show that the net efficiency of the combination is given by
ε net = ε 1 + ε 2 − ε 1ε 2 .
Picture the Problem We can express the net efficiency of the two engines in
terms of W1, W2, and Qh and then use ε1 = W1/Qh and ε2 = W2/Qm to eliminate W1,
W2, Qh, and Qm.
W1 + W2
Qh
Express the net efficiency of the
two heat engines connected in
series:
ε net =
Express the efficiencies of engines 1
and 2:
ε1 =
Solve for W1 and W2 and substitute
to obtain:
ε net =
Express the efficiency of engine 1 in
terms of Qm and Qh:
ε1 = 1 −
Substitute for Qm/Qh and simplify to
obtain:
ε net = ε 1 + (1 − ε 1 )ε 2 = ε 1 + ε 2 − ε 1ε 2
W1
W
and ε 2 = 2
Qh
Qm
ε1Qh + ε 2Qm
Qh
= ε1 +
Qm
ε2
Qh
Qm
Q
⇒ m = 1 − ε1
Qh
Qh
88 ••• Suppose that two heat engines are connected in series, such that the
heat released by the first engine is used as the heat absorbed by the second
engine, as shown in Figure 19-22. Suppose that each engine is an ideal reversible
heat engine. Engine 1 operates between temperatures Th and Tm and Engine 2
operates between Tm and Tc, where Th > Tm > Tc. Show that the net efficiency of
T
the combination is given by ε net = 1 − c . (Note that this result means that two
Th
The Second Law of Thermodynamics 1949
reversible heat engines operating ″in series″ are equivalent to one reversible heat
engine operating between the hottest and coldest reservoirs.)
Picture the Problem We can express the net efficiency of the two engines in
terms of W1, W2, and Qh and then use ε1 = W1/Qh and ε2 = W2/Qm to eliminate W1,
W2, Qh, and Qm. Finally, we can substitute the expressions for the efficiencies of
the ideal reversible engines to obtain ε net = 1 − Tc Th .
Express the efficiencies of ideal
reversible engines 1 and 2:
Tm
Th
(1)
Tc
Tm
(2)
W1 + W2
Qh
(3)
ε1 = 1 −
and
ε2 = 1−
The net efficiency of the two engines
connected in series is given by:
ε net =
Express the efficiencies of engines 1
and 2:
ε1 =
Solve for W1 and W2 and substitute
in equation (3) to obtain:
ε net =
Express the efficiency of engine 1 in
terms of Qm and Qh:
ε1 = 1 −
Substitute for
Qm
to obtain:
Qh
Substitute for ε1 and ε2 and simplify
to obtain:
W1
W
and ε 2 = 2
Qh
Qm
ε 1Qh + ε 2Qm
Qh
= ε1 +
Qm
ε2
Qh
Q
Qm
⇒ m = 1 − ε1
Qh
Qh
ε net = ε1 + (1 − ε1 )ε 2
Tm ⎛ Tm ⎞ ⎛ Tc ⎞
+ ⎜ ⎟ ⎜1 − ⎟
Th ⎜⎝ Th ⎟⎠ ⎜⎝ Tm ⎟⎠
T
T
T
T
= 1− m + m − c = 1− c
Th Th Th
Th
ε net = 1 −
89 ••• [SSM] The English mathematician and philosopher Bertrand Russell
(1872-1970) once said that if a million monkeys were given a million typewriters
and typed away at random for a million years, they would produce all of
Shakespeare’s works. Let us limit ourselves to the following fragment of
Shakespeare (Julius Caesar III:ii):
1950 Chapter 19
Friends, Romans, countrymen! Lend me your ears.
I come to bury Caesar, not to praise him.
The evil that men do lives on after them,
The good is oft interred with the bones.
So let it be with Caesar.
The noble Brutus hath told you that Caesar was ambitious,
And, if so, it were a grievous fault,
And grievously hath Caesar answered it . . .
Even with this small fragment, it will take a lot longer than a million years! By
what factor (roughly speaking) was Russell in error? Make any reasonable
assumptions you want. (You can even assume that the monkeys are immortal.)
Picture the Problem There are 26 letters and four punctuation marks (space,
comma, period, and exclamation point) used in the English language, disregarding
capitalization, so we have a grand total of 30 characters to choose from. This
fragment is 330 characters (including spaces) long; there are then 30330 different
possible arrangements of the character set to form a fragment this long. We can
use this number of possible arrangements to express the probability that one
monkey will write out this passage and then an estimate of a monkey’s typing
speed to approximate the time required for one million monkeys to type the
passage from Shakespeare.
Assuming the monkeys type at
random, express the probability P
that one monkey will write out this
passage:
Use the approximation
30 ≈ 1000 = 101.5 to obtain:
Assuming the monkeys can type at a
rate of 1 character per second, it
would take about 330 s to write a
passage of length equal to the
quotation from Shakespeare. Find
the time T required for a million
monkeys to type this particular
passage by accident:
Express the ratio of T to Russell’s
estimate:
P=
P=
T=
1
30330
1
10
(1.5 )(330 )
=
1
= 10 −495
495
10
(330 s )(10495 )
106
⎛
⎞
1y
⎟
= 3.30 × 10 491 s ⎜⎜
7 ⎟
⎝ 3.16 × 10 s ⎠
(
)
≈ 10484 y
T
=
10484 y
= 10478
6
10 y
TRussell
or
T ≈ 10 478 TRussell