Thermodynamics T8
Second Law of Thermodynamics
Learning Goal:
To understand the implications of the second law of thermodynamics.
The second law of thermodynamics explains the direction in which the thermodynamic processes tend to go. That is,
it limits the types of final states of the system that naturally evolve from a given initial state. The second law has
many practical applications. For example it explains the limits of efficiency for heat engines and refrigerators. To
develop a better understanding of this law, try these conceptual questions.
Part A
The thermodynamic processes that occur in nature ____________.
ANSWER:
convert thermal energy into mechanical energy
lead to a more ordered state
cannot be reversed
do not conserve energy
Correct
Only infinitely slow, or quasi-equilibrium, processes can be reversible; such processes exist only in the
imaginations of scientists.
One of the ways to state the second law of thermodynamics is as follows:
Any process occurring in a closed system either increases the entropy (disorder) of the system or leaves it
constant. For irreveresible processes, the entropy increases; for the reversible ones, the entropy remains
constant.
Part B
According to the second law of thermodynamics, it is impossible for ____________.
ANSWER:
heat energy to flow from a colder body to a hotter body
an ideal heat engine to have the efficiency of 99%
an ideal heat engine to have non-zero power.
a physical process to yield more energy than what is put in
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Correct
The ideal engine follows a reversible cycle--therefore, an infinitely slow one. If the work is being done at the
infinitely slow rate, the power of such an engine is zero.
An alternative way to state the second law of thermodynamics is as follows:
It is impossible to construct a cyclical heat engine whose sole effect is the continuous transfer of heat
energy from a colder object to a hotter one.
This statement is known as Clausius statement of the second law. Note the word "sole." Of course, it is
possible to construct a machine in which a heat flow from a colder to a hotter object is accompanied by
another process, such as work input.
Part C
If the coefficient of performance of a refrigerator is 1, which the following statements is true?
Hint 1. Definition of the coefficient of performance of a refrigerator
The coefficient of performance of a refrigerator is defined as the heat extracted from the colder region per
unit of work done.
ANSWER:
The temperature outside equals the temperature inside of the refrigerator.
The rate at which heat is removed from the inside equals the rate at which heat is delivered outside.
The power consumed by the refrigerator equals the rate at which heat is removed from the inside.
The power consumed by the refrigerator equals the rate at which heat is delivered to the outside.
Correct
Part D
To increase the efficiency of an ideal heat engine, one must increase which of the following?
Hint 1. Formula for the efficiency of an ideal engine
The efficiency e of an ideal (Carnot) engine that absorbs heat Qh from a reservoir at a temperature Th ,
and releases heat Qc to another reservoir at a temperature Tc , while doing work Wout = |Qh | − |Qc |,
is
e=
ANSWER:
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Wout
Qh
=1−
Tc
.
Th
the amount of heat consumed per second
the temperature of the cold reservoir
the temperature of the hot reservoir
the size of the cold reservoir
the size of the hot reservoir
Correct
Part E
How would you increase the coefficient of performance of an ideal refrigerator?
Hint 1. Graphical approach to the problem
The figure shows a pV diagram for a
refrigerator undergoing an ideal (Carnot) cycle.
To increase the efficiency, you would have to
decrease the work input required (i.e., the area
of the shaded region) for a given Qc . How
would you achieve this?
ANSWER:
Increase the mechanical work input.
Decrease the outside temperature.
Decrease the inside temperature.
Increase the outside temperature.
Correct
The efficiency e of an ideal/Carnot refrigerator that absorbs heat Qc from a reservoir at a temperature Tc ,
and releases heat Qh to another reservoir at a temperature Th , with a work input |Win | = |Qc | − |Qh |, is
Qc
Tc
=
.
T
|Win |
h −Tc
From this equation, you can see that decreasing Th and/or increasing Tc lead to an increase in the
efficiency e.
e=
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Part F
Why must every heat engine have a cold reservoir?
ANSWER:
Because it is impossible for even a perfect engine to convert heat entirely into mechanical work.
Because the cold reservoir keeps the engine from overheating.
Because the cold reservoir keeps the engine from overcooling.
Because the cold reservoir increases the power of the engine.
Correct
Another way to state the second law of thermodynamics is as follows:
It is impossible to construct a cyclical heat engine whose sole effect is absorption of energy from the hot
reservoir and the performance of the equal amount of work.
This statement is known as the Kelvin-Planck statement of the second law. Note the word "sole."
You have now seen three different statements of the second law. Understanding the equivalence of these
three statements is important. However, it is not a trivial matter: Your textbook and discussions should help
you to get a better grasp of this equivalency.
Carnot Cycle
After Count Rumford (Benjamin Thompson) and James Prescott Joule had shown the equivalence of mechanical
energy and heat, it was natural that engineers believed it possible to make a "heat engine" (e.g., a steam engine)
that would convert heat completely into mechanical energy. Sadi Carnot considered a hypothetical piston engine that
contained n moles of an ideal gas, showing first that it was reversible, and most importantly that—regardless of the
specific heat of the gas—it had limited efficiency, defined as e = W /Qh , where W is the net work done by the
engine and Qh is the quantity of heat put into the engine at a (high) temperature Th . Furthermore, he showed that
the engine must necessarily put an amount of heat |Qc | back into a heat reservoir at a lower temperature Tc . The
cycle associated with a Carnot engine is known as a Carnot cycle. A pV plot of the Carnot cycle is shown in the
figure . The working gas first expands isothermally from
state A to state B, absorbing heat Qh from a reservoir at
temperature Th . The gas then expands adiabatically until
it reaches a temperature Tc , in state C. The gas is
compressed isothermally to state D, giving off heat |Qc |.
Finally, the gas is adiabatically compressed to state A, its
original state.
Part A
Which of the following statements are true?
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Check all that apply.
Hint 1. Heat flow in an adiabatic process
An adiabatic process is one in which heat does not flow into or out of the gas.
ANSWER:
For the gas to do positive work, the cycle must be traversed in a clockwise manner.
Positive heat is added to the gas as it proceeds from state C to state D.
The net work done by the gas is proportional to the area inside the closed curve.
The heat transferred as the gas proceeds from state B to state C is greater than the heat transferred
as the gas proceeds from state D to state A.
Correct
Part B
Find the total work W done by the gas after it completes a single Carnot cycle.
Express the work in terms of any or all of the quantities Qh , Th , |Qc |, and Tc .
Hint 1. How to approach the problem
Find the total amount of heat added during the entire cycle and the change in internal energy of the gas
over the entire cycle. Then apply the first law of thermodynamics: dQ = dU + dW .
Hint 2. Compute the change in internal energy
What is the net change in the gas's internal energy U after one complete cycle?
ANSWER:
change in U = 0
Correct
ANSWER:
W = Qh − Qc
Correct
Part C
Suppose there are n moles of the ideal gas, and the volumes of the gas in states A and B are, respectively, VA
and VB . Find Qh , the heat absorbed by the gas as it expands from state A to state B.
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Express the heat absorbed by the gas in terms of n, VA , VB , the temperature of the hot reservoir, Th ,
and the gas constant R.
Hint 1. General method of finding Qh
First, find the work WAB done by the gas as it expands from state A to state B. Then, use the first law of
thermodynamics to relate WAB to both Qh and the change in the gas's internal energy. Finally, recall
that this process is isothermal; what does this tell you about the change in the gas's internal energy?
Hint 2. Find the work done by the gas
What is WAB , the work done by the gas as it expands from state A to state B?
Express the work in terms of n, VA , VB , the temperature of the hot reservoir Th , and the gas
constant R.
Hint 1. How to find the work done by the gas
To find WAB , the net work done by the gas as it proceeds from state A to state B, you need to
integrate dW = p dV from state A to state B.
Hint 2. Express p in terms of V
Use the ideal gas equation of state to find an expression for the pressure p in terms of n, R, Th ,
and V .
ANSWER:
p(V ) =
nRTh
V
Hint 3. Express the integral over dV
Given that p(V ) ∝ 1/V and that
V
VB
what is the integral ∫ V
A
dV /V
WAB = ∫ V B p(V ) dV
A
,
?
Express your answer in terms of VA and VB .
ANSWER:
V
∫V B
A
dV
=
V
ln( VB )
V
A
ANSWER:
WAB = nRT h ln( B )
VA
V
Hint 3. Relation between Qh and WAB
Because the process is isothermal, the internal energy of the gas does not change. Therefore, the work
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done by the gas will equal the net heat flow into the gas: Qh = WAB .
ANSWER:
Qh = nRT h ln(
VB
)
VA
Correct
Note that VB > VA . The quantity Qh is therefore positive, indicating that the thermal energy is added to the
gas.
Part D
The volume of the gas in state C is VC , and its volume in state D is VD . Find |Qc |, the magnitude of the heat that
flows out of the gas as it proceeds from state C to state D.
Express your answer in terms of n, VC , VD , Tc (the temperature of the cold reservoir), and R.
Hint 1. How to approach the problem
First, find the work WCD done by the gas as it expands from state C to state D. Then, use the first law of
thermodynamics to relate WCD to both Qc and the change in the gas's internal energy. Finally, recall that
this process is isothermal; what does this tell you about the change in the gas's internal energy?
ANSWER:
|Qc | =
nRT c ln( VC )
V
D
Correct
Observe that the three parts together imply that WBC + WDA = 0. This is because BC and DA are
adiabatic processes. So using the first law, WBC = −ΔUBC = −nCV (Th − Tc ), whereas
WDA = −ΔUDA = −nCV (Tc − Th ). So WBC = −WDA , or WBC + WDA = 0. This is a general
result: Any two adiabatic processes operating between the same two temperatures result in the same
amount of work, regardless of the pressure and volume differences. Note also that the sign of Qc is actually
negative, because thermal energy is leaving the gas; however, the energy entering the reservoir is the
quantity of interest.
Part E
Now, by considering the adiabatic processes (from B to C and from D to A), find the ratio VC /VD in terms of VA
and VB .
Hint 1. How to approach the problem
Suppose the ratio of the gas's specific heats is denoted by γ = Cp /Cv . Along adiabatic curves, you
know that pV γ = C, where C is some constant. Rewrite this equation in terms of T and V instead of p
and V . Then use this new equation to relate the temperature and volume at the end points of the two
adiabatic legs of the Carnot cycle. This will give you two equations that you can solve for VC /VD .
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Hint 2. Rewrite pV γ in terms of T and V
Use the ideal gas equation of state to eliminate p from the expression pV γ .
Express your answer in terms of γ, n, R, and the temperature T .
ANSWER:
pV γ = nRT V (γ−1)
Hint 3. Express Th and VB in terms of Tc and VC
States B and C are connected by an adiabatic expansion. Use the result found in the previous hint to find
(γ−1)
an expression for Th VB
.
Express your answer in terms of Tc , VC , and γ.
ANSWER:
(γ−1)
T h VB
(γ−1)
)
= T c VC (
Hint 4. Express Th and VA in terms of Tc and VD
States D and A are connected by an adiabatic expansion. Use the result found in Hint 2 to find an
(γ−1)
expression for Th VA
.
Express your answer in terms of Tc , VD , and γ.
ANSWER:
(γ−1)
T h VA
(γ−1)
)
= T c VD (
Hint 5. Solving for VC /VD in terms of VA and VB
Combine the results of the previous two hints, eliminating the temperatures Th and Tc . (One way to do
this is to divide one equation by the other.) This should allow you to solve for VC /VD in terms of VA and
VB .
ANSWER:
V C /V D =
VB
VA
Correct
Part F
Using your expressions for Qh and |Qc | (found in Parts C and D), and your result from Part E, find a simplified
expression for |Qc |/Qh .
No volume variables should appear in your expression, nor should any constants (e.g., n or R).
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ANSWER:
T
|Qc |/Qh = T c
h
Correct
Part G
The efficiency of any engine is, by definition, e = W /Qh . Carnot proved that no engine can have an efficiency
greater than that of a Carnot engine. Find the efficiency eCarnot of a Carnot engine.
Express the efficiency in terms of Th and Tc .
Hint 1. Express the efficiency in terms of Qh and |Qc |
Using your result from Part B, find the efficiency e (of any engine) in terms of the engine's heat input and
output.
Express your answer in terms of Qh and |Qc |.
ANSWER:
e= 1−
∣Qc ∣
Qh
ANSWER:
eCarnot = 1 − T c
h
T
Correct
Because Tc is generally fixed (e.g., the cold reservoir for power plants is often a river or a lake), engineers,
trying to increase efficiency, have always sought to raise the upper temperature Th . This explains why
(historically) there were some spectacular explosions of boilers used for steam power.
Exercise 20.16
An ice-making machine operates in a Carnot cycle. It takes heat from water at 0.0 ∘ C and rejects heat to a room of
temperature 25.0 ∘ C. Suppose that a mass 79.5 kg of water at 0.0 ∘ C is converted to ice at 0.0 ∘ C.
Part A
How much heat is rejected to the room?
ANSWER:
|QH | = 2.90×107 J
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Correct
Part B
How much energy must be supplied to the device?
ANSWER:
|W | = 2.43×106 J
Correct
Exercise 20.19
A Carnot heat engine has a thermal efficiency of 0.570 and the temperature of its hot reservoir is 830 K .
Part A
If 2700 J of heat is rejected to the cold reservoir in one cycle, what is the work output of the engine during one
cycle?
ANSWER:
W = 3580 J
Correct
Exercise 20.20
A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir
consisting of a large tub of ice and water. In five minutes of operation of the engine, the heat rejected by the engine
melts 0.0400 kg of ice.
Part A
During this time, how much work W is performed by the engine?
ANSWER:
W = 4890 J
Correct
Does Entropy Really Always Increase?
An aluminum bar of mass 2.00 kg at 300 ∘ C is thrown into a lake. The temperature of the water in the lake is 15.0
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∘
C; the specific heat capacity of aluminum is 900 J/(kg ⋅ K) .
Part A
The bar eventually reaches thermal equilibrium with the lake. What is the entropy change ΔSlake of the lake?
Assume that the lake is so large that its temperature remains virtually constant.
Express your answer numerically in joules per kelvin
Hint 1. How to approach the problem
You can calculate the entropy change of the lake using the formula for the entropy change in an
isothermal process. Note that the amount of heat transferred to the water in the lake is equal to the
amount of heat lost by the bar.
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Hint 2. Find the heat absorbed by the lake
Find ΔQlake , the amount of heat absorbed by the lake.
Express your answer numerically in joules.
Hint 1. Find the temperature of thermal equilibrium
What is the temperature of thermal equilibrium Te of the aluminum bar and the lake?
Express your answer numerically in kelvins.
ANSWER:
Te = 288 K
Hint 2. Find the energy change for the aluminum bar
What is the change of heat energy ΔQbar for the metal bar?
Express your answer numerically in joules.
Hint 1. Heat and the temperature change
The heat Q absorbed or given up by an object is given by
Q = mc(Tfinal − Tinitial ),
where m is the mass of the object, Tinitial is the initial temperature, Tfinal is the final
temperature, and c is the specific heat of the substance.
ANSWER:
ΔQbar = −5.13×105 J
ANSWER:
ΔQlake = 5.13×105 J
Correct
Hint 3. Entropy change in an isothermal process
When there is a heat transfer of ΔQ to a substance at constant temperature T , the entropy change ΔS
of the substance is given by
ΔS = ΔQ
,
T
where T is absolute temperature.
ANSWER:
ΔSlake = 1780 J/K
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Correct
Part B
Has the entropy of the aluminum bar decreased or increased?
Hint 1. How to approach the question
Since the temperature of the aluminum bar changes during the cooling process, the exact entropy
change of the bar cannot be easily calculated. However, you can use the interpretation of entropy as a
measure of randomness to determine whether its change is positive or negative.
ANSWER:
Since the entropy change of a system is always positive, we can deduce that the entropy of the
aluminum bar has increased.
Since the final lower temperature of the bar means lower average speed of molecular motion, we can
deduce that the entropy of the bar has decreased.
We don't have enough information to determine whether the entropy of the aluminum bar has
decreased or increased.
Correct
Part C
Since the aluminum bar is not an isolated system, the second law of thermodynamics cannot be applied to the
bar alone. Rather, it should be applied to the bar in combination with its surroundings (the lake).
Assume that the entropy change of the bar is -73.5 J/K , what is the change in total entropy ΔStotal ?
Express your answer numerically in joules per Kelvin
Hint 1. Total change of entropy
The change in total entropy is defined as the sum of the entropy changes of the system and the
surroundings.
ANSWER:
ΔStotal = 1710 J/K
Correct
Even though the aluminum bar lowers its entropy, the total entropy change of the bar and its surroundings
(the water in the lake) is positive, and the total entropy increases.
Part D
The second law of thermodynamics states that spontaneous processes tend to be accompanied by entropy
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increase. Consider, however, the following spontaneous processes:
the growth of plants from simple seeds to well-organized systems
the growth of a fertilized egg from a single cell to a complex adult organism
the formation of snowflakes from molecules of liquid water with random motion to a highly ordered
crystal
the growth of organized knowledge over time
In all these cases, systems evolve to a state of less disorder and lower entropy, apparently violating the second
law of thermodynamics. Could we, then, consider them as processes occurring in systems that are not isolated?
ANSWER:
True
False
Correct
All the processes listed above require energy input to occur just as a refrigerator requires electrical energy
to run. Systems can become more ordered and lower their entropy as time passes. However, this can
happen only as the entropy of the environment increases, just as we found out in the case of the hot
aluminum bar cooling down in the lake.
Exercise 20.22
A 4.50 kg block of ice at 0 ∘ C falls into the ocean and melts. The average temperature of the ocean is 3.50 ∘ C,
including all the deep water.
Part A
By how much does the melting of this ice change the entropy of the world?
ANSWER:
ΔS = 70.0 J/K
Part B
Does it make it larger or smaller?
Hint 1.
Do you think that the ocean will change temperature appreciably as the ice melts?
ANSWER:
Larger.
Smaller.
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Correct
Exercise 20.29
Part A
Calculate the change in entropy when 1.00 kg of water at 100 ∘ C is vaporized and converted to steam at 100
∘
C. Assume that the heat of vaporization of water is 2256 × 103 J/kg.
ANSWER:
ΔS = 6050 J/K
Correct
Part B
Calculate the change in entropy when 1.00 kg of ice is melted at 0 ∘ C. Assume that the heat of fusion of water
is Lf = 3.34 × 10 5 J/kg.
ANSWER:
ΔS = 1220 J/K
Correct
Part C
Is the change entropy greater for melting or for vaporization?
ANSWER:
the change entropy greater for melting
the change entropy greater for vaporization
Correct
Exercise 20.33
Two moles of an ideal gas occupy a volume V. The gas expands isothermally and reversibly to a volume 3V.
Part A
Is the velocity distribution changed by the isothermal expansion?
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ANSWER:
yes
no
Correct
Part B
Explain.
ANSWER:
3592 Character(s) remaining
The KE=1/2*m*v^2=3/2*k*T as the expansion is isothermic
T will not change therefore KE will not change. If KE
remains the same as does the mass v will have to remain
constant as well for the KE to not change.
Part C
Use the equation ΔS
= k ln
w2
w1 to calculate the change in entropy of the gas.
ANSWER:
ΔS = 18.3 J/K
Correct
Part D
Use the equation ΔS
=
Q
to calculate the change in entropy of the gas.
T
ANSWER:
ΔS = 18.3 J/K
Correct
Part E
Compare the result in part (d) to that obtained in part (c).
ANSWER:
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the same
greater
less
Correct
Entropy Change in a Free Expansion: A Microscopic View
A thin partition divides a thermally insulated vessel into a lower compartment of volume V and an upper
compartment of volume 2V . The lower compartment contains n moles of an ideal gas; the upper part is evacuated.
Part A
When the partition is removed, the gas expands and fills both compartments. How many moles n of gas were
initially contained in the lower compartment if the entropy change of the gas in this free-expansion process is
17.28 J/K?
Express your answer to three significant figures.
Hint 1. How to approach the problem
When the partition is removed, the gas expands freely and no work is done. Thus the velocities of the
gas molecules are unaffected. However, now each molecule has a bigger volume in which it can move
and the number of possible positions of each molecule has increased. By considering the change in
volume of the gas after the expansion, you can determine the number of possible microscopic states.
This information allows you to determine the change in entropy of the gas as a function of the number of
moles in the gas, and finally to find the number of moles in the gas.
Hint 2. Find the number of possible microscopic states of a gas after a free expansion
Consider a thermally insulated box such as the one described in the introduction. Assume that the
smaller compartment contains X molecules of gas, whereas the other is evacuated. When the partition is
removed the gas expands and occupies the entire box. The number of microscopic states of the system
when the gas initially occupies a volume V is wi . Find the number of microscopic states wf when the
gas occupies the entire box (volume 3V ).
Express your answer in terms of wi and X.
ANSWER:
wf = 3X wi
Hint 3. Find the microscopic expression for the change in entropy of a gas
Consider a gas that undergoes a thermodynamic process such that the gas is taken from an initial
macroscopic state, where the possible microscopic states are wi , to a final macroscopic state
characterized by wf possible microscopic states. What is the change in entropy δS of the gas after this
process?
Express your answer in terms of some or all of the variables wi , wf , and the Boltzmann constant
kB .
Hint 1. Microscopic expression for entropy
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The entropy S of a macroscopic state with w possible microscopic states is
S = kB ln w,
where kB is the Boltzmann constant.
ANSWER:
δS = kB ln( wi )
wf
Hint 4. Relating kB to the gas constant
Recall the relation R = kB NA , where R is the gas constant, kB is the Boltzmann constant, and NA is
Avogadro's number.
ANSWER:
n = 1.89 mol
Correct
± From Hot to Cool: The Second Law of Thermodynamics
Learning Goal:
To understand the meaning and applications of the second law of thermodynamics, to understand the meaning of
entropy, and perform some basic calculations involving entropy changes.
The first law of thermodynamics (which states that energy is conserved) does not specify the direction in which
thermodynamic processes in nature can spontaneously occur. For example, imagine an object initially at rest
suddenly taking off along a rough horizontal surface and speeding up (gaining kinetic energy) while cooling down
(losing thermal energy). Although such a process would not violate conservation of energy, it is, of course,
impossible and could never take place spontaneously.
The second law of thermodynamics dictates which processes in nature may occur spontaneously and which ones
may not. The second law can be stated in many ways, one of which uses the concept of entropy.
Entropy
Entropy can be thought of as a measure of a system's disorder: A lower degree of disorder implies lower entropy,
and vice versa. For example, a highly ordered ice crystal has a relatively low entropy, whereas the same amount of
water in a much less ordered state, such as water vapor, has a much higher entropy. Entropy is usually denoted by
S, and has units of energy divided by temperature (J/K). For an isothermal process (the temperature of the system
remains constant as it exchanges heat with its surroundings), the change in a system's entropy is given by
Q
,
T
where Q is the amount of heat involved in the process and T is the absolute temperature of the system. The heat Q
is positive if thermal energy is absorbed by the system from its surroundings, and is negative if thermal energy is
transferred from the system to its surroundings.
ΔS =
Using the idea of entropy, the second law can be stated as follows:
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The entropy of an isolated system may not decrease. It either increases as the system approaches equilibrium, or
stays constant if the system is already in equilibrium.
Any process that would tend to decrease the entropy of an isolated system could never occur spontaneously in
nature. For a system that is not isolated, however, the entropy can increase, stay the same, or decrease.
Part A
What happens to the entropy of a bucket of water as it is cooled down (but not frozen)?
ANSWER:
It increases.
It decreases.
It stays the same.
Correct
Presumably, the bucket is not isolated: Heat must be transferred to another object, which is most likely at a
lower temperature than that of the bucket.
Part B
What happens to the entropy of a cube of ice as it is melted?
ANSWER:
It increases.
It decreases.
It stays the same.
Correct
Part C
What happens to the entropy of a piece of wood as it is burned?
ANSWER:
It increases.
It decreases.
It stays the same.
Correct
When a solid object is turned into a gas, the degree of disorder increases, so the entropy increases.
Let us try some calculations now.
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Part D
An object at 20∘ C absorbs 25.0 J of heat. What is the change in entropy ΔS of the object?
Express your answer numerically in joules per kelvin.
ANSWER:
ΔS = 8.53×10−2 J/K
Correct
Part E
An object at 500 K dissipates 25.0 kJ of heat into the surroundings. What is the change in entropy ΔS of the
object? Assume that the temperature of the object does not change appreciably in the process.
Express your answer numerically in joules per kelvin.
ANSWER:
ΔS = -50 J/K
Correct
Part F
An object at 400 K absorbs 25.0 kJ of heat from the surroundings. What is the change in entropy ΔS of the
object? Assume that the temperature of the object does not change appreciably in the process.
Express your answer numerically in joules per kelvin.
ANSWER:
ΔS = 62.5 J/K
Correct
Part G
Two objects form a closed system. One object, which is at 400 K, absorbs 25.0 kJ of heat from the other
object,which is at 500 K. What is the net change in entropy ΔSsys of the system? Assume that the
temperatures of the objects do not change appreciably in the process.
Express your answer numerically in joules per kelvin.
ANSWER:
ΔSsys = 12.5 J/K
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Correct
Note that the net entropy change is positive as the heat is transferred from the hotter object to the colder
one. If heat were transferred in the other direction, the change in entropy would have been negative; that is,
the entropy of the system would have decreased. This observation, not surprisingly, is in full accord with the
second law of thermodynamics.
Exercise 20.18
A certain brand of freezer is advertised to use 880 kW ⋅ h of energy per year.
Part A
Assuming the freezer operates for 6 hours each day, how much power does it require while operating?
Express your answer using two significant figures.
ANSWER:
P = 400 W
Part B
If the freezer keeps its interior at a temperature of -4.0 ∘ C in a 20.0 ∘ C room, what is its theoretical maximum
performance coefficient?
ANSWER:
K = 11.2
Correct
Part C
What is the theoretical maximum amount of ice this freezer could make in an hour, starting with water at 20.0
∘
C?
ANSWER:
m = 38.6 kg