PSE9e ISM chapter22 final tủ tài liệu training

22
Heat Engines, Entropy, and the
Second Law of Thermodynamics
CHAPTER OUTLINE
22.1

Heat Engines and the Second Law of Thermodynamics

22.2

Heat Pumps and Refrigerators

22.3

Reversible and Irreversible Processes

22.4

The Carnot Engine

22.5

Gasoline and Diesel Engines

22.6

Entropy

22.7

Changes in Entropy for Thermodynamic Systems

22.8

Entropy and the Second Law

* An asterisk indicates a question or problem new to this edition.

ANSWERS TO OBJECTIVE QUESTIONS
OQ22.1

Answer (d). The second law says that you must put in some work to
pump heat from a lower-temperature to a higher-temperature
location. But it can be very little work if the two temperatures are
very nearly equal.

OQ22.2

Answer (d). Heat input will not necessarily produce an entropy
increase, because a heat input could go on simultaneously with a
larger work output, to carry the gas to a lower-temperature, lowerentropy final state. Work input will not necessarily produce an
entropy increase, because work input could go on simultaneously
with heat output to carry the gas to a lower-volume, lower-entropy
final state. Either temperature increase at constant volume, or
volume increase at constant temperature, or simultaneous increases
in both temperature and volume, will necessarily end in a higherentropy final state.
1147

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1148
OQ22.3

Heat Engines, Entropy, and the Second Law of Thermodynamics
Answer (c). The coefficient of performance of this refrigerator is

COP =

Qc
W

=

115 kJ
= 6.39
18.0 kJ

OQ22.4

Answer (c). Choice (c) is a statement of the first law of
thermodynamics, not the second law. Choices (a), (b), (d), and (e) are
alternative statements of the second law, (a) being the Kelvin-Planck
formulation, (b) the Carnot statement, (d) the Clausius statement,
and (e) summarizes the primary consequence of all these various
statements.

OQ22.5

(i) Answer (b). (ii) Answer (a). (iii) Answer (b). (iv) Answer (a). (v)
Answer (c). (vi) Answer (a). For any cyclic process the total input

energy must be equal to the total output energy. This is a
consequence of the first law of thermodynamics. It is satisfied by
processes (ii), (iv), (v), and (vi) but not by processes (i) and (iii). The
second law says that a cyclic process that takes in energy by heat
must put out some of the energy by heat. This is not satisfied for
process (v).

OQ22.6

Answer (a). The air conditioner operates on a cyclic process so the
change in the internal energy of the refrigerant is zero. Then, the
conservation of energy gives the thermal energy exhausted to the
room as Qh = Qc + Weng, where Qc is the thermal energy the air
conditioner removes from the room and Weng is the work done to
operate the device. Since Weng > 0, the air conditioner is returning
more thermal energy to the room than it is removing, so the average
temperature in the room will increase.

OQ22.7

Answer (c). The maximum theoretical efficiency (the Carnot
efficiency) of a device operating between absolute temperatures Tc <
Th is ec = 1 − Tc/Th. For the given steam turbine, this is

ec = 1 −

3.0 × 102 K
= 0.33 or 33%.
450 K

OQ22.8

Answer (d). The whole Universe must have an entropy change of
zero or more. The environment around the system comprises the rest
of the Universe, and must have an entropy change of +8.0 J/K, or
more.

OQ22.9

Answer: E > D > C > B > A. Recall that for and ideal gas, PV = nRT,
and Cv = 3R/2 and Cp = 5R/2.
Process A: isobaric, volume V goes to 0.5V, so temperature T goes to
5
0.5T, dQ = nCpdT, so dS = nCpdT/T; therefore ΔS = − nR ln 2.
2

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Chapter 22

1149

Process B: isothermal, volume V goes to 0.5V, so temperature T is
constant and pressure P goes to 2P, dQ = pdV = nRT(dV/V), so dS =
nR(dV/V); therefore ΔS = –nR ln 2.
Process C: adiabatic, Q = 0; therefore ΔS = 0.
Process D: isovolumetric, pressure P goes to 2P, so temperature T
3
goes to 2T, dQ = nCvdT, so dS = n Cv dT/T; therefore ΔS = nR ln 2.

2
Process E: isobaric, volume V goes to 2V; therefore ΔS =
OQ22.10

5
nR ln 2.
2

Answer (b). From conservation of energy, the energy input to the
engine must be

Qh = Weng + Qc = 15.0 kJ + 37.0 kJ = 52.0 kJ
so the efficiency is

e=
OQ22.11

Weng
Qc

=

15.0 kJ
= 0.288 or 28.8%.
52.0 kJ

Answer (b). In the reversible adiabatic expansion OA, the gas does
work against a piston, takes in no energy by heat, and so drops in
internal energy and in temperature. In the free adiabatic expansion
OB, there is no piston, no work output, constant internal energy, and

constant temperature for the ideal gas. Point A is at a lower
temperature than O and point C is at an even lower temperature. The
only point that could possibly have the same temperature as O is
point B.

ANSWERS TO CONCEPTUAL QUESTIONS
CQ22.1

(a)

The reduced flow rate of ‘cooling water’ reduces the amount of
heat exhaust Qc that the plant can put out each second. Even
with constant efficiency, the rate at which the turbines can take
in heat is reduced and so is the rate at which they can put out
work to the generators. If anything, the efficiency will drop,
because the smaller amount of water carrying the heat exhaust
will tend to run hotter. The steam going through the turbines
will undergo a smaller temperature change. Thus there are two
reasons for the work output to drop.

(b)

The engineer’s version of events, as seen from inside the plant,
is complete and correct. Hot steam pushes hard on the front of a
turbine blade. Still-warm steam pushes less hard on the back of

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1150

Heat Engines, Entropy, and the Second Law of Thermodynamics
the blade, which turns in response to the pressure difference.
Higher temperature at the heat exhaust port in the lake works
its way back to a corresponding higher temperature of the
steam leaving a turbine blade, a smaller temperature drop
across the blade, and a lower work output.

CQ22.2

One: Energy flows by heat from a hot bowl of chili into the cooler
surrounding air. Heat lost by the hot stuff is equal to heat gained by
the cold stuff, but the entropy decrease of the hot stuff is less than the
entropy increase of the cold stuff.
Two: As you inflate a soft car tire at a service station, air from a tank
at high pressure expands to fill a larger volume. That air increases in
entropy and the surrounding atmosphere undergoes no significant
entropy change.
Three: The brakes of your car get warm as you come to a stop. The
shoes and drums increase in entropy and nothing loses energy by
heat, so nothing decreases in entropy.

CQ22.3

No. The first law of thermodynamics is a statement about energy
conservation, while the second is a statement about stable thermal
equilibrium. They are by no means mutually exclusive. For the
particular case of a cycling heat engine, the first law implies
Qh = Weng + Qc , and the second law implies Qc > 0 .

CQ22.4

Take an automobile as an example. According to the first law or the
idea of energy conservation, it must take in all the energy it puts out.
Its energy source is chemical energy in gasoline. During the
combustion process, some of that energy goes into moving the
pistons and eventually into the mechanical motion of the car. The
chemical potential energy turning into internal energy can be
modeled as energy input by heat. The second law says that not all of
the energy input can become output mechanical energy. Much of the
input energy must and does become energy output by heat, which,
through the cooling system, is dissipated into the atmosphere.
Moreover, there are numerous places where friction, both mechanical
and fluid, turns mechanical energy into internal energy. In even the
most efficient internal combustion engine cars, less than 30% of the
energy from the fuel actually goes into moving the car. The rest ends
up as useless internal energy in the atmosphere.

CQ22.5

Either statement can be considered an instructive analogy. We
choose to take the first view. All processes require energy, either as
energy content or as energy input. The kinetic energy that it
possessed at its formation continues to make the Earth go around.
Energy released by nuclear reactions in the core of the Sun drives
weather on the Earth and essentially all processes in the biosphere.

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Chapter 22

1151

The energy intensity of sunlight controls how lush a forest or jungle
can be and how warm a planet is. Continuous energy input is not
required for the motion of the planet. Continuous energy input is
required for life because energy tends to be continuously degraded,
as heat flows into lower-temperature sinks. The continuously
increasing entropy of the Universe is the index to energy-transfers
completed.
CQ22.6

(a) A slice of hot pizza cools off. Road friction brings a skidding car
to a stop. A cup falls to the floor and shatters. Your cat dies. Any
process is irreversible if it looks funny or frightening when shown in
a videotape running backwards. (b) The free flight of a projectile is
nearly reversible.

CQ22.7

(a)

When the two sides of the semiconductor are at different
temperatures, an electric potential (voltage) is generated across
the material, which can drive electric current through an
external circuit. The two cups at 50°C contain the same amount
of internal energy as the pair of hot and cold cups. But no
energy flows by heat through the converter bridging between
them and no voltage is generated across the semiconductors.

(b)

A heat engine must put out exhaust energy by heat. The cold
cup provides a sink to absorb output or wasted energy by heat,
which has nowhere to go between two cups of equally warm
water.

CQ22.8

A higher steam temperature means that more energy can be
extracted from the steam. For a constant temperature heat sink at Tc,
and steam at Th, the efficiency of the power plant goes as
Th − Tc
T
= 1 − c and is maximized for a high Th.
Th
Th

CQ22.9

(a)

For an expanding ideal gas at constant temperature, the internal
energy stays constant. The gas must absorb by heat the same
amount of energy that it puts out by work. Then its entropy
⎛V ⎞
ΔQ
change is ΔS =
= nR ln ⎜ 2 ⎟ .

T
⎝ V1 ⎠

(b)

For a reversible adiabatic expansion, ΔQ = 0 and ΔS = 0. An
ideal gas undergoing an irreversible adiabatic expansion can
have any positive value for ΔS up to the value given in part (a).

CQ22.10

No. Your roommate creates “order” locally, but as she works, she
transfers energy by heat to the room, causing the net entropy to
increase. An analogy used by Carnot is instructive: A waterfall
continuously converts mechanical energy into internal energy. It

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1152

Heat Engines, Entropy, and the Second Law of Thermodynamics
continuously creates entropy as the motion of the falling water turns
into molecular motion at the bottom of the falls. We humans put
turbines into the waterfall, diverting some of the energy stream to
our use. Water flows spontaneously from high to low elevation and
energy spontaneously flows by heat from high to low temperature.
Into the great flow of solar radiation from Sun to Earth, living things
put themselves. They live on energy flow, more than just on energy.
A basking snake diverts energy from a high-temperature source (the

Sun) through itself temporarily, before the energy inevitably is
radiated from the body of the snake to a low-temperature sink (outer
space). A tree builds cellulose molecules and we build libraries and
babies who look like their grandmothers, all out of a thin diverted
stream in the universal flow of energy. We do not violate the second
law, for we build local reductions in the entropy of one thing within
the inexorable increase in the total entropy of the Universe.

CQ22.11

No. An engine with no thermal pollution would absorb energy from
a reservoir and convert it completely into work; this is a clear
violation of the second law of thermodynamics.

CQ22.12

(a) Shaking opens up spaces between jellybeans. The smaller ones
more often can fall down into spaces below them. (b) The
accumulation of larger candies on top and smaller ones on the
bottom implies a small decrease in one contribution to the total
entropy, but the second law is not violated. The total entropy
increases as the system warms up, its increase in internal energy
coming from the work put into shaking the box and also from a small
decrease in gravitational potential energy as the beans settle
compactly together.

CQ22.13

First, the efficiency of the automobile engine cannot exceed the
Carnot efficiency: it is limited by the temperature of burning fuel and

the temperature of the environment into which the exhaust is
dumped. Second, the engine block cannot be allowed to go over a
certain temperature. Third, any practical engine has friction,
incomplete burning of fuel, and limits set by timing and energy
transfer by heat.

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Chapter 22

1153

SOLUTIONS TO END-OF-CHAPTER PROBLEMS
Section 22.1

P22.1

(a)

Heat Engines and the
Second Law of Thermodynamics
 
We have e =

Weng
Qh

= 1−

Qc
Qh

→

With Qc = 8 000 J, we have Qh =
(b)

Qc
Qh

Qc
1− e

= 1− e

=

→

Qh =

Qc
1− e

8 000 J
=  10.7 kJ
1 − 0.250

The work per cycle is

Weng = Qh – Qc = 2 667 J
From the definition of output power,
P=

Weng
Δt

we have the time for one cycle:

Δt =
P22.2

(a)

Weng
2 667 J
=
= 0.533 s
P
5 000 J/s

The efficiency of a heat engine is e = Wenv Qh , where Wenv is the
work done by the engine and Qh is the energy absorbed from
the higher temperature reservoir. Thus, if Wenv = Qh 4 , the
efficiency is e = 1 4 = 0.25 or 25% .

(b)

From conservation of energy, the energy exhausted to the lower

temperature reservoir is Qc = Qh − Wenv . Therefore, if
Wenv = Qh 4 , we have Qc = 3 Qh 4 or Qc Qh = 3 4 .

P22.3

(a)

The efficiency of the engine is
e=

(b)

Weng
Qh

=

25.0 J
= 0.069 4 or 6.94%
360 J

The energy expelled to the cold reservoir during each cycle is

Qc = Qh − Weng = 360 J − 25.0 J = 335 J

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1154
P22.4

Heat Engines, Entropy, and the Second Law of Thermodynamics
The engine’s output work we identify with the kinetic energy of the
bullet:

Weng = K =
e=

1 2 1
2
mv = ( 0.002 4 kg ) ( 320 m s ) = 123 J
2
2

Weng
Qh
Weng

123 J
= 1.12 × 10 4 J
e
0.011
Qh = Weng + Qc
Qh =

=

The energy exhaust is

Qc = Qh − Weng = 1.12 × 10 4 J − 123 J = 1.10 × 10 4 J

Q = mcΔT
ΔT =
P22.5

(a)

Q 1.10 × 10 4 J ⋅ kg°C
=
= 13.7°C
mc
(1.80 kg)(448 J)

The engine’s efficiency is given by
e=

Weng
Qh

=

Qh − Qc
Q
1.20 kJ
= 1− c = 1−
Qh
Qh
1.70 kJ

= 0.294 (or 29.4%)

(b)

During each cycle, the work done by the engine is

Weng = Qh − Qc = 1.70 kJ − 1.20 kJ = 5.00 × 102 J
(c)

The power transferred out of the engine is

P=
P22.6

(a)

Weng
Δt

=

5.00 × 102 J
= 1.67 × 103 W = 1.67 kW
0.300 s

The input energy each hour is

(7.89 × 10

3

⎛ 60 min ⎞

J revolution ) ( 2 500 rev min ) ⎜
⎝ 1 h ⎟⎠
= 1.18 × 109 J h

⎛
1L
⎞
= 29.4 L h .
implying fuel input ( 1.18 × 109 J h ) ⎜
7 ⎟
⎝ 4.03 × 10 J ⎠

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Chapter 22
(b)

1155

Qh = Weng + Qc . For a continuous-transfer process we may divide
by time to have

Qh Weng Qc
=
+
Δt
Δt
Δt
Useful power output =

Weng
Δt

=

Qh Qc
−
Δt Δt

⎛ 7.89 × 103 J 4.58 × 103 J ⎞ ⎛ 2 500 rev ⎞ ⎛ 1 min ⎞
=⎜
−
⎜
⎟⎜
⎟
⎝ revolution revolution ⎟⎠ ⎝ 1 min ⎠ ⎝ 60 s ⎠
= 1.38 × 105 W
⎛ 1 hp ⎞
Peng = ( 1.38 × 105 W ) ⎜
= 185 hp
⎝ 746 W ⎟⎠

(c)

(d)
P22.7

Peng

⎛ 1.38 × 105 J s ⎞ ⎛ 1 rev ⎞
= τω ⇒ τ =
=⎜
⎜
⎟ = 527 N ⋅ m
ω
⎝ 2 500 rev 60 s ⎟⎠ ⎝ 2π rad ⎠
Peng

⎛ 4.58 × 103 J ⎞ ⎛ 2 500 rev ⎞
5
=
⎜
⎟ = 1.91 × 10 W
Δt ⎜⎝ revolution ⎟⎠ ⎝ 60 s ⎠

Qc

The energy to melt a mass ΔmHg of Hg is Qc = mHg L f . The energy
absorbed to freeze ΔmAl of aluminum is Qh = mAl L f . The efficiency is

e = 1−

ΔmHg LHg
Qc
(15.0 g )(1.18 × 104 J kg )
= 1−
= 1−
Qh
ΔmAlLAl

(1.00 g )( 3.97 × 105 J kg )

= 0.554 = 55.4%


 

Section 22.2
P22.8

P22.9

Heat Pumps and Refrigerators

COP ( refrigerator ) =

Qc
W

(a)

If Qc = 120 J and COP = 5.00, then W = 24.0 J .

(b)

Qh = Qc + W = 120 J + 24 J = 144 J

(a)

The work done on the refrigerant in each cycle is

W = QH − QL = 625 kJ − 550 kJ = 75.0 kJ

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1156

Heat Engines, Entropy, and the Second Law of Thermodynamics
(b)

The coefficient of performance of a refrigerator is:

COP =

QL
QL
=
W QH − QL

Solving numerically:

COP =
P22.10

(a)

QL
QL
550 kJ

=
=
= 7.33
W QH − QL 625 kJ − 550 kJ

The coefficient of performance of a heat pump is COP = Qh W ,
where Qh is the thermal energy delivered to the warm space and
W is the work input required to operate the heat pump.
Therefore,

Qh = W ⋅ COP = ( P ⋅ Δt ) ⋅ COP
⎡⎛
⎛ 3 600 s ⎞ ⎤
J⎞
8
= ⎢⎜ 7.03 × 103 ⎟ 8.00 h ⎜
⎥ 3.80 = 7.69 × 10 J
⎟
s⎠
⎝ 1 h ⎠⎦
⎣⎝

(

(b)

The energy extracted from the cold space (outside air) is

Qc = Qh − W = Qh −
or

*P22.11

)

Qh

1 ⎞
⎛
= Qh ⎜ 1 −
⎟
⎝
COP
COP ⎠

1 ⎞
⎛
Qc = ( 7.69 × 108 J ) ⎜ 1 −
= 5.67 × 108 J
⎟
⎝
3.80 ⎠

COP = 3.00 =

Qc
Q
. Therefore, W = c .
W
3.00

The heat removed each minute is

QC
= ( 0.030 0 kg ) ( 4 186 J kg °C ) ( 22.0°C )
t
+ ( 0.030 0 kg ) ( 3.33 × 105 J kg )

+ ( 0.030 0 kg ) ( 2 090 J kg °C ) ( 20.0°C )

= 1.40 × 10 4 J min = 233 J/s
Thus, the work done per second = P =
P22.12

(a)

233 J/s
= 77.8 W .
3.00

The coefficient of performance of a heat pump is

COPh.p. =

QH
QH
=
W QH − QL

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Chapter 22

1157

Because work (energy) is power times time (W = PΔt ) , the
equation above may be rearranged to obtain the heat added to the
home:

QH = COP ⋅W = COP ⋅ PΔt
= ( 4.20 )( 1.75 × 103 J/s )( 3600 s ) = 2.65 × 107 J
(b)

The coefficient of performance of a refrigerator or air conditioner
is

COPrefr. =

QL
QL
=
W QH − QL

and can be written in terms of the coefficient of performance of a
heat pump because:

W = QH − QL :
COPrefr. =

QL QH − W

QH
W
=
=
+
W QH − QL QH − QL QH − QL

Where the first term on the far right is identically the coefficient
of performance of the heat pump, and the second term is
identically one (because W = QH − QL ). Thus,

COPrefr. =

QH
W
+
= COPh.p. − 1 = ( 4.20 ) − 1
QH − QL QH − QL

= 3.20
P22.13

(a)

The energy use by the freezer each day is
⎛
kWh ⎞ ⎛ 3.60 × 106 J ⎞ ⎛ 1 y ⎞
W = P ⋅ Δt = ⎜ 457
⋅ (1 d )
⎟

y ⎠ ⎜⎝ 1 kWh ⎟⎠ ⎜⎝ 365 d ⎟⎠
⎝
= 4.51 × 106 J

(b) From the definition of the coefficient of performance for a
refrigerator, ( COP )R = Qc W , the thermal energy removed from
the cold space each day is

Qc = ( COP )R ⋅ W = 6.30 ( 4.51 × 106 J ) = 2.84 × 107 J

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1158

Heat Engines, Entropy, and the Second Law of Thermodynamics
(c) The water must be cooled 20.0°C before it will start to freeze, so
the thermal energy that must be removed from mass m of water
to freeze it is Qc = mcw ΔT + mL f . The mass of water that can be
frozen each day is then
m=

Qc
2.84 × 107 J
=
cw ΔT + L f ( 4186 J kg ⋅°C )( 20.0°C ) + 3.33 × 105 J kg

= 68.1 kg

 

Section 22.3

Reversible and Irreversible Processes

Section 22.4

The Carnot Engine

P22.14

The maximum possible efficiency for a heat engine operating between
reservoirs with absolute temperatures of Tc = 25° + 273 = 298 K and
Th = 375° + 273 = 648 K is the Carnot efficiency:

ec = 1 −
P22.15

Tc
298 K
= 1−
= 0.540 ( or 54.0% )
Th
648 K

We use the Carnot expression for maximum possible efficiency, and
the definition of efficiency to find the useful output. The engine is a
steam turbine in an electric generating station with
Tc = 430°C = 703 K

Th = 1 870°C = 2 143 K

and

ΔT 1 440 K
=
= 0.672 = 67.2%
Th
2 143 K

(a)

eC =

(b)

e = Weng/ Qh = 0.420

and

Qh = 1.40 × 105 J

for one second of operation, so

Weng = 0.420 Qh = 5.88 × 10 4 J
and the power is

P=
P22.16

Weng
Δt

=

5.88 × 10 4 J
= 58.8 kW
1s

The efficiency of a Carnot engine operating between these
temperatures is

eC  = 1 − 

Tc
273 K
 = 1 − 
 = 0.068 3 = 6.83%
Th
293 K

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Chapter 22

1159

Therefore, there is no way that the inventor’s engine can have an

efficiency of 0.110 = 11.0%.
P22.17

e=

Weng
Qh

= ec = 1 −
Weng

Tc
Th

→

Weng / Δt
Qh / Δt

P
Qh Δt

= 1−

Tc
Th

1.50 × 105 W ) ( 3 600 s )
(
PΔt

=
=
= 8.70 × 108 J
1 − (Tc / Th )
1 − ( 293 K / 773 K )

(a)

Qh =

(b)

Qc = Qh − Weng = Qh − PΔt

e

=

= 8.70 × 108 J − ( 1.50 × 105 W ) ( 3 600 s )
= 3.30 × 108 J

P22.18

e=

Weng
Qh

= ec = 1 −

Weng

→

Weng / Δt
Qh / Δt

=

P
Qh Δt

= 1−

Tc
Th

⎛ Th ⎞
PΔt
= PΔt ⎜
1 − (Tc / Th )
⎝ Th − Tc ⎟⎠

(a)

Qh =

(b)

⎛ Th ⎞

Qc = Qh − Weng = Qh − PΔt = PΔt ⎜
− PΔt
⎝ Th − Tc ⎟⎠

e

=

Tc
Th

⎛ T − (Th − Tc ) ⎞
⎛ Th
⎞
⎛ Tc ⎞
= PΔt ⎜
− 1⎟ = PΔt ⎜ h
=
PΔt
⎜⎝ T − T ⎟⎠
⎝ Th − Tc
⎠
⎝ Th − Tc ⎟⎠
h
c

P22.19

( COP )refrig =

P22.20

(a)

Tc
270 K
=
= 9.00
ΔT 30.0 K

For a complete cycle, ΔEint = 0 and
⎡ (Qh )
⎤
W = Qh − Qc = Qc ⎢
− 1⎥
⎢⎣ Qc
⎥⎦

The text shows that for a Carnot cycle (and only for a reversible
T − Tc
Q
T
Qc .
cycle), h = h . Therefore, W = h
Tc
Qc
Tc
(b)

We have the definition of the coefficient of performance for a

Q
refrigerator, COP = c . Using the result from part (a), this
W
Tc
becomes COP =
.
Th − Tc

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1160

Heat Engines, Entropy, and the Second Law of Thermodynamics

Qc + W

P22.21

( COP )heat pump =

P22.22

( COP )Carnot refrig =

W

=

Th

295 K
=
= 11.8
ΔT
25 K

Q
Tc
4.00 K
=
= 0.013 8 = c
ΔT 289 K
W

∴W = 72.2 J per 1 J of energy removed by heat.
P22.23

We wish to evaluate COP = Qc W for a refrigerator, which is a Carnot
engine run in reverse. For a Carnot engine,

⎫
⎪ 1 Qc + W Qc
=
+1
W
W ⎬→ =
e=
=
e
W

W
⎪
Qh
Qc + W ⎭
Qh = Qc + W

which gives

COP =

Qc
W

=

1
−1
e

Therefore,
COP =

*P22.24

1
1
−1=
− 1 = 1.86 .
e
0.350

The Carnot summer efficiency is

ec , s = 1 −

( 273 K + 20.0°C )
Tc
= 1−
= 0.530
( 273 K + 350°C )
Th

And in winter,
ec , w = 1 −

283
= 0.546
623

Then the actual winter efficiency is
0.320

P22.25

(a)

( )

0.546
= 0.330

0.530

or

33.0%

The absolute temperature of the cold reservoir is
Tc = 20.0° + 273 = 293 K. If the Carnot efficiency is to be
eC = 0.650, it is necessary that

1−

Tc
= 0.650
Th

or

Tc
= 0.350
Th

and

Th =

Tc
0.35

Thus,

Th =

293 K
= 837 K
0.35

or

Th = 837 − 273 = 564°C

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 22

P22.26

1161

(b)

No. A real engine will always have an efficiency less than the
Carnot efficiency because it operates in an irreversible manner.

(a)

eC = 1 −

(b)

We differentiate eC = 1 − Tc /Th to find

Tc
350 K
= 1−
= 0.300
Th
500 K

dec
T
350 K
−3
= 0 − Tc ( −1)Th−2 = c2 =
K −1
2 = 1.40 × 10
dTh
Th ( 500 K )
(c)

We differentiate eC = 1 – Tc/Th to find

dec
1
1
= 0− = −
= −2.00 × 10−3 K −1
dTc
Th
500 K

(d)
P22.27

No. The derivative in part (c) depends only on Th .

Isothermal expansion at Th = 523 K
Isothermal compression at Tc = 323 K
Gas absorbs 1 200 J during expansion.
(a)

ec = 1 −

For a Carnot cycle,

For any engine,

e=

Weng
Qh

= 1−

Tc
Th

Qc
Qh

Therefore, for a Carnot engine, 1 −

Q
Tc
= 1− c
Th
Qh

Then we have
⎛T ⎞
⎛ 323 K ⎞
Qc = Qh ⎜ c ⎟ = (1 200 J) ⎜
= 741 J
⎝ 523 K ⎟⎠
⎝ Th ⎠

(b)

The work we can calculate as

Weng = Qh − Qc = ( 1 200 J − 741 J ) = 459 J
P22.28

(a)

emax = 1 −

(b)

P=

Weng
Δt

Tc
278 K
= 1−
= 5.12 × 10−2 = 5.12%
Th
293 K

= 75.0 × 106 J s

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1162

Heat Engines, Entropy, and the Second Law of Thermodynamics

(

)

Therefore, Weng = 75.0 × 106 J s ( 3 600 s h ) = 2.70 × 1011 J h .
From e =

Weng
Qh

Qh =

(c)

*P22.29

(a)

we find

Weng
e

=

2.70 × 1011 J h
= 5.27 × 1012 J h = 5.27 TJ h
−2
5.12 × 10

As fossil-fuel prices rise, this way to use solar energy will become
a good buy.
With reservoirs at absolute temperatures of Tc = 80.0°C + 273 =
353 K and Th = 350°C + 273 = 623 K, the Carnot efficiency is

eC = 1 −

Tc
353 K
= 1−
= 0.433
Th

623 K

( or 43.3% )

so the maximum power output is

Pmax =
(b)

Weng

From e = 1 −

Δt

=

eC Qh 0.433 ( 21.0 kJ )
=
= 9.10 kW
Δt
1.00 s

Qc
, the energy expelled by heat each cycle is
Qh

Qc = Qh ( 1 − e ) = (21.0 kJ)( 1 − 0.433 ) = 11.9 kJ
P22.30

(a)

e=

Weng1 + Weng 2
Q1h

=

e1Q1h + e2Q2 h
Q1h

Now, Q2h = Q1c = Q1h − Weng1 = Q1h − e1Q1h,
so
e=

(b)

e = e1 + e 2 − e1 e 2 = 1 −
=2−

(c)

e1Q1h + e2 (Q1h − e1Q1h )
= e1 + e 2 − e1 e 2
Q1h

Ti
T ⎛
T ⎞⎛

T⎞
+ 1− c − ⎜1− i ⎟ ⎜1− c ⎟
Th
Ti ⎝
Th ⎠ ⎝
Ti ⎠

Ti Tc
T T T
T
− − 1+ i + c − c = 1− c
Th Ti
Th Ti Th
Th

The combination of reversible engines is itself a reversible engine
so it has the Carnot efficiency. No improvement in net efficiency
has resulted.

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Chapter 22

(d) With Weng2 = Weng1 , e =
1−

⎛
Tc
T⎞

= 2⎜1 − i ⎟
Th
Th ⎠
⎝

0−

Tc
2T
= 1− i
Th
Th

Weng1 + Weng2
Q1h

=

2Weng1
Q1h

1163

= 2e1

2Ti = Th + Tc
Ti =

(e)

1
(Th + Tc )
2

e1 = e 2 = 1 −

Ti
T
= 1− c
Th
Ti

Ti2 = TcTh
Ti = (ThTc )

12

γ

P22.31

(a)

γ

⎛ Pf Vf ⎞
⎛ PV ⎞
In an adiabatic process, Pf Vfγ = PiViγ . Also, ⎜
=⎜ i i⎟ .
⎟

⎝ Ti ⎠
⎝ Tf ⎠

⎛ Pf ⎞
Dividing the second equation by the first yields T f = Ti ⎜ ⎟
⎝P⎠

(γ −1) γ

.

i

Since γ =

5
γ −1 2
= = 0.400 and we have
for argon,
3
γ
5

⎛ 300 × 103 Pa ⎞
T f = ( 1 073 K ) ⎜
⎝ 1.50 × 106 Pa ⎟⎠
(b)

0.400

= 564 K

ΔEint = nCV ΔT = Q − Weng = 0 − Weng , so Weng = −nCV ΔT,
and the power output is (suppressing the units of R)

P=

=

Weng
Δt

=

−nCV ΔT
Δt

( −80.0 kg )(1 mol

3
0.039 9 kg ) ⎛ ⎞ ( 8.314 )( 564 K − 1 073 K )
⎝ 2⎠
60.0 s

P = 2.12 × 105 W = 212 kW
(c)

eC = 1 −

Tc

564 K
= 1−
= 0.475
Th
1 073 K

or

47.5%

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1164
P22.32

Heat Engines, Entropy, and the Second Law of Thermodynamics
(a)

First, consider the adiabatic process D → A:

PDVDγ = PAVAγ

so
γ

53
⎛ VA ⎞
⎛ 10.0 L ⎞
PD = PA ⎜ ⎟ = ( 1 400 kPa ) ⎜

= 712 kPa
⎝ 15.0 L ⎟⎠
⎝V ⎠
D

⎛ nRTD ⎞ γ ⎛ nRTA ⎞ γ
Also, ⎜
VD = ⎜
VA ,
⎝ VD ⎟⎠
⎝ VA ⎟⎠

⎛V ⎞
or TD = TA ⎜ A ⎟
⎝ VD ⎠

γ −1

⎛ 10.0 ⎞
= ( 720 K ) ⎜
⎝ 15.0 ⎟⎠

23

= 549 K .

Now, consider the isothermal process C → D: TC = TD = 549 K
⎛V ⎞ ⎡ ⎛V ⎞
PC = PD ⎜ D ⎟ = ⎢ PA ⎜ A ⎟
⎝ VC ⎠ ⎢⎣ ⎝ VD ⎠

γ

⎤⎛ V ⎞
P Vγ
⎥ ⎜ D ⎟ = A γA−1
⎥⎦ ⎝ VC ⎠ VCVD

(1 400 kPa) (10.0 L )5 3
( 24.0 L ) (15.0 L )2 3

PC =

= 445 kPa

Next, consider the adiabatic process B → C: PBVBγ = PCVCγ
But, PC =

PAVAγ
from above. Also considering the isothermal
VCVDγ −1

⎛V ⎞
process, PB = PA ⎜ A ⎟ .
⎝ VB ⎠
⎛ P Vγ ⎞
⎛V ⎞
Hence, PA ⎜ A ⎟ VBγ = ⎜ A γA−1 ⎟ VCγ , which reduces to
⎝ VB ⎠
⎝ VCVD ⎠

VB =

VAVC ( 10.0 L ) ( 24.0 L )
=
= 16.0 L
VD
15.0 L

⎛V ⎞
⎛ 10.0 L ⎞
Finally, PB = PA ⎜ A ⎟ = ( 1 400 kPa ) ⎜
= 875 kPa .
⎝ 16.0 L ⎟⎠
⎝ VB ⎠

State

P (kPa)

V (L)

T (K)

A

1 400

10.0

720

B

875

16.0

720

C

445

24.0

549

D

712

15.0

549

TABLE P22.32(a)
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 22
(b)

1165

For the isothermal process A → B: ΔEint = nCV ΔT = 0 so
⎛V ⎞
Q = −W = nRT ln ⎜ B ⎟
⎝ VA ⎠
⎛ 16.0 L ⎞
= ( 2.34 mol ) ( 8.314 J mol ⋅ K ) ( 720 K ) ln ⎜
⎝ 10.0 L ⎟⎠
= +6.58 kJ

For the adiabatic process B → C: Q = 0
ΔEint = nCV (TC − TB )
⎡3
⎤
= ( 2.34 mol ) ⎢ ( 8.314 J mol ⋅ K ) ⎥ ( 549 K − 720 K )
⎣2
⎦
= −4.98 kJ

and W = −Q + ΔEint = 0 + ( −4.98 kJ ) = −4.98 kJ .
For the isothermal process C → D: ΔEint = nCV ΔT = 0 and
⎛V ⎞
Q = −W = nRT ln ⎜ D ⎟
⎝ VC ⎠
⎛ 15.0 L ⎞
= ( 2.34 mol ) ( 8.314 J mol ⋅ K ) ( 549 K ) ln ⎜

⎝ 24.0 L ⎟⎠
= −5.02 kJ

Finally, for the adiabatic process D → A: Q = 0
ΔEint = nCV (TA − TD )
⎡3
⎤
= ( 2.34 mol ) ⎢ ( 8.314 J mol ⋅ K ) ⎥ ( 720 K − 549 K )
2
⎣
⎦
= +4.98 kJ

and W = −Q + ΔEint = 0 + 4.98 kJ = +4.98 kJ

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1166

Heat Engines, Entropy, and the Second Law of Thermodynamics

Process

Q (kJ)

W (kJ)

Δ Eint (kJ)

A→B

+6.58

–6.58

0

B→C

0

–4.98

–4.98

C→D

–5.02

+5.02

0

D→ A

0

+4.98

+4.98

ABCDA

+1.56

–1.56

0

TABLE P22.32(b)
The work done by the engine is the negative of the work input.
The output work Weng is given by the work column in TABLE
P22.32(b) with all signs reversed.

P22.33

Weng

e=

(d)

ec = 1 −

(a)

“The actual efficiency is two thirds the Carnot efficiency” reads as
an equation

Qh

=

−WABCD 1.56 kJ
=
= 0.237
QA→B
6.58 kJ

(c)

Tc
549 K
= 1−
= 0.237
Th
720 K

Weng
Qh

=

Weng
Qc + Weng

=

or

or

23.7%

23.7%

2⎛
T ⎞ 2 Th − Tc
1− c ⎟ =
⎜
3⎝
Th ⎠ 3 Th

All the T’s represent absolute temperatures. Then

Qc + Weng
Weng
Qc = Weng

=

1.5 Th
Th − Tc

→

Qc
Weng

=

1.5 Th
1.5 Th − Th + Tc
−1=
Th − Tc
Th − Tc

Weng 0.5 Th + Tc
Q
0.5 Th + Tc
→ c =
Th − Tc
Δt
Δt Th − Tc

⎛ 0.5Th  + 383 ⎞
Qc
 = 1.40 ⎜
, where Qc /Δt is in megawatts and T is
Δt
⎝ Th  − 383 ⎟⎠
in kelvins.

(b)

The exhaust power decreases as the firebox temperature increases.

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Chapter 22

(c)

1167

⎛ 0.5 Th + 383 K ⎞
Qc
= ( 1.40 MW ) ⎜
Δt
⎝ Th − 383 K ⎟⎠
⎛ 0.5(1 073 K) + 383 K ⎞
= ( 1.40 MW ) ⎜
= 1.87 MW
⎝ 1 073 K − 383 K ⎟⎠

(d) We require

Qc
Δt

=

⎛ 0.5 Th + 383 K ⎞
1
(1.87 MW ) = (1.40 MW ) ⎜
2
⎝ Th − 383 K ⎟⎠

0.5 Th + 383 K
= 0.666
Th − 383 K
0.5 Th + 383 K = 0.666Th − 255 K
Th = 638 K/0.166 = 3.84 × 103 K
(e)

P22.34

The minimum possible heat exhaust power is approached as the
firebox temperature goes to infinity, and it is Qc / Δt =
1.40 MW(0.5/1) = 0.700 MW. The heat exhaust power cannot be
as small as (1/4)(1.87 MW) = 0.466 MW. So no answer exists. The
energy exhaust cannot be that small.

We determine the power required from
Qc
W

= COPc ( refrigerator ) =

Qc / Δt
Tc
=
Th − Tc
W / Δt

0.150 W 260 K
=
W / Δt

40.0 K
W
⎛ 40.0 K ⎞
P=
= ( 0.150 W ) ⎜
= 23.1 mW
⎝ 260 K ⎟⎠
Δt

P22.35

The coefficient of performance of the device is
COP = 0.100 COPCarnot cycle
or

⎛Q ⎞
⎛
⎞
1
= 0.100 ⎜ h ⎟
= 0.100 ⎜
W
⎝ Carnot efficiency ⎟⎠
⎝ W ⎠ Carnot cycle

Qh

⎛ Th ⎞
293 K
⎛

⎞
= 0.100 ⎜
= 0.100 ⎜
⎟ = 1.17
⎟
⎝
W
293 K − 268 K ⎠
⎝ Th − Tc ⎠

Qh

Thus, 1.17 joules of energy enter the room by heat for each joule of
work done.

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1168

Heat Engines, Entropy, and the Second Law of Thermodynamics

Section 22.5
P22.36

Gasoline and Diesel Engines

Compression ratio = 6.00, γ = 1.40
(a)

⎛V ⎞
Efficiency of an Otto engine: e = 1 − ⎜ 2 ⎟
⎝V ⎠

γ −1

1

⎛ 1 ⎞
e = 1− ⎜
⎝ 6.00 ⎟⎠

P22.37

0.400

= 51.2%

(b)

If actual efficiency e’ = 15.0%, the fraction of fuel wasted is
(assuming complete combustion of the air-fuel mixture)
e − e′ = 36.2% .

(a)

For adiabatic expansion, PiViγ = Pf Vfγ . Therefore,
γ

1.40

⎛ Vi ⎞
⎛ 50.0 cm 3 ⎞
6
Pf = Pi ⎜ ⎟ = (3.00 × 10 Pa) ⎜
⎝ 300 cm 3 ⎟⎠
⎝ Vf ⎠

= 2.44 × 105 Pa
(b)

Since Q = 0, we have Weng = Q − ΔE = −nCV ΔT = −nCV (T f − Ti ).
From γ =
CV =

CP CV + R
=
, we get (γ − 1)CV = R, so that
CV
CV
R
= 2.50 R
1.40 – 1

The work done by the gas in expanding is then

Weng = n(2.50 R)(Ti − T f ) = 2.50PiVi − 2.50Pf Vf
= 2.50 ⎡⎣( 3.00 × 106 Pa ) ( 50.0 × 10−6 m 3 )

− ( 2.44 × 105 Pa ) ( 300 × 10−6 m 3 ) ⎤⎦

= 192 J
P22.38

The energy transferred by heat over the
paths CD and BA is zero since they are
adiabatic.
Over path BC: QBC = nCP (TC − TB) > 0
Over path DA: QDA = nCV (TA − TD) < 0
Therefore, Qc = QDA and Qh = QBC.
ANS. FIG. P22.38

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Chapter 22

1169

The efficiency is then
e = 1−

Section 22.6
P22.39

Qc
(T − TA ) CV = 1 − 1 ⎛ TD − TA ⎞
= 1− D
Qh
γ ⎜⎝ TC − TB ⎟⎠
(TC − TB ) CP

Entropy

Each marble is returned to the bag before the next is drawn, so the
probability of drawing a red one is the same as drawing a green one.
(a)
Result

Possible Combinations

Total

All red

RRR

1

2R, 1G

RRG, RGR, GRR

3

1R, 2G

RGG, GRG, GGR

3

All green

GGG

1
TABLE P22.39(a)

(b)
Result

Possible Combinations

Total

All red

RRRRR

1

4R, 1G

RRRRG, RRRGR, RRGRR, RGRRR,
GRRRR

5

3R, 2G

RRRGG, RRGRG, RGRRG, GRRRG,

RRGGR, RGRGR, GRRGR, RGGRR,
GRGRR, GGRRR

10

2R, 3G

GGGRR, GGRGR, GRGGR, RGGGR,
GGRRG, GRGRG, RGGRG, GRRGG,
RGRGG, RRGGG

10

1R, 4G

RGGGG, GRGGG, GGRGG, GGGRG,
GGGGR

5

All green

GGGGG

1
TABLE P22.39(b)

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1170
P22.40

Heat Engines, Entropy, and the Second Law of Thermodynamics
(a)

The table is shown in TABLE P22.40 below.

(b)

On the basis of the table, the most probable recorded result of a
toss is 2 heads and 2 tails .
Result

Possible Combinations

Total

All heads

HHHH

1

3H, 1T

THHH, HTHH, HHTH, HHHT

4

2H, 2T

TTHH, THTH, THHT, HTTH, HTHT,
HHTT

6

1H, 3T

HTTT, THTT, TTHT, TTTH

4

All tails

TTTT

1
TABLE P22.40

P22.41

(a)

A 12 can only be obtained one way, as 6 + 6.

(b)

A 7 can be obtained six ways: 6 + 1, 5 + 2, 4 + 3, 3 + 4, 2 + 5,
and 1 + 6.

 

Section 22.7

Changes in Entropy for Thermodynamic Systems

Section 22.8

Entropy and the Second Law

P22.42

For a freezing process,
5
ΔQ − ( 0.500 kg ) ( 3.33 × 10 J kg )
ΔS =
=
= −610 J K
T
273 K

P22.43

The hot water has negative energy input by heat, given by Q = mcΔT.
The surrounding room has positive energy input of this same number
of joules, which we can write as Qroom = ( mc ΔT )water . Imagine the room
absorbing this energy reversibly by heat, from a stove at 20.001°C. Then
its entropy increase is Qroom/T:

Qr mcw ΔT ( 0.125 kg ) ( 4186 J kg ⋅°C )( 80°C )
=
=
T
T
293 K
= 143 J K

ΔS =

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Chapter 22
*P22.44

1171

ciron = 448 J kg ⋅ °C ; c water = 4 186 J kg ⋅ °C
From Qcold = −Qhot :

( 4.00 kg ) ( 4 186

J kg ⋅ °C ) (T f − 10.0°C )

= − ( 1.00 kg ) ( 448 J kg ⋅ °C ) (T f − 900°C )

which yields Tf = 33.2°C = 306.3 K. Then,

ΔS =

306.3 K

∫

283 K

c water mwater dT 306.3 K ciron miron dT
+ ∫
T
T
1 173 K

ΔS = c water mwater ln

(

)

⎛ 306.3 K ⎞
306.3 K
+ ciron miron ln ⎜
⎝ 1 173 K ⎟⎠
283 K

ΔS = ( 4 186 J kg ⋅ K ) ( 4.00 kg ) ( 0.078 7 )

+ ( 448 J kg ⋅ K ) ( 1.00 kg ) ( −1.34 )

ΔS = 717 J K
*P22.45

The car ends up in the same thermodynamic state as it started, so it
undergoes zero changes in entropy. The original kinetic energy of the
car is transferred by heat to the surrounding air, adding to the internal
energy of the air. Its change in entropy is
1
1
mv 2
(1 500 kg )( 20.0 m/s )2
ΔS = 2
= 2
= 1.02 kJ K
T
293 K

P22.46

The total momentum before collision is zero, so the combined mass
must be at rest after the collision. The energy dissipated by heat equals
the total initial kinetic energy,
2
⎛1
⎞
Q = 2 ⎜ mv 2 ⎟ = ( 2000 kg ) ( 20.0 m s ) = 8.00 × 105 J = 800 kJ
⎝2
⎠

With the environment at an absolute temperature of T = 23 + 273 = 296

K, the change in entropy is
ΔS =

P22.47

ΔQr 800 kJ
=
= 2.70 kJ K
T
296 K

The potential energy lost by the log is eventually transferred by heat
into thermal energy of the environment, so Q = mgh, and the change in
entropy is

ΔS =

2
Q mgh ( 70.0 kg ) ( 9.80 m s ) ( 25.0 m )
=
=
= 57.2 J K
T
T
300 K

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