1.
22
Heat Engines, Entropy, and the Second
Law of Thermodynamics
ANSWERS TO QUESTIONS
CHAPTER OUTLINE
22.1
Heat Engines and the Second
Law of Thermodynamics
Heat Pumps and Refrigerators
Reversible and Irreversible
Processes
The Carnot Engine
Gasoline and Diesel Engines
Entropy
Entropy Changes in Irreversible
Processes
Entropy on a Microscopic Scale
22.4
22.5
22.6
22.7
22.8
Q22.1
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.
*Q22.2
22.2
22.3
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, vi, vii but not
by processes i, iii, viii. 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 processes v, vii, and viii. Thus the answers are
(i) b (ii) a (iii) b (iv) a (v) c (vi) a (vii) c
(viii) d.
Q22.3
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
Q22.4
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.
Q22.5
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 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 heat. 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 heat in the atmosphere.
571
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2.
572
Chapter 22
*Q22.6
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. The points O and B are on a hyperbolic isotherm. The points O
and A are on an adiabat, steeper than an isotherm by the factor γ.
Q22.7
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. The free ﬂight of a projectile is nearly reversible.
Q22.8
(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.
*Q22.9
(i)
Answer (a). The air conditioner operating in a closed room takes in energy by electric transmission and turns it all into energy put out by heat. That is its whole net effect.
(ii) Answer (b). The frozen stuff absorbs energy by heat from the air. But if you ﬁll the ice
trays with tap water and put them back into the freezer, the refrigerator will pump more
heat into the air than it extracts from the water to make it freeze.
*Q22.10 (i) Answer (d). (ii) 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.
Q22.11
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.
Q22.12 (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
⎛V ⎞
∆Q
entropy change is ∆ S =
= nR ln ⎜ 2 ⎟
T
⎝ V1 ⎠
(b)
For a reversible adiabatic expansion ∆Q = 0, and ∆S = 0. An ideal gas undergoing an irreversible adiabatic expansion can have any positive value for ∆S up to the value given in part (a).
*Q22.13 Answer (f). 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 Jր K,
or more.
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3.
Heat Engines, Entropy, and the Second Law of Thermodynamics
*Q22.14 (i)
(ii)
573
Consider the area that ﬁts under each of the arrows, between its line segment and the
horizontal axis. Count it as positive for arrows to the right, zero for vertical arrows, and
negative for arrows tending left. Then E > F > G > H = D > A > B > C.
The thin blue hyperbolic lines are isotherms. Each is a set of points representing states with
the same internal energy for the ideal gas simple. An arrow tending farther from the origin
than the BE hyperbola represents a process for which internal energy increases. So we have
D = E > C > B = F > G > A = H.
(iii) The arrows C and G are along an adiabat. Visualize or sketch in a set of these curves, uniformly steeper than the blue isotherms. The energy input by heat is determined by how far
above the starting adiabat the process arrow ends. We have E > D > F > C = G > B > H > A.
*Q22.15 Processes C and G are adiabatic. They can be carried out reversibly. Along these arrows entropy
does not change. Visualize or sketch in a set of these adiabatic curves, uniformly steeper than the
blue isotherms. The entropy change is determined by how far above the starting adiabat the process arrow ends. We have E > D > F > C = G > B > H > A.
*Q22.16 (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 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.
*Q22.17 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,
lower-entropy 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 more disordered, higher-entropy ﬁnal state.
Q22.18 An analogy used by Carnot is instructive: A waterfall continuously converts mechanical energy
into internal energy. It continuously creates entropy as the organized motion of the falling water
turns into disorganized molecular motion. 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 lowtemperature sink (outer space). A tree builds organized 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 crashing down to disorder. 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. Your roommate’s exercise puts energy into the room by heat.
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4.
574
Chapter 22
Q22.19 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 which
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. 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. Arnold Sommerfeld suggested the idea for this question.
Q22.20 Shaking opens up spaces between jellybeans. The smaller ones more often can fall down into
spaces below them. The accumulation of larger candies on top and smaller ones on the bottom
implies a small increase in order, 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 bit of gravitational energy loss as the beans settle compactly together.
SOLUTIONS TO PROBLEMS
Section 22.1
Heat Engines and the Second Law of Thermodynamics
Weng
*P22.2
e=
Qc = Qh − Weng = 360 J − 25.0 J = 335 J
Qh
=
25.0 J
= 0.069 4 or 6.94%
360 J
(a)
(b)
P22.1
The engine’s output work we identify with the kinetic energy of the bullet:
Weng = K =
e=
1
1
2
mv 2 = 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 = mc∆T
T
∆T =
P22.3
(a)
We have e =
Weng
Qh
=
Q 1.10 × 10 4 J kg°C
=
= 13.7°C
mc
1.80 kg 448 J
Qh − Qc
Q
= 1 − c = 0.250
Qh
Qh
with Qc = 8 000 J, we have Qh = 10.7 kJ
(b)
Weng = Qh − Qc = 2 667 J
and from P =
13794_22_ch22_p571-600.indd 574
Weng
∆t
, we have ∆t =
Weng
P
=
2 667 J
= 0.533 s
5 000 J s
1/8/07 7:53:27 PM
5.
Heat Engines, Entropy, and the Second Law of Thermodynamics
P22.4
(a)
575
The input energy each hour is
J revolution ) ( 2 500 rev min )
60 min
= 1.18 × 10 9 J h
1h
1L
⎛
⎞ = 29.4 L h
implying fuel input (1.18 × 10 9 J h )
⎝ 4.03 × 10 7 J ⎠
( 7.89 × 10
(b)
3
Qh = Weng + Qc . For a continuous-transfer process we may divide by time to have
Qh Weng Qc
=
+
∆t
∆t
∆t
Useful power output =
Weng
∆t
=
Qh Qc
−
∆t ∆t
⎛ 7.89 × 10 3 J 4.58 × 10 3 J ⎞ 2 500 rev 1 min
=⎜
−
= 1.38 × 10 5 W
⎝ revolution
⎠
revolution ⎟ 1 min 60 s
e
P
eng
P
⎛ 1 hp ⎞
= 1.38 × 10 5 W ⎜
= 185 hp
⎝ 746 W ⎟
⎠
1.38 × 10 5 J s ⎛ 1 rev ⎞
= 527 N ⋅ m
⎝
⎠
( 2 500 rev 60 s ) ⎜ 2π rad ⎟
(c)
P
(d)
Qc 4.58 × 10 J ⎛ 2 500 rev ⎞
=
= 1.91 × 10 5 W
∆t
revolution ⎝ 60 s ⎠
eng
= τω ⇒ τ =
eng
ω
=
3
P22.5
The heat to melt 15.0 g of Hg is
Qc = mL f = (15 × 10 −3 kg ) (1.18 × 10 4 J kg ) = 177 J
The energy absorbed to freeze 1.00 g of aluminum is
Qh = mL f = (10 −3 kg ) ( 3.97 × 10 5 J/ kg ) = 397 J
Weng = Qh − Qc = 220 J
and the work output is
e=
The theoretical (Carnot) efﬁciency is
Section 22.2
P22.6
Weng
Qh
=
220 J
= 0.554, or 55.4%
397 J
Th − Tc 933 K − 243.1 K
=
= 0.749 = 74.9%
Th
933 K
Heat Pumps and Refrigerators
COP ( refrigerator ) =
Qc
W
(a)
If Qc = 120 J and COP = 5.00, then W = 24.0 J
(b)
Heat expelled = Heat removed + Work done.
Qh = Qc + W = 120 J + 24 J = 144 J
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6.
576
P22.7
Chapter 22
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 ) + ( 0.030 0 kg ) ( 3.33 × 10 5 J kg )
t
+ ( 0.030 0 kg ) ( 2 090 J kg °C ) ( 20.0°C ) = 1.40 × 10 4 J min
Qc
= 233 J s
t
or,
Thus, the work done per second is
P
=
233 J s
= 77.8 W
3.00
(a)
⎛ 10.0 Btu ⎞ ⎛ 1055 J ⎞ ⎛ 1 h ⎞ ⎛ 1 W ⎞ = 2.93
⎝
⎠
⎠⎝
⎝
h ⋅ W ⎠ ⎝ 1 Btu ⎠ ⎜ 3 600 s ⎟ ⎜ 1 J s ⎟
(b)
P22.8
The energy extracted by heat from the cold side divided by required work input is by
( COP )refrigerator
deﬁnition the coefﬁcient of performance for a refrigerator:
(c)
With EER 5, 5
Btu 10 000 Btu h
:
=
P
h⋅W
Energy purchased is
P = 10 000
Btu h
= 2 000 W = 2.00 kW
5 Btu h ⋅ W
P ∆t = ( 2.00 kW )(1 500 h ) = 3.00 × 10
3
kWh
Cost = ( 3.00 × 10 3 kWh ) ( 0.100 $ kWh ) = $300
With EER 10, 10
Btu 10 000 Btu h
=
:
h⋅W
P
Energy purchased is
P = 10 000
Btu h
= 1 000 W = 1.00 kW
10 Btu h ⋅ W
P ∆t = (1.00 kW )(1 500 h ) = 1.50 × 10
3
kWh
Cost = (1.50 × 10 3 kWh ) ( 0.100 $ kWh ) = $150
Thus, the cost for air conditioning is half as much for an air conditioner with EER 10
compared with an air conditioner with EER 5.
Section 22.3
Reversible and Irreversible Processes
Section 22.4
The Carnot Engine
P22.9
Tc = 703 K
Th = 2 143 K
∆T 1 440
=
= 67.2%
Th
2 143
(a)
ec =
(b)
Qh = 1.40 × 10 5 J, Weng = 0.420 Qh
P
13794_22_ch22_p571-600.indd 576
=
Weng
∆t
=
5.88 × 10 4 J
= 58.8 kW
1s
1/8/07 7:53:28 PM
7.
Heat Engines, Entropy, and the Second Law of Thermodynamics
P22.10
When e = ec, 1−
(a)
Qh =
(W
Tc Weng
=
Th
Qh
eng
)
ր ∆t ∆t
1 − (Tc րTh )
=
Weng ր ∆t
and
(1.50 × 10
Qh ր ∆t
5
= 1−
577
Tc
Th
W ) ( 3 600 s )
1 − 293ր773
Qh = 8.70 × 108 J = 870 MJ
(b)
⎛ Weng ⎞
Qc = Qh − ⎜
∆t = 8.70 × 108 − (1.50 × 10 5 ) ( 3 600 ) = 3.30 × 108 J = 330 MJ
⎝ ∆t ⎟
⎠
*P22.11 We use amounts of energy to ﬁnd the actual efﬁciency. Qh = Qc + Weng = 20 kJ + 1.5 kJ = 21.5 kJ
e = WengրQh = 1.5 kJր21.5 kJ = 0.0698
We use temperatures to ﬁnd the Carnot efﬁciency of a reversible engine eC = 1 − TcրTh =
1 – 373 Kր453 K = 0.177
The actual efficiency of 0.0698 is less than four-tenths of the Carnot efficiency of 0.177.
n
*P22.12 (a)
(b)
eC = 1 − TcրTh = 1 − 350ր500 = 0.300
In eC = 1 − TcրTh we differentiate to ﬁnd deC րdTh = 0 – Tc(−1)Th–2 = TcրTh2 = 350ր5002 =
1.40 × 10 −3 This is the increase of efﬁciency per degree of increase in the temperature of the
hot reservoir.
(c) In eC = 1 − TcրTh we differentiate to ﬁnd deC րdTc = 0 − 1րTh = −1ր500 = –2.00 × 10–3
Then deC ր(−dTc) = +2.00 × 10 −3 This is the increase of efﬁciency per degree of decrease
in the temperature of the cold reservoir. Note that it is a better deal to cool the exhaust than
to supercharge the ﬁrebox.
Isothermal expansion at
Th = 523 K
Isothermal compression at
P22.13
Tc = 323 K
Gas absorbs 1 200 J during expansion.
(a)
⎛T ⎞
323 ⎞
Qc = Qh ⎜ c ⎟ = 1 200 J ⎛
= 741 J
⎝ 523 ⎠
⎝ Th ⎠
(b)
Weng = Qh − Qc = (1 200 − 741) J = 459 J
ec ,s = 1 −
Tc
( 273 + 20 ) K
= 1−
= 0.530
Th
( 273 + 350 ) K
ec ,w = 1 −
283
= 0.546
623
Then the actual winter efﬁciency is
13794_22_ch22_p571-600.indd 577
The Carnot summer efﬁciency is
And in winter,
P22.14
0.546 ⎞
0.320 ⎛
= 0.330
⎝ 0.530 ⎠
or 33.0%
1/8/07 7:53:29 PM
8.
578
Chapter 22
γ
P22.15
(a)
⎛ Pf V f ⎞
⎛ PVi ⎞
i
In an adiabatic process, Pf V fγ = PViγ . Also, ⎜
i
⎟ =⎜ T ⎟
Tf ⎠
⎝ i ⎠
⎝
⎛ Pf ⎞
Dividing the second equation by the ﬁrst yields T f = Ti ⎜ ⎟
⎝ Pi ⎠
γ −1 2
5
Since γ = for Argon,
= = 0.400 and we have
γ
5
3
⎛ 300 × 10 3 Pa ⎞
T f = (1 073 K ) ⎜
⎝ 1.50 × 10 6 Pa ⎟
⎠
(b)
= 564 K
− nCV ∆T
=
or
t
t
( −80.0 kg)(1 mol/0.0399 kg)( 3 )(8.314 J mol ⋅ K )( 564 − 1 073) K
2
=
60.0 s
= 2.12 × 10 5 W = 212 kW
Tc
564 K
= 1−
= 0.475 or 47.5%
Th
1 073 K
(c)
eC = 1 −
(a)
emax = 1 −
(b)
P
=
Tc
278
= 1−
= 5.12 × 10 −2 = 5.12%
Th
293
Weng
= 75.0 × 10 6 J s
∆t
Weng = ( 75.0 × 10 6 J s ) ( 3 600 s h ) = 2.70 × 1011 J h
Therefore,
From e =
*P22.17 (a)
0. 400
Weng
=
P
(c)
(γ −1) γ
∆Eint = nCV ∆T = Q − Weng = 0 − Weng, so Weng = − nCV ∆T ,
and the power output is
P
P22.16
γ
Weng
Qh
we ﬁnd
Qh =
Weng
e
=
2.70 × 1011 J h
= 5.27 × 1012 J h = 5.27 TJ h
5.12 × 10 −2
As fossil-fuel prices rise, this way to use solar energy will become a good buy.
e=
Weng1 + Weng2
Q1h
=
e1Q1h + e2Q2 h
Q1h
Now
Q2 h = Q1c = Q1h − Weng1 = Q1h − e1Q1h
So
e=
e1Q1h + e2 (Q1h − e1Q1h )
= e1 + e2 − e1e2
Q1h
continued on next page
13794_22_ch22_p571-600.indd 578
1/8/07 7:53:31 PM
9.
Heat Engines, Entropy, and the Second Law of Thermodynamics
(b)
e = e1 + e2 − e1e2 = 1 −
Ti
T
+1− c
Th
Ti
= 2−
579
⎛ T ⎞⎛ T ⎞
− ⎜1− i ⎟ ⎜1− c ⎟
⎝ 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.
(c)
Weng1 + Weng2
With Weng2 = Weng1, e =
1−
2Weng1
Q1h
= 2e1
⎛ T ⎞
Tc
= 2 ⎜1 − i ⎟
Th
⎝ Th ⎠
0−
Q1h
=
2T
Tc
= 1− i
Th
Th
2Ti = Th + Tc
Ti =
(d)
1
(Th + Tc )
2
Ti
T
= 1− c
Th
Ti
e1 = e2 = 1 −
Ti 2 = TcTh
Ti = (ThTc )
12
*P22.18 (a)
“The actual efﬁciency is two thirds the Carnot efﬁciency” reads as an equation
Weng
Qh
=
Weng
Qc + Weng
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
0.5 Th + Tc
Th − Tc
Qc
Weng
Qc
∆t
=
1.5 Th
1.5 Th − Th + Tc
−1=
Th − Tc
Th − Tc
=
Weng 0.5 Th + Tc
0.5 Th + 383 K
= 1.40 MW
∆t Th − Tc
Th − 383 K
The dominating Th in the bottom of this fraction means that the exhaust power decreases
as the f irebox temperature increases.
(b)
Qc
0.5 Th + 383 K
0.5(1073 K) + 383 K
(
= 1.40 MW
= 1.40 MW
= 1.87 MW
∆t
(1073 − 383) K
Th − 383 K
(c)
We require
Qc 1
0.5 Th + 383 K
= 2 1.87 MW = 1.40 MW
Th − 383 K
∆t
6
0.5 Th + 383 K = 0.666Th − 255 K
(d)
13794_22_ch22_p571-600.indd 579
0.5 Th + 383 K
= 0.666
Th − 383 K
Th = 638 K/0.166 = 3.84 × 10 3 K
The minimum possible heat exhaust power is approached as the ﬁrebox temperature goes to
inﬁnity, and it is |Qc|ր∆t = 1.40 MW( 0.5ր1) = 0.7 MW. The heat exhaust power cannot be as
small as (1ր4)(1.87 MW) = 0.466 MW. So
no answer exists. The energy exhaust cannot be that small.
1/8/07 7:53:32 PM
10.
580
Chapter 22
P22.19
P22.20
(a)
Tc
270
=
= 9.00
∆T 30.0
( COP )refrig =
First, consider the adiabatic process D → A:
γ
γ
γ
PDVD = PAVA
so
53
⎛V ⎞
10.0 L ⎞
PD = PA ⎜ A ⎟ = 1 400 kPa ⎛
= 712 kPa
⎝ 15.0 L ⎠
⎝ VD ⎠
Also
⎛ nRTD ⎞ γ ⎛ nRTA ⎞ γ
⎜ V ⎟ VD = ⎜ V ⎟ VA
⎝ D ⎠
⎝ A ⎠
or
⎛V ⎞
TD = TA ⎜ A ⎟
⎝V ⎠
γ −1
D
10.0 ⎞
= 720 K ⎛
⎝ 15.0 ⎠
23
= 549 K
Now, consider the isothermal process C → D: TC = TD = 549 K
γ
⎛V ⎞ ⎡ ⎛ V ⎞ ⎤⎛V ⎞
P Vγ
PC = PD ⎜ D ⎟ = ⎢ PA ⎜ A ⎟ ⎥ ⎜ D ⎟ = A γA−1
⎝ VC ⎠ ⎢ ⎝ VD ⎠ ⎥ ⎝ VC ⎠ VCVD
⎣
⎦
1 400 kPa (10.0 L )
= 445 kPa
23
24.0 L (15.0 L )
53
PC =
γ
Next, consider the adiabatic process B → C: PBVBγ = PCVC
But, PC =
γ
⎛V ⎞
PAVA
from above. Also considering the isothermal process, PB = PA ⎜ A ⎟
γ −1
VCVD
⎝ VB ⎠
⎛V ⎞ γ ⎛ P Vγ ⎞ γ
VV
10.0 L ( 24.0 L )
Hence, PA ⎜ A ⎟ VB = ⎜ A γA−1 ⎟ VC which reduces to VB = A C =
= 16.0 L
VD
15.0 L
⎝ VB ⎠
⎝ VCVD ⎠
⎛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
continued on next page
13794_22_ch22_p571-600.indd 580
1/8/07 7:53:33 PM
11.
Heat Engines, Entropy, and the Second Law of Thermodynamics
(b)
For the isothermal process A → B:
581
∆Eint = nCV ∆T = 0
⎛V ⎞
⎛ 16.0 ⎞ = +6.58 kJ
so Q = −W = nRT ln ⎜ B ⎟ = 2.34 mol (8.314 J mol ⋅ K ) ( 720 K ) ln
⎝ 10.0 ⎠
⎝ VA ⎠
For the adiabatic process B → C :
Q= 0
3
∆Eint = nCV (TC − TB ) = 2.34 mol ⎡ (8.314 J mol ⋅ K ) ⎤ ( 549 − 720 ) K = −4.98 kJ
⎢2
⎥
⎦
⎣
and W = −Q + ∆Eint = 0 + ( −4.98 kJ ) = −4.98 kJ
For the isothermal process C → D:
∆Eint = nCV ∆T = 0
⎛V ⎞
15.0 ⎞
and Q = −W = nRT ln ⎜ D ⎟ = 2.34 mol (8.314 J mol ⋅ K ) ( 549 K ) ln ⎛
= −5.02 kJ
⎝ 24.0 ⎠
⎝ VC ⎠
Finally, for the adiabatic process D → A:
Q= 0
3
∆Eint = nCV (TA − TD ) = 2.34 mol ⎡ (8.314 J mol ⋅ K ) ⎤ ( 720 − 549 ) K = +4.98 kJ
⎢2
⎥
⎦
⎣
and W = −Q + ∆Eint = 0 + 4.98 kJ = +4.98 kJ
Process
Q (kJ)
W (kJ)
∆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
The work done by the engine is the negative of the work input. The output work Weng is
given by the work column in the table with all signs reversed.
(c)
e=
Weng
Qh
ec = 1 −
P22.21
(a)
=
−WABCD 1.56 kJ
=
= 0.237 or 23.7%
QA→ B
6.58 kJ
Tc
549
= 1−
= 0.237 or 23.7%
Th
720
For a complete cycle, ∆Eint = 0
and
⎡ (Qh )
⎤
W = Qh − Qc = Qc ⎢
− 1⎥
⎢ Qc
⎥
⎣
⎦
The text shows that for a Carnot cycle (and only for a reversible cycle)
Qh Th
=
Qc Tc
Therefore,
⎡ T − Tc ⎤
W = Qc ⎢ h
⎥
⎣ Tc ⎦
(b)
We have the deﬁnition of the coefﬁcient of performance for a refrigerator,
Using the result from part (a), this becomes
13794_22_ch22_p571-600.indd 581
COP =
Tc
Th − Tc
COP =
Qc
W
1/8/07 7:53:33 PM
12.
582
Chapter 22
Qc + W
T
295
= h =
= 11.8
W
∆T
25
P22.22
( COP )heat pump =
P22.23
( COP )Carnot refrig =
Q
4.00
Tc
=
= 0.013 8 = c
∆T 289
W
∴ W = 72.2 J per 1 J energy removed by heat.
P22.24
COP = 0.100 COPCarnot cycle
or
⎛Q ⎞
Qh
⎛
1
⎞
= 0.100 ⎜ h ⎟
= 0.100 ⎜
⎝ Carnot efficiency ⎟
⎠
W
⎝ W ⎠ Carnot cycle
Qh
⎛ Th ⎞
293 K
⎛
⎞
= 0.100 ⎜
⎟ = 0.100 ⎝ 293 K − 268 K ⎠ = 1.17
W
⎝ Th − Tc ⎠
FIG. P22.24
Thus, 1.17 joules of energy enter the room by heat for each joule of work done.
t
*P22.25
Qc
Qc ր ∆t
Tc
= COPC ( refrigerator ) =
=
W
Th − Tc
W ր ∆t
0.150 W 260 K
=
W ր∆t
40.0 K
W
⎛ 40.0 K ⎞
= 0.150 W ⎜
= 23.1 mW
⎝ 260 K ⎟
⎠
∆t
P
P22.26
=
e=
W
= 0.350
Qh
W = 0.350Qh
Qh = W + Qc
Qc = 0.650Qh
COP ( refrigerator ) =
Qc 0.650Qh
=
= 1.86
W 0.350Qh
Section 22.5
P22.27
(a)
Gasoline and Diesel Engines
PViγ = Pf V fγ
i
γ
1.40
⎛V ⎞
⎛ 50.0 cm 3 ⎞
Pf = Pi ⎜ i ⎟ = ( 3.00 × 10 6 Pa ) ⎜
= 244 kPa
⎝ 300 cm 3 ⎟
⎠
⎝ Vf ⎠
Vi
(b)
W = ∫ PdV
Vi
V
P = Pi ⎛ i ⎞
⎝V⎠
γ
Integrating,
⎡ ⎛ V ⎞ γ −1 ⎤
⎛ 1 ⎞
W =⎜
PVi ⎢1 − ⎜ i ⎟ ⎥
i
⎝ γ − 1⎟
⎠
⎢ ⎝ Vf ⎠ ⎥
⎦
⎣
⎡ ⎛ 50.0 cm 3 ⎞ 0.400 ⎤
= ( 2.50 ) ( 3.00 × 10 6 Pa ) ( 5.00 × 10 −5 m 3 ) ⎢1 − ⎜
⎥
3 ⎟
⎢
⎥
⎣ ⎝ 300 cm ⎠
⎦
= 192 J
13794_22_ch22_p571-600.indd 582
1/8/07 7:53:35 PM
13.
Heat Engines, Entropy, and the Second Law of Thermodynamics
P22.28
583
(a), (b) The quantity of gas is
n=
3
−6
3
PAVA (100 × 10 Pa ) ( 500 × 10 m )
=
= 0.020 5 mol
RTA
(8.314 J mol ⋅ K ) ( 293 K )
Eint, A =
5
5
5
nRTA = PAVA = (100 × 10 3 Pa ) ( 500 × 10 −6 m 3 ) = 125 J
2
2
2
γ
⎛V ⎞
1.40
In process AB, PB = PA ⎜ A ⎟ = (100 × 10 3 Pa ) (8.00 ) = 1.84 × 10 6 Pa
⎝V ⎠
B
TB =
6
−6
3
PBVB (1.84 × 10 Pa ) ( 500 × 10 m 8.00 )
=
= 673 K
nR
( 0.020 5 mol ) (8.314 J mol ⋅ K )
Eint, B =
so
5
5
nRTB = ( 0.020 5 mol ) (8.314 J mol ⋅ K ) ( 673 K ) = 287 J
2
2
∆Eint, AB = 287 J − 125 J = 162 J = Q − Wout = 0 − Wout
WAB = −162 J
Process BC takes us to:
PC =
nRTC ( 0.020 5 mol ) (8.314 J mol ⋅ K ) (1 023 K )
=
= 2.79 × 10 6 Pa
VC
62.5 × 10 −6 m 3
Eint, C =
5
5
nRTC = ( 0.020 5 mol ) (8.314 J mol ⋅ K ) (1 023 K ) = 436 J
2
2
Eint, BC = 436 J − 287 J = 149 J = Q − Wout = Q − 0
QBC = 149 J
In process CD:
γ
1.40
⎛V ⎞
1 ⎞
PD = PC ⎜ C ⎟ = ( 2.79 × 10 6 Pa ) ⎛
= 1.52 × 10 5 Pa
⎝ 8.00 ⎠
⎝ VD ⎠
TD =
5
−6
3
PDVD (1.52 × 10 Pa ) ( 500 × 10 m )
=
= 445 K
nR
( 0.020 5 mol ) (8.314 J mol ⋅ K )
Eint, D =
5
5
nRTD = ( 0.020 5 mol ) (8.314 J mol ⋅ K ) ( 445 K ) = 190 J
2
2
∆Eint, CD = 190 J − 436 J = −246 J = Q − Wout = 0 − Wout
WCD = 246 J
and
∆Eint, DA = Eint, A − Eint, D = 125 J − 190 J = −65.0 J = Q − Wout = Q − 0
QDA = −65.0 J
For the entire cycle, ∆Eint, net = 162 J + 149 − 246 − 65.0 = 0 . The net work is
Weng = −162 J + 0 + 246 J + 0 = 84.3 J
Qnet = 0 + 149 J + 0 − 65.0 J = 84.3 J
continued on next page
13794_22_ch22_p571-600.indd 583
1/8/07 7:53:36 PM
14.
584
Chapter 22
The tables look like:
State
T (K)
P (kPa)
V (cm3)
Eint (J)
A
293
100
500
125
B
673
1 840
62.5
287
C
1 023
2 790
62.5
436
D
445
152
500
190
A
293
100
500
125
Process
Q (J)
output W (J)
∆Eint (J)
AB
0
−162
162
BC
149
0
149
CD
0
246
−246
DA
−65.0
0
84.3
84.3
ABCDA
−65.0
0
(c)
The input energy is Qh = 149 J , the waste is Qc = 65.0 J , and Weng = 84.3 J .
(d)
The efﬁciency is: e =
(e)
f
Let f represent the angular speed of the crankshaft. Then is the frequency at which we
2
obtain work in the amount of 84.3 Jրcycle:
Weng
Qh
=
84.3 J
= 0.565
149 J
f
1 000 J s = ⎛ ⎞ (84.3 J cycle )
⎝ 2⎠
f =
P22.29
2 000 J s
= 23.7 rev s = 1.42 × 10 3 rev min
84.3 J cycle
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 ⎠
(b)
Section 22.6
P22.30
0.400
= 51.2%
If actual efﬁciency e′ = 15.0% losses in system are e − e′ = 36.2%
Entropy
For a freezing process,
∆S =
5
∆Q − ( 0.500 kg ) ( 3.33 × 10 J kg )
=
= −610 J K
T
273 K
*P22.31 The process of raising the temperature of the sample in this way is reversible, because an inﬁnitesimal change would make δ negative, and energy would ﬂow out instead of in. Then we may
ﬁnd the entropy change of the sample as
Tf
Tf
Ti
Ti
∆S = ∫ dS = ∫
13794_22_ch22_p571-600.indd 584
dQ T f mcdT
=
= mc ln T
T ∫Ti T
Tf
Ti
= mc ⎡ ln T f − ln Ti ⎤ = mc ln(T f րTi )
⎣
⎦
1/8/07 7:53:37 PM
15.
Heat Engines, Entropy, and the Second Law of Thermodynamics
*P22.32 (a)
(b)
585
The process is isobaric because it takes place under constant atmospheric pressure. As
described by Newton’s third law, the stewing syrup must exert the same force on the air
as the air exerts on it. The heating process is not adiabatic (because energy goes in by
heat), isothermal (T goes up), isovolumetric (it likely expands a bit), cyclic (it is different at the end), or isentropic (entropy increases). It could be made as nearly reversible
as you wish, by not using a kitchen stove but a heater kept always just incrementally
higher in temperature than the syrup. The process would then also be eternal, and
impractical for food production.
The ﬁnal temperature is
100 − 0°C ⎞
220°F = 212°F + 8°F = 100°C + 8°F ⎛
= 104°C
⎝ 212 − 32°F ⎠
For the mixture,
Q = m1c1∆T + m2 c2 ∆T
= ( 900 g 1 cal g ⋅ °C + 930 g 0.299 cal g ⋅ °C ) (104.4°C − 23°C )
= 9.59 × 10 4 cal = 4.02 × 10 5 J
(c)
Consider the reversible heating process described in part (a):
f
∆S =
dQ
∫i T =
f
∫
( m1c1 + m2 c2 ) dT
T
i
= ( m1c1 + m2 c2 ) ln
Tf
Ti
4.186 J ⎞ ⎛ 1°C ⎞ ⎛ 273 + 104 ⎞
= [ 900 (1) + 930 ( 0.299 )] ( cal °C ) ⎛
ln
⎝ 1 cal ⎠ ⎝ 1 K ⎠ ⎝ 273 + 23 ⎠
= ( 4 930 J K ) 0.243 = 1.20 × 10 3 J K
f
P22.33
∆S =
dQ
∫i T =
Tf
∫
Ti
⎛ Tf ⎞
mcdT
= mc ln ⎜ ⎟
T
⎝ Ti ⎠
353 ⎞
∆S = 250 g (1.00 cal g ⋅ °C ) ln ⎛
= 46.6 cal K = 195 J K
⎝ 293 ⎠
Section 22.7
Entropy Changes in Irreversible Processes
Q2 Q1 ⎛ 1 000 1 000 ⎞
−
=
−
J K = 3.27 J K
⎝
⎠
T2 T1 ⎜ 290 5 700 ⎟
P22.34
∆S =
P22.35
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
∆S =
13794_22_ch22_p571-600.indd 585
1
2
mv 2 750 ( 20.0 )
=
T
293
2
J K = 1.02 kJ K
1/8/07 7:53:38 PM
16.
586
Chapter 22
*P22.36 Deﬁne T1 = Temp Cream = 5.00°C = 278 K. Deﬁne T2 = Temp Coffee = 60.0°C = 333 K
(20.0 g)T1 + (200 g)T2
= 55.0°C = 328 K
220 g
The ﬁnal temperature of the mixture is:
Tf =
The entropy change due to this mixing is
∆S = ( 20.0 g ) ∫
Tf
T1
T f c dT
cV dT
+ ( 200 g ) ∫ V
T2
T
T
⎛ Tf ⎞
⎛ Tf ⎞
328 ⎞
328 ⎞
∆S = (84.0 J K ) ln ⎜ ⎟ + (840 J K ) ln ⎜ ⎟ = (84.0 K J ) ln ⎛
+ (840 J K ) ln ⎛
⎝ 333 ⎠
⎝ 278 ⎠
⎝ T1 ⎠
⎝ T2 ⎠
∆S = +1.18 J K
P22.37
Sitting here writing, I convert chemical energy, in ordered molecules in food, into internal energy
that leaves my body by heat into the room-temperature surroundings. My rate of energy output is
equal to my metabolic rate,
2 500 kcal d =
2 500 × 10 3 cal ⎛ 4.186 J ⎞
= 120 W
86 400 s ⎝ 1 cal ⎠
My body is in steady state, changing little in entropy, as the environment increases in entropy at
the rate
∆S Q T Q ∆t 120 W
=
=
=
= 0.4 W K ~ 1 W K
∆t
∆t
T
293 K
When using powerful appliances or an automobile, my personal contribution to entropy production is much greater than the above estimate, based only on metabolism.
P22.38
ciron = 448 J kg ⋅ °C ;
cwater = 4 186 J kg ⋅ °C
)
(
(
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
)
T f = 33.2°C = 306.2 K
306.2 K
∆S =
∫
283 K
306.2 K
cwater mwater dT
c m dT
+ ∫ iron iron
T
T
1173 K
⎛ 306.2 ⎞
306.2 ⎞
+ c m ln
∆S = cwater mwater ln ⎛
⎝ 283 ⎠ iron iron ⎜ 1173 ⎟
⎠
⎝
∆S = ( 4 186 J kg ⋅ K ) ( 4.00 kg ) ( 0.078 8 ) + ( 448 J kg ⋅ K ) (1.00 kg ) ( −1.34 )
∆S = 718 J K
P22.39
⎛ Vf ⎞
∆S = nR ln ⎜ ⎟ = R ln 2 = 5.76 J K
⎝ Vi ⎠
There is no change in temperature for an ideal gas.
FIG. P22.39
13794_22_ch22_p571-600.indd 586
1/8/07 7:53:38 PM
17.
Heat Engines, Entropy, and the Second Law of Thermodynamics
P22.40
587
⎛ Vf ⎞
∆S = nR ln ⎜ ⎟ = ( 0.044 0 ) ( 2 ) R ln 2
⎝ Vi ⎠
∆S = 0.088 0 (8.314 ) ln 2 = 0.507 J K
FIG. P22.40
Section 22.8
Entropy on a Microscopic Scale
P22.42
(a)
A 12 can only be obtained one way, as 6 + 6
(b)
P22.41
A 7 can be obtained six ways: 6 + 1, 5 + 2, 4 + 3, 3 + 4, 2 + 5, 1 + 6
(a)
The table is shown below. On the basis of the table, the most probable recorded result of a
toss is 2 heads and 2 tails .
(b)
The most ordered state is the least likely macrostate. Thus, on the basis of the table this is
either all heads or all tails .
(c)
The most disordered is the most likely macrostate. Thus, this is 2 heads and 2 tails .
Result
All heads
3H, 1T
2H, 2T
1H, 3T
All tails
Total
1
4
6
4
1
(a)
Result
All red
2R, 1G
1R, 2G
All green
Possible Combinations
RRR
RRG, RGR, GRR
RGG, GRG, GGR
GGG
Total
1
3
3
1
(b)
P22.43
Possible Combinations
HHHH
THHH, HTHH, HHTH, HHHT
TTHH, THTH, THHT, HTTH, HTHT, HHTT
HTTT, THTT, TTHT, TTTH
TTTT
Result
All red
4R, 1G
3R, 2G
2R, 3G
1R, 4G
All green
13794_22_ch22_p571-600.indd 587
Possible Combinations
RRRRR
RRRRG, RRRGR, RRGRR, RGRRR, GRRRR
RRRGG, RRGRG, RGRRG, GRRRG, RRGGR,
RGRGR, GRRGR, RGGRR, GRGRR, GGRRR
GGGRR, GGRGR, GRGGR, RGGGR, GGRRG,
GRGRG, RGGRG, GRRGG, RGRGG, RRGGG
RGGGG, GRGGG, GGRGG, GGGRG, GGGGR
GGGGG
Total
1
5
10
10
5
1
1/8/07 7:53:40 PM
18.
588
Chapter 22
Additional Problems
P22.44
The conversion of gravitational potential energy into kinetic energy as the water falls is reversible. But the subsequent conversion into internal energy is not. We imagine arriving at the same
ﬁnal state by adding energy by heat, in amount mgy, to the water from a stove at a temperature
inﬁnitesimally above 20.0°C. Then,
3
3
2
dQ Q mgy 5 000 m (1 000 kg m ) ( 9.80 m s ) ( 50.0 m )
= =
=
= 8.36 × 10 6 J K
∫T T T
293 K
T
300 K
*P22.45 For the Carnot engine, ec = 1 − c = 1 −
= 0.600
Th
750 K
∆S =
Weng
Also,
ec =
so
Qh =
and
Qc = Qh − Weng = 250 J − 150 J = 100 J
(a)
Qh =
Weng
eS
=
Qh
Weng
ec
=
150 J
= 250 J
0.600
FIG. P22.45
150 J
= 214 J
0.700
Qc = Qh − Weng = 214 J − 150 J = 64.3 J
(b)
Qh,net = 214 J − 250 J = −35.7 J
Qc,net = 64.3 J − 100 J = −35.7 J
The net ﬂow of energy by heat from the cold to the
hot reservoir without work input, is impossible.
For engine S:
Qc = Qh − Weng =
so
(c)
Weng =
and
(d)
Weng
eS
− Weng
Qc
100 J
= 1
= 233 J
− 1 0.700 − 1
FIG. P22.45(b)
1
eS
Qh = Qc + Weng = 233 J + 100 J = 333 J
Qh,net = 333 J − 250 J = 83.3 J
Wnet = 233 J − 150 J = 83.3 J
Qc,net = 0
The output of 83.3 J of energy from the
heat engine by work in a cyclic process
without any exhaust by heat is impossible.
FIG. P22.45(d)
continued on next page
13794_22_ch22_p571-600.indd 588
1/8/07 7:53:42 PM
19.
Heat Engines, Entropy, and the Second Law of Thermodynamics
Both engines operate in cycles, so
∆SS = ∆SCarnot = 0
For the reservoirs,
(e)
∆Sh = −
589
Qh
Q
and ∆Sc = + c
Th
Tc
Thus,
∆Stotal = ∆SS + ∆SCarnot + ∆Sh + ∆Sc = 0 + 0 −
83.3 J
0
+
= −0.111 J K
750 K 300 K
A decrease in total entropy is impossible.
*P22.46 (a)
Let state i represent the gas before its compression and state f afterwards, V f =
diatomic ideal gas,
Cv =
5
7
R, C p = R,
2
2
γ =
and
Cp
CV
Vi
. For a
8
= 1.40. Next,
PViγ = Pf V fγ
i
γ
⎛V ⎞
Pf = Pi ⎜ i ⎟ = Pi 81.40 = 18.4 Pi
⎝ Vf ⎠
PiVi = nRTi
Pf V f =
so
T f = 2.30Ti
18.4 PVi
i
= 2.30 PVi = 2.30 nRTi = nRT f
i
8
(
)
5
5
5
∆Eint = nCV ∆T = n R T f − Ti = nR1.30Ti = 1.30 PiVi
2
2
2
5
= 1.30 (1.013 × 10 5 N m 2 ) 0.12 × 10 −3 m 3 = 39.4 J
2
Since the process is adiabatic,
(b)
Q=0
∆Eint = Q + W gives W = 39.4 J
and
1
1
2
MR 2 = 5.1 kg ( 0.085 m ) = 0.018 4 kg ⋅ m 2
2
2
We want the ﬂywheel to do work 39.4 J, so the work on the ﬂywheel should be −39.4 J:
The moment of inertia of the wheel is I =
K rot i + W = K rot f
1 2
Iω i − 39.4 J = 0
2
⎛ 2 ( 39.4 J ) ⎞
ωi = ⎜
⎝ 0.018 4 kg ⋅ m 2 ⎟
⎠
(c)
12
= 65.4 rad s
Now we want W = 0.05K rot i
1
39.4 J = 0.05 0.018 4 kg ⋅ m 2ω i2
2
⎛
2 ( 789 J ) ⎞
ω =⎜
⎝ 0.018 4 kg ⋅ m 2 ⎟
⎠
13794_22_ch22_p571-600.indd 589
12
= 293 rad s
1/8/07 7:53:43 PM
20.
590
P22.47
Chapter 22
(a)
H ET
so if all the electric energy is converted into internal energy, the steady-state
∆t
condition of the house is described by H ET = Q .
P
electric
=
Therefore,
P
(b)
For a heat pump,
Q
= 5 000 W
∆t
T
295 K
= h =
= 10.92
27 K
∆T
electric
( COP )Carnot
=
Qh
Q ∆t
= h
W
W ∆t
Therefore, to bring 5 000 W of energy into the house only requires input power
Pheat pump = W = Qh ∆t = 5 000 W = 763 W
∆t
COP
6.56
−1 000 J
∆Shot =
600 K
+750 J
∆Scold =
350 K
Actual COP = 0.6 (10.92 ) = 6.55 =
P22.48
(a)
∆SU = ∆Shot + ∆Scold = 0.476 J K
(b)
ec = 1 −
T1
= 0.417
T2
Weng = ec Qh = 0.417 (1 000 J ) = 417 J
(c)
Wnet = 417 J − 250 J = 167 J
T1 ∆SU = 350 K ( 0.476 J K ) = 167 J
⎛V ⎞
Q = nRT ln ⎜ 2 ⎟
⎝ V1 ⎠
Q1 = nR ( 3Ti ) ln 2
and
(a)
For an isothermal process,
Therefore,
P22.49
1
Q3 = nR (Ti ) ln ⎛ ⎞
⎝ 2⎠
3
nR (Ti − 3Ti )
2
3
= nR ( 3Ti − Ti )
2
For the constant volume processes, Q2 = ∆Eint, 2 =
and
Q4 = ∆Eint, 4
The net energy by heat transferred is then
Q = Q1 + Q2 + Q3 + Q4
or
(b)
FIG. P22.49
Q = 2nRTi ln 2
A positive value for heat represents energy transferred into the system.
Therefore,
Qh = Q1 + Q4 = 3nRTi (1 + ln 2 )
Since the change in temperature for the complete cycle is zero,
∆Eint = 0 and Weng = Q
Therefore, the efﬁciency is
13794_22_ch22_p571-600.indd 590
ec =
Weng
Qh
=
Q
2 ln 2
=
= 0.273
Qh 3 (1 + ln 2 )
1/8/07 7:53:44 PM
21.
Heat Engines, Entropy, and the Second Law of Thermodynamics
P22.50
(a)
591
5
35.0°F = ( 35.0 − 32.0 ) °C = (1.67 + 273.15) K = 274.82 K
9
5
98.6°F = ( 98.6 − 32.0 ) °C = ( 37.0 + 273.15) K = 310.15 K
9
310.15
∆Sice water =
dQ
dT
⎛ 310.15 ⎞
∫ T = ( 453.6 g) (1.00 cal g ⋅ K ) × 274∫.82 T = 453.6 ln ⎝ 274.82 ⎠ = 54.86 cal K
∆Sbody = −
Q
( 310.15 − 274.82 )
= − ( 453.6 ) (1.00 )
= −51.67 cal K
310.15
Tbody
∆Ssystem = 54.86 − 51.67 = 3.19 cal K
(b)
( 453.6 ) (1) (TF − 274.82 ) = ( 70.0 × 10 3 ) (1) ( 310.15 − TF )
Thus,
( 70.0 + 0.453 6) × 103 TF = ⎡( 70.0 ) ( 310.15) + ( 0.453 6) ( 274.82)⎤ × 103
⎣
⎦
and TF = 309.92 K = 36.77°C = 98.19°F
309.92 ⎞
∆Sice water = 453.6 ln ⎛
= 54.52 cal K
′
⎝ 274.82 ⎠
310.15 ⎞
∆Sbody = − ( 70.0 × 10 3 ) ln ⎛
= −51.93 cal K
′
⎝ 309.92 ⎠
∆Ssys = 54.52 − 51.93 = 2.59 cal K
′
This is significantly less than the estimate in part (a).
e
P22.51
ec = 1 −
Tc Weng Weng / ∆t
:
=
=
Th
Qh
Qh / ∆t
Qh = Weng + Qc :
Qc = mc∆T :
Qh
P = P Th
=
∆t (1 − Tc Th ) Th − Tc
Weng
Qc
Q
= h −
∆t
∆t
∆t
Qc
P Th − P = P Tc
=
Th − Tc
∆t Th − Tc
Qc ⎛ ∆m ⎞
P Tc
c∆T =
=⎜
⎝ ∆t ⎟
⎠
Th − Tc
∆t
∆m
P Tc
=
∆t (Th − Tc ) c∆T
(1.00 × 109 W ) (300 K ) = 5.97 × 10 4 kg s
∆m
=
∆t 200 K ( 4 186 J kg ⋅ °C ) ( 6.00°C )
13794_22_ch22_p571-600.indd 591
1/8/07 7:53:46 PM
22.
592
P22.52
Chapter 22
ec = 1 −
Tc Weng Weng / ∆t
=
=
Th
Qh
Qh / ∆t
Qh
P
P Th
=
=
∆t 1 − (Tc / Th ) Th − Tc
Qc ⎛ Qh ⎞
P Tc
=
−P =
Th − Tc
∆t ⎜ ∆t ⎟
⎝
⎠
Qc = mc∆T , where c is the speciﬁc heat of water.
Therefore,
Qc ⎛ ∆m ⎞
P Tc
=⎜
⎟ c∆T =
Th − Tc
∆t ⎝ ∆t ⎠
and
∆m
=
∆t
PT
c
(Th − Tc ) c∆T
We test for dimensional correctness by identifying the units of the right-hand side:
W ⋅ °C
( J s ) kg
=
= kg s , as on the left hand side. Think of yourself as a power-company
°C ( J kg ⋅ °C ) °C
J
engineer arranging to have enough cooling water to carry off your thermal pollution. If the plant
power P increases, the required ﬂow rate increases in direct proportion. If environmental regulations require a smaller temperature change ∆T, then the required ﬂow rate increases again, now in
inverse proportion. Next note that Th is in the bottom of the fraction. This means that if you can
run the reactor core or ﬁrebox hotter, the required coolant ﬂow rate decreases! If the turbines take
in steam at higher temperature, they can be made more efﬁcient to reduce waste heat output.
13794_22_ch22_p571-600.indd 592
1/8/07 7:53:47 PM
23.
Heat Engines, Entropy, and the Second Law of Thermodynamics
P22.53
593
Like a refrigerator, an air conditioner has as its purpose the removal of energy by heat from the
cold reservoir.
Tc
280 K
=
= 14.0
Th − Tc
20 K
Its ideal COP is
COPCarnot =
(a)
0.400 (14.0 ) = 5.60 =
Its actual COP is
5.60
Qh
Q
Q
− 5.60 c = c
∆t
∆t
∆t
5.60 (10.0 kW ) = 6.60
Weng
Qc
Qc ∆t
=
Qh − Qc
Qh ∆t − Qc ∆t
=
Qc
∆t
and
Qc
= 8.48 kW
∆t
Qh Qc
−
= 10.0 kW − 8.48 kW = 1.52 kW
∆t
∆t
(b)
Qh = Weng + Qc :
(c)
The air conditioner operates in a cycle, so the entropy of the working ﬂuid does not change.
The hot reservoir increases in entropy by
∆t
Qh (10.0 × 10 3 J s ) ( 3 600 s )
=
= 1.20 × 10 5 J K
300 K
Th
The cold room decreases in entropy by
∆S = −
Qc
(8.48 × 103 J s ) (3 600 s ) = −1.09 × 105 J K
=−
Tc
280 K
The net entropy change is positive, as it must be:
+1.20 × 10 5 J K − 1.09 × 10 5 J K = 1.09 × 10 4 J K
Tc
280 K
=
= 11.2
Th − Tc
25 K
The new ideal COP is
COPCarnot =
We suppose the actual COP is
(d)
0.400 (11.2 ) = 4.48
As a fraction of the original 5.60, this is
drop by 20.0% .
13794_22_ch22_p571-600.indd 593
4.48
= 0.800, so the fractional change is to
5.60
1/8/07 7:53:47 PM
24.
594
Chapter 22
*P22.54 (a)
For the isothermal process AB, the work on the gas is
⎛V ⎞
WAB = − PAVA ln ⎜ B ⎟
⎝ VA ⎠
50.0 ⎞
WAB = −5 (1.013 × 10 5 Pa ) (10.0 × 10 −3 m 3 ) ln ⎛
⎝ 10.0 ⎠
WAB = −8.15 × 10 3 J
where we have used
1.00 atm = 1.013 × 10 5 Pa
and
1.00 L = 1.00 × 10 −3 m 3
FIG. P22.54
WBC = − PB ∆V = − (1.013 × 10 5 Pa ) ⎡(10.0 − 50.0 ) × 10 −3 ⎤ m 3 = +4.05 × 10 3 J
⎣
⎦
WCA = 0 and Weng = −WAB − WBC = 4.10 × 10 3 J = 4.10 kJ
Since AB is an isothermal process,
∆Eint, AB = 0
and
(b)
QAB = −WAB = 8.15 × 10 3 J
For an ideal monatomic gas,
CV =
3R
5R
and CP =
2
2
)(
(
)
1.013 × 10 5 50.0 × 10 −3
PBVB
5.06 × 10 3
TB = TA =
=
=
R
nR
R
PCVC (1.013 × 10 ) (10.0 × 10
=
nR
R
5
Also,
TC =
−3
) = 1.01 × 10
3
R
⎛ 3 ⎞ ⎛ 5.06 × 10 3 − 1.01 × 10 3 ⎞
QCA = nCV ∆T = 1.00 ⎜ R⎟ ⎜
⎟
R
⎝ 2 ⎠⎝
⎠
= 6.08 kJ
so the total energy absorbed by heat is QAB + QCA = 8.15 kJ + 6.08 kJ = 14.2 kJ
(c)
5
5
QBC = nCP ∆T = ( nR∆T ) = PB ∆VBC
2
2
5
(1.013 × 105 ) ⎡(10.0 − 50.0 ) × 10 −3 ⎤ = −1.01 × 10 4 J = −10.1 kJ
⎣
⎦
2
Weng
Weng
4.10 × 10 3 J
e=
=
=
= 0.288 or 28.8%
Qh
QAB + QCA 1.42 × 10 4 J
QBC =
(d)
(e)
A Carnot engine operating between Thot = TA = 5060րR and Tcold = TC = 1010րR has
efﬁciency 1 − Tc րTh = 1 − 1ր5 = 80.0%.
The three-process engine considered in this problem has much lower efficiency.
13794_22_ch22_p571-600.indd 594
1/8/07 7:53:48 PM
25.
Heat Engines, Entropy, and the Second Law of Thermodynamics
PVi = nRTi
i
and
n = 1.00 mol
At point B,
3PVi = nRTB
i
so
TB = 3Ti
At point C,
(3Pi ) ( 2Vi ) = nRTC
and
TC = 6Ti
At point D,
Pi ( 2Vi ) = nRTD
so
595
TD = 2Ti
*P22.55 At point A,
The heat for each step in the cycle is found using CV =
and CP =
5R
:
2
3R
2
QAB = nCV ( 3Ti − Ti ) = 3nRTi
FIG. P22.55
QBC = nCP ( 6Ti − 3Ti ) = 7.50 nRTi
QCD = nCV ( 2Ti − 6Ti ) = −6nRTi
QDA = nCP (Ti − 2Ti ) = −2.50 nRTi
Qentering = Qh = QAB + QBC = 10.5nRTi
(a)
Therefore,
(b)
Qleaving = Qc = QCD + QDA = 8.50 nRTi
(c)
Actual efﬁciency,
e=
(d)
Carnot efﬁciency,
ec = 1 −
Qh − Qc
= 0.190
Qh
Tc
T
= 1 − i = 0.833
Th
6Ti
The Carnot efficiency is much higher.
f
P22.56
∆S =
f
f
dQ
nCP dT
−1
∫i T = ∫i T = nCP ∫i T dT = nCP ln T
Tf
Ti
⎛ Tf ⎞
= nCP ln T f − ln Ti = nCP ln ⎜ ⎟
⎝ Ti ⎠
(
)
⎛ PV f nR ⎞
= nCP ln 3
∆S = nCP ln ⎜
⎠
⎝ nR PVi ⎟
13794_22_ch22_p571-600.indd 595
1/8/07 7:53:49 PM
26.
596
P22.57
Chapter 22
(a)
The ideal gas at constant temperature keeps constant internal energy. As it puts out energy
by work in expanding it must take in an equal amount of energy by heat. Thus its entropy
increases. Let Pi , Vi, Ti represent the state of the gas before the isothermal expansion. Let
PC, VC, Ti represent the state after this process, so that PiVi = PCVC. Let Pi, 3Vi, Tf represent
the state after the adiabatic compression.
Then
γ
PCVC = Pi ( 3Vi )
Substituting
PC =
gives
γ
PViVC −1 = Pi ( 3γ Viγ )
i
Then
γ
VC −1 = 3γ Viγ −1 and
γ
PVi
i
VC
VC
= 3γ
Vi
(γ −1)
The work output in the isothermal expansion is
C
C
⎛V ⎞
W = ∫ PdV = nRTi ∫ V −1dV = nRTi ln ⎜ C ⎟ = nRTi ln 3γ
⎝ Vi ⎠
i
i
This is also the input heat, so the entropy change is
(
∆S =
(γ − 1) CV
and
CP =
γR
γ −1
R
γ −1
The pair of processes considered here carries the gas from the initial state in Problem 56
to the ﬁnal state there. Entropy is a function of state. Entropy change does not depend
on path. Therefore the entropy change in Problem 56 equals ∆Sisothermal + ∆Sadiabatic in this
problem. Since ∆Sadiabatic = 0, the answers to Problems 56 and 57(a) must be the same.
Vf
W=
∫ PdV = nRT
Vi
13794_22_ch22_p571-600.indd 596
= R , CV =
∆S = nCP ln 3
Then the result is
(b)
i
CP = γ CV = CV + R
we have
*P22.58 (a)
) = nRT ⎛ γ γ− 1⎞ ln 3
⎜
⎟
⎝
⎠
⎛ γ ⎞
Q
= nR ⎜
ln 3
T
⎝ γ − 1⎟
⎠
Since
(b)
(γ −1)
2Vi
∫
Vi
⎛ 2V ⎞
dV
= (1.00 ) RT ln ⎜ i ⎟ = RT ln 2
V
⎝ Vi ⎠
While it lasts, this process does convert all of the energy input into work output. But
the gas sample is in a different state at the end than it was at the beginning. The process
cannot be done over unless the gas is recompressed by a work input. To be practical, a
heat engine must operate in a cycle. The second law refers to a heat engine operating in
a cycle, so this process is consistent with the second law of thermodynamics.
1/8/07 7:53:50 PM
27.
Heat Engines, Entropy, and the Second Law of Thermodynamics
P22.59
The heat transfer over the paths CD and BA is zero since
they are adiabatic.
Over path BC:
Qc = QDA
Adiabatic
Processes
QDA = nCV (TA − TD ) < 0
Therefore,
P
QBC = nCP (TC − TB ) > 0
Over path DA:
597
and
B
C
D
Qh = QBC
A
The efﬁciency is then
e = 1−
Qc
(T − TA ) CV
= 1− D
Qh
(TC − TB ) CP
Vi
V
FIG. P22.59
1 ⎡ T − TA ⎤
e = 1− ⎢ D
γ ⎣ TC − TB ⎥
⎦
P22.60
3Vi
Simply evaluate the maximum (Carnot) efﬁciency.
eC =
∆T 4.00 K
=
= 0.014 4
Th
277 K
The proposal does not merit serious consideration. Operating between these temperatures, this
device could not attain so high an efﬁciency.
*P22.61 (a)
20.0°C
Tf ⎤
⎡ Tf
= 1.00 kg ( 4.19 kJ kg ⋅ K ) ⎢ ln + ln ⎥
T1
T2 ⎦
⎣
⎛ 293 293 ⎞
⋅
= ( 4.19 kJ K ) ln ⎜
⎝ 283 303 ⎟
⎠
Tf
(c)
∆S = +4.88 J K
(d)
13794_22_ch22_p571-600.indd 597
∆S = mc ln
Yes, the mixing is irreversible. Entropy has increased.
s
T1
+ mc ln
Tf
(b)
T2
1/8/07 7:53:51 PM
28.
598
P22.62
Chapter 22
(a)
Use the equation of state for an ideal gas
nRT
P
1.00 (8.314 ) ( 600 )
VA =
= 1.97 × 10 −3 m 3
9
25.0 (1.013 × 10 5 )
V=
VC =
1.00 (8.314 ) ( 400 )
= 32.8 × 10 −3 m 3
8
1.013 × 10 5
FIG. P22.62
Since AB is isothermal,
PAVA = PBVB
and since BC is adiabatic,
γ
PBVBγ = PCVC
Combining these expressions,
⎡⎛ P ⎞ V γ ⎤
VB = ⎢⎜ C ⎟ C ⎥
⎣⎝ PA ⎠ VA ⎦
1 (γ −1)
⎡⎛ 1.00 ⎞ ( 32.8 × 10 −3 m 3 )1.40 ⎤
⎥
=⎢
−3
3
⎥
⎢⎝ 25.0 ⎠ 1.97 × 10 m
⎦
⎣
(1 0.400 )
⎡ 25.0 ⎞ (1.97 × 10 −3 m 3 )1.40 ⎤
⎥
= ⎢⎛
−3
3
⎢⎝ 1.00 ⎠ 32.8 × 10 m ⎥
⎣
⎦
(1 0.400 )
VB = 11.9 × 10 −3 m 3
1 (γ −1)
Similarly,
⎡⎛ P ⎞ V γ ⎤
VD = ⎢⎜ A ⎟ A ⎥
⎣⎝ PC ⎠ VC ⎦
or
VD = 5.44 × 10 −3 m 3
Since AB is isothermal,
PAVA = PBVB
and
⎛V ⎞
⎛ 1.97 × 10 −3 m 3 ⎞
PB = PA ⎜ A ⎟ = 25.0 atm ⎜
= 4.14 atm
⎝ 11.9 × 10 −3 m 3 ⎟
⎠
⎝ VB ⎠
Also, CD is an isothermal and
⎛V ⎞
⎛ 32.8 × 10 −3 m 3 ⎞
PD = PC ⎜ C ⎟ = 1.00 atm ⎜
= 6.03 atm
⎝ 5.44 × 10 −3 m 3 ⎟
⎠
⎝ VD ⎠
Solving part (c) before part (b):
ec = 1 −
Tc
400 K
= 1−
= 0.333
Th
600 K
(c)
For this Carnot cycle,
(b)
Energy is added by heat to the gas during the process AB. For the isothermal process,
∆Eint = 0.
⎛V ⎞
and the ﬁrst law gives
QAB = −WAB = nRTh ln ⎜ B ⎟
⎝ VA ⎠
⎛ 11.9 ⎞
or
Qh = QAB = 1.00 mol (8.314 J mol ⋅ K ) ( 600 K ) ln ⎜
⎝ 1.97 ⎟
⎠
= 8.97 kJ
Then, from
e=
Weng
Qh
the net work done per cycle is Weng = ec Qh = 0.333 (8.97 kJ ) = 2.99 kJ
13794_22_ch22_p571-600.indd 598
1/8/07 7:53:52 PM
29.
Heat Engines, Entropy, and the Second Law of Thermodynamics
599
ANSWERS TO EVEN PROBLEMS
P22.2
13.7°C
P22.4
(a) 29.4 Lրh
P22.6
(a) 24.0 J
P22.8
(a) 2.93 (b) coefﬁcient of performance for a refrigerator (c) The cost for air conditioning is
half as much for an air conditioner with EER 10 compared with an air conditioner with EER 5.
P22.10
(a) 870 MJ
P22.12
(a) 0.300
P22.14
33.0%
P22.16
(a) 5.12% (b) 5.27 TJրh
a good buy.
P22.18
(a) |Qc|ր∆t = 700 kW(Th + 766 K)ր(Th − 383 K) The exhaust power decreases as the ﬁrebox temperature increases. (b) 1.87 MW (c) 3.84 × 103 K (d) No answer exists. The energy exhaust
cannot be that small.
P22.20
(a) State
A
B
C
D
(b)
(c) 527 N . m
(b) 185 hp
(d)1.91 × 105 W
(b) 144 J
(b) 330 MJ
(b) 1.40 × 10−3 (c) 2.00 × 10−3
Process
A→B
B→C
C→D
D→A
ABCDA
P (kPa)
1 400
875
445
712
(c) As fossil-fuel prices rise, this way to use solar energy will become
V (L) T (K)
10.0 720
16.0 720
24.0 549
15.0 549
Q (kJ)
6.58
0
−5.02
0
1.56
W (kJ)
−6.58
−4.98
5.02
4.98
−1.56
∆Eint (kJ)
0
−4.98
0
4.98
0
(c) 23.7%; see the solution
P22.22
11.8
P22.24
1.17 J
P22.26
1.86
P22.28
(a), (b) see the solution
(e) 1.42 × 103 revրmin
P22.30
−610 JրK
13794_22_ch22_p571-600.indd 599
(c) Qh = 149 J; Qc = 65.0 J; Weng = 84.3 J
(d) 56.5%
1/8/07 7:53:53 PM
30.
600
Chapter 22
P22.32
(a) The process is isobaric because it takes place under constant atmospheric pressure. The
heating process is not adiabatic (because energy goes in by heat), isothermal (T goes up),
isovolumetric (it likely expands a bit), cyclic (it is different at the end), or isentropic (entropy
increases). It could be made as nearly reversible as you wish, by not using a kitchen stove but a
heater kept always just incrementally higher in temperature than the syrup. (b) 402 kJ
(c) 1.20 kJրK
P22.34
3.27 JրK
P22.36
+1.18 JրK
P22.38
718 JրK
P22.40
0.507 JրK
P22.42
(a) 2 heads and 2 tails
P22.44
8.36 MJրK
P22.46
(a) 39.4 J
P22.48
(a) 0.476 JրK
(b) 417 J
P22.50
(a) 3.19 calրK
(b) 98.19ºF, 2.59 calրK This is signiﬁcantly less than the estimate in part (a).
P22.52
(Th − Tc ) c∆T
P22.54
(a) 4.10 kJ (b) 14.2 kJ (c) 10.1 kJ (d) 28.8% (e) The three-process engine considered in
this problem has much lower efﬁciency than the Carnot efﬁciency.
P22.56
nCpln3
P22.58
(a) see the solution (b) While it lasts, this process does convert all of the energy input into work
output. But the gas sample is in a different state at the end than it was at the beginning. The
process cannot be done over unless the gas is recompressed by a work input. To be practical, a
heat engine must operate in a cycle. The second law refers to a heat engine operating in a cycle,
so this process is consistent with the second law of thermodynamics.
P22.60
The proposal does not merit serious consideration. Operating between these temperatures, this
device could not attain so high an efﬁciency.
P22.62
(a)
(b) All heads or all tails
(b) 65.4 radրs = 625 revրmin
(c) 293 radրs = 2 790 revրmin
(c) Wnet = T1 ∆ SU = 167 J
PT
c
P, atm
A
V, L
25.0
1.97
B
4.14
11.9
C
1.00
32.8
D
6.03
(b) 2.99kJ
13794_22_ch22_p571-600.indd 600
(c) 2 heads and 2 tails
5.44
(c) 33.3%
1/8/07 7:53:53 PM
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