Chapter 20
Entropy and the Second Law of Thermodynamics
In this chapter we will introduce the second law of
thermodynamics. The following topics will be covered:
Reversible processes
Entropy
The Carnot engine
Refrigerators
Real engines
(20–1)
Irreversible Processes
One-way processes that cannot be reversed by only small changes in their environment
are called irreversible. Irreversible processes are so common that if they were to occur
spontaneously in the opposite way (the "wrong" way) we would be astonished. Yet
none of these events violate the first law of thermodynamics. As an example, imagine
that we put our hands around a hot cup of coffee. Experience tells us that our hands
will get warmer. We would be astonished if our hands got cooler even though
such an event obeys the first law of thermodynamics. Thus, changes in the energy of a
closed system do not set the direction of an irreversible process.
This is defined by what we will call the "change in entropy" (S ) of the system,
which we define on the next slide.
The entropy postulate is stated as follows:
If an irreversible process occurs in a closed system, the entropy of the system
always increases; it never decreases.
(20–2)
f
dQ
S  S f  Si  
T
i
f
Change in Entropy
S  S f  Si  
i
dQ
T
Consider the free expansion of a gas shown in the figure.
The initial and final states  Pi , Vi  and  Pf , V f
 are shown
on the P  V diagram below. Even though the initial and
final states are well defined, we do not have intermediate
equilibrium states that take us from  Pi , Vi  to  Pf , V f  .
During the free expansion the temperature does not change:
Ti  T f .
In order to define the entropy change S for an
irreversible process that takes us from an initial
state i to a final state f of a system, we find a
reversible process that connects states i and f .
We then calculate :
f
dQ
S  S f  S i  
T
i
SI units for S : J/K
(20–3)
In the free expansion of the example, Ti  T f .
We thus replace the free expansion with an
isothermal expansion that connects states
 Pi ,Vi  and  Pf ,V f  .
From the first law of thermodynamics we have:
dEint  dQ  dW  dQ  dEint  dW  nCV dT  PdV 
dQ
dV
dT
P
 nCV
.
T
V
T
From ideal gas law we have: pV  nRT  pdV  nR
dV
.
V
f
f
f
Vf
Tf
dQ
dV
dT
S  
nR 
nCV 
 nR ln  nCV ln
T
V
T
Vi
Ti
i
i
i
The change in entropy depends only on the properties
of the initial and final states. It does not depend on
how the system changes from the initial to the final state.
(20–4)
Example 1:
An isothermal process is one in which Ti = Tf which
implies ln (Tf /Ti) = 0. Therefore, Eq. 20-4 leads to
 Vf 
22.0
S = nR ln    n =
= 2.75 mol.
8.31 ln  3.4/1.3
 Vi 
S  0
The Second Law of Thermodynamics
Consider now the reverse of the isothermal
process shown on the previous page.
f
In this case the gas will give back heat Q
i
to the reservoir at temperature T .
The change in entropy of the gas Sgas  
Q
T
.
The change in entropy of the reservoir S res 
Q
T
.
The net change in entropy of the gas-reservoir
closed system S  Sgas  Sres  0.
In a process that occurs in a closed system, the entropy of the
system increases for irreversible processes and remains
constant for reversible processes. The entropy never decreases.
The second law of thermodynamics can be written as S  0.
(20–5)
Example 2:
(a) The energy absorbed as heat is given by Eq. 19-14. Using Table 19-3, we find

J 
4
Q = cmT =  386
2.00
kg
75
K
=
5.79

10
J




kg  K 

We use the following relation derived in Sample Problem 20-2:
 Tf
S = mc ln 
 Ti

.

(b) With Tf = 373.15 K and Ti = 298.15 K, we obtain

J   373.15 
S =  2.00 kg   386
 ln 
 = 173 J/K.
kg  K   298.15 

Example 3:
ma ca Tai  T f  = mwcw T f  Twi   T f =
Tf 
ma caTai + mwcwTwi
.
ma ca + mwcw
 0.200 kg  900 J/kg  K 100C    0.0500 kg  4190 J/kg  K  20C 
 0.200 kg  900 J/kg  K    0.0500 kg  4190 J/kg  K 
 57.0C  330 K.
(b) Now temperatures must be given in Kelvins: Tai = 393 K, Twi = 293 K, and Tf = 330
K. For the aluminum, dQ = macadT and the change in entropy is
Tf
T f dT
dQ
 330 K 
Sa  
 ma ca 
 ma ca ln
  0.200 kg  900 J/kg  K  ln 

Tai T
T
Tai
 373 K 
 22.1 J/K.
(c) The entropy change for the water is
Sw  
T f dT
Tf
 330 K 
dQ
 mwcw 
 mwcw ln
 (0.0500 kg) (4190 J kg.K) ln 

Twi T
T
Twi
 293K 
 24.9 J K.
(d) The change in the total entropy of the aluminum-water system is
DS = DSa + DSw = -22.1 J/K + 24.9 J/K = +2.8 J/K.
Example 5:
Engines
A heat engine is a device that extracts heat from its environment
and does work. Every engine has a working substance. In a steam
engine the working substance is water. In a car engine the working
substance is an air-gasoline mixture. The engine operates on a
cycle; it passes through a series of thermodynamic processes and
returns again and again to each state in the cycle.
Ideal Engines
In what follows, we will examine "ideal engines" in which the
working substance is an ideal gas. Furthermore, all processes are
reversible and there is no friction or turbulence.
(20–6)
The Carnot Engine
One such ideal engine is the Carnot engine. It operates between
two reservoirs: one at higher temperature TH and the other at
lower temperature TL . The Carnot engine cycle is shown in the
P  V diagram.
The engine starts at point a and undergoes an isothermal expansion a → b
at temperature TH . During this process it absorbs an amount of heat QH
at temperature TH from the high-temperature reservoir. The gas then
undergoes an adiabatic expansion b → c and its temperature drops to TL .
The gas then is isothermally compressed from c  d . During this process
it delivers an amount of heat QL at temperature TL to the low-temperature
reservoir. Finally, the gas undergoes adiabatic compression d  a and its
temperature rises back to TH . During processes ab and bc the gas does
positive work on its environment. During processes cd and da the
environment does work on the gas. The net work W per cycle is equal
to the area enclosed by the curve abcda .
(20–7)
Efficiency of a Carnot Engine
  1
TL
TH
In the T -S diagram to the left, the temperature is plotted as a
function of the entropy S during one cycle of the Carnot engine.
The net work W done by the engine can be determined from the
first law of thermodynamics: Eint  Q  W  0. Since the
working substance returns to its original state there is no
change in its internal energy. Thus W  Q  QH  QL .
The change in entropy S 
QL

QL
QH

TL
TH
Since TH  TL  QH  QL
QH
TH

QL
TL
0
QH
TH

QL
TL
(eq. 1).
More energy is extracted from the high-temperature
reservoir than is delivered to the low-temperature reservoir. The efficiency  of the
Carnot engine is defined as  
From eq. 1 we have:   1 
W
Q  QL
Q
energy we get

 H
 1 L .
energy we pay for QH
QH
QH
TL
.
TH
(20–8)
The Perfect Engine
The efficiency  of the Carnot engine is:   1 
QL
QH
 1
TL
 1.
TH
It can be shown that no real engine can have an efficiency greater
than that of a Carnot engine operating between the same reservoirs.
Improvements in the efficiency of an engine can be achieved by
minimizing QL . A perfect engine (see lower figure) would have
QL  0. Such an engine would have  = 1. Unfortunately,
a perfect engine can be achieved either by having TL  0 or
TH  . Both requirements are impossible to meet.
The second law of thermodynamics can be stated as follows:
No series of processes is possible whose sole result is the transfer of heat from
a thermal reservoir and the complete conversion of this heat to work.
(20–9)
Example 7:
Example
A particular engine has a power output of 5000 W and an
efficiency of 25%. If the engine expels 8000 J of heat in each
cycle, find (a) the heat absorbed in each cycle and (b) the time
for each cycle
P  5000W
e  0.25
Qc  8000 J
QC
e  1
QH
0.25  1 
QH 
8000
QH
10,667 J
W  QH  QC  W  QH  8000
W
P
2667 J
W
W
 5000 
t
t
t
0.53 s
Example
The efficiency of a Carnot engine is 30%. The engine absorbs 800 J of
heat per cycle from a hot temperature reservoir at 500 K.
Determine (a) the heat expelled per cycle and (b) the temperature
of the cold reservoir
W
W
e
 0.30 
QH
800 J
W
W  QH  QC  W  800  QC
QC 
560 J
TC
TC
eC  1 
 0.30  1 
TH
500
TC 
350 K
240 J
Example
An automobile engine has an efficiency of 22.0% and produces 2510 J of work.
How much heat is rejected by the engine?
QH  W  QC
QC  QH  W
1 
QC 
 W  W   1
e
e 
W
 1

 2510 J 
 1  8900 J
 0.220 
e
W
QH
QH 
W
e
Refrigerators
A refrigerator is an engine that uses work to transfer heat from a
low-temperature reservoir to a high-temperature reservoir
as the engine repeats a set series of thermodynamic processes.
In the household refrigerator the work is provided by the
compressor and it transfers heat from the food storage area
 low-temperature reservoir  to the surrounding air
 high-temperature reservoir  . In an air conditioner
the low-temperature reservoir is the room that is being cooled.
The high-temperature reservoir is the outdoor air. An ideal
refrigerator is one in which all the processes are reversible and
no energy is lost to friction or turbulence.
One such refrigerator is a Carnot engine that operates
in the reverse order:
a → d → c → b → a This is known as a Carnot refrigerator.
(20–10)
TL
K
TH  TL
Efficiency of a Refrigerator
The coefficient of performance of a refrigerator is given
by the equation
QL
heat we want to transfer QL
K


.
work we pay for
W
QH  QL
For a Carnot refrigerator we have:
K 
QL
QH

TL
TH
TL
.
TH  TL
A perfect refrigerator would be one that does not
use any work (see lower figure). For such an engine
S  
Q Q

 0 because TH  TL . A perfect
TL TH
refrigerator violates the second law of thermodynamics
and does not exist. No series of processes is possible
whose sole result is the transfer of heat from a low temperature to a high - temperature reservoir.
(20–11)
Efficiency of a Real Engine
 X  C
In this section we will prove that the efficiency
 X of a real engine X cannot be larger than the
efficiency of a Carnot engine  C . Let's assume for
a moment that  X   C .
We couple the engine X with a Carnot refrigerator C as shown in the figure.
They both operate between the same high- and low-temperature reservoirs. The engine
provides the work necessary to operate the refrigerator so that no net work is being
W
W
used by the engine-refrigerator combination:  X   C 

 QH  QH .
QH
QH
Since W  0  QH  QL  QH  QL  QH  QH  QL  QL  Q.
By virtue of QH  QH  Q  0. Thus the net result of the engine-refrigerator
combination is to transfer heat Q from the low-temperature to the high-temperature
reservoir and thus it is a perfect refrigerator, which violates the second law
of thermodynamics. Thus
 X  C .
(20–12)
Example 10:
Example 11:
Example 12:
Example 13:
Example 14:
Example 15:
Example 16:
Example 17:
Example 18:
Example 19:
Example 20:
Example 21:
Example 22:
Example 23:
Example 24:
Example 25:
Example 26:
Example 27:
Example 28:
Example 29:
Example 30:
Example 31:
Example 32:
Example 33:
Example 34:
Example 35:
P38
P74