Download chapter 21 thermodynamics

CHAPTER 21
THERMODYNAMICS
Select odd-numbered solutions, marked with a dagger (†), appear in the Student Solutions
Manual, available for purchase. Answers to all solutions below are underscored.
†21-1.
The change in internal energy for this ideal monatomic gas is the difference between the heat
absorbed Q by the gas and the work that it does on the piston. The heat input into the system is
Q = nCp ∆T = (1.0 mole) ( 20.8 J/(mole i K) )( 90 K ) = 1.9 × 103 J
where cp is given in Table 20.5. The change in the internal energy is then,
∆E = Q − W = nCp ∆T − W
= (1.0 mole) ( 20.8 J/(mole i K) )( 90 K ) − 800 J = 1.1 × 103 J
21-2.
Because there is no temperature change and no pressure change, the process results in no change
in the internal energy of the gas. Therefore, the amount of heat transferred to the system must be
equal to the work done by the gas as it expands aginst the piston.
Q −W = 0
Q = W = p∆V = (1.01 × 105 N/m 2 )(1.50 × 10−5 m3 ) = 1.52 J
†21-3.
(a) In step A, the gas isovolumetrically (∆V = 0) expands, so the
work done by the gas W = p∆V = 0 J
The heat absorbed by the ideal gas is Q = ncV∆T. From the ideal
gas law, we have
nRT1 nRT2
p1V1 = nRT1 or V1 =
=
p1
p2
since the volume is constant. Solving for ∆T = T2 − T1, gives
⎛p
⎞
∆T = T1 ⎜ 2 − 1 ⎟
⎝ p1
⎠
B
P2
P1
A
V1
V2
The system is initially at standard temperature and pressure (STP), T1 = 273 K and P1 = 1.0 atm.
One mole of an ideal gas at STP has a volume of 22.4 liters. The heat absorbed in step A is
⎛p
⎞
⎛ 2p
⎞
1 mole
Q = nCVT1 ⎜ 2 − 1 ⎟ = 4.0 liters ( 22.4
12.5 J/(mole i K) ) (273 K) ⎜ 1 − 1⎟ = 610 J
liters ) (
⎝ p1
⎠
⎝ p1
⎠
The internal energy in this case is equal to the heat absorbed by the system,
∆E = Q − W = Q = 610 J
(b) In step B, the gas isobarically (∆P = 0) expands by ∆V = 4.0 liters (10−3 m3/liter) = 4.0 × 10−3
m3, so the work done by the gas is
W = p∆V = (2.02 × 105 N/m2)(4.0 × 10−3 m3) = 808 J = 810 J
The amount of heat absorbed by the system in this process is again Q = nCV∆T. From the ideal
gas law, we have
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CHAPTER
p2V1 = nRT2
and
p2V1
nR
and
21
p2V2 = nRT3
p2V2
nR
pV
pV
p
∆T = T3 − T2 = 2 2 − 2 1 = 2 ∆V
nR
nR
nR
5
2.02 × 10 N/m 2
=
4.0 × 10−3 m3 ) = 544.5 K
(
1 mole
4.0 liters ( 22.4 liters ) ( 8.31 J/(mole i K) )
T2 =
Then,
21-4.
T3 =
1 mole
Q = nCp ∆T = 4.0 liters ( 22.4
20.8 J/(mole i K) ) (544.5 K) = 2022 J = 2.0 × 103 J
liters ) (
The internal energy in this case is
∆E = Q − W = Q = 2.0 × 103 J − 810 J = 1.2 × 103 J
The free expansion of the air is considered an adiabatic process (Q = 0), so the change in the
internal energy is equal to the work done by the gas against the atmosphere, ∆E = Q − W = −1800
J. The change in internal energy is also ∆E = nCV∆T (see section 20.6). Therefore,
nCV ∆T = −1800 J/mole
∆T =
1800 J/mole
−1800 J/mole
=
= −86.5 K
nCV
(1.0 mole) ( 20.8 J/(mole i K) )
T2 = T1 + ∆T = 293 K − 86.5 K = 206.5 K or − 66.5°C
†21-5.
21-6.
In an adiabatic process, heat is neither absorbed nor emitted from a system; Q = 0. Since work is
being done on the gas instead of by the gas, the internal energy of the system increases
∆E = Q − W = −W
Since work is done on the gas, the work done has a negative value and the internal energy is
increasing in this situation. According to section 20.6, the change in internal energy is also equal
to nCV∆T. Therefore,
∆E = nCV ∆T = ( 0.0300 mole )( 20.8 J/(mole i K) )( (790 − 40) K ) = 470 J
The work done is W = −470 J.
Since all of the heat generated in the collision with the floor stays in the lead ball, Q = 0 J. Work
is done on the ball by the floor causing a change in its kinetic energy and causing an increase in
the internal energy of the ball. In the instant before impact, the ball has a kinetic energy K1 equal
to its initial energy, which was its gravitational potential energy. Since its final speed is zero, its
final kinetic energy is zero.
W = ∆K = K 2 − K1 = 0 − mgh = − ( 0.25 kg ) ( 9.81 m/s 2 ) ( 0.80 m ) = −1.96 J
The increase in the internal energy of the ball is ∆E = −W = 1.96 J
This increase in the internal energy will increase the temperature of the ball by
∆E = mC ∆T
∆E
1.96 J
∆T =
=
= 0.060°C
mC (0.25 kg) (130 J/(kg i °C) )
†21-7.
(a) Assuming the gas obeys the ideal gas law, pV = nRT,
5
2
3
pV (1.01 × 10 N/m )( 0.100 m )
n=
=
= 4.29 moles
RT
( 8.31 J/(mole i K) ) (283 K)
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CHAPTER
21
(b) The work done by the gas as it expands is
W = p∆V = (1.01 × 105 N/m 2 )( (0.110 − 0.100)m3 ) = 1010 J
(c) The change in internal energy is ∆E = Q − W = 3500 J − 1010 J = 2490 J
(d) Consider section 20.5 and equations 20.30 to 20.33, in particular. The internal energy of the
gas is ∆E = fnR∆T , where f is a fraction related to the nature of the gas (monatomic or diatomic).
For the situation given,
∆E
2490 J
5
f =
=
=
nR∆T (4.29 moles) ( 8.31 J/(mole i K) ) (28 K) 2
21-8.
Therefore, CV = 5R/2 and the gas is diatomic.
(a) The initial volume of the water is 1.00 liter = 1.00 × 10−3 m3. Assuming the water vapor
obeys the ideal gas law, the volume of the steam vapor at 1 atm is given by
⎛ 1 mole ⎞
1.00 kg ⎜
⎟ ( 8.31 J/(mole i K) )( 373 K )
0.018 kg ⎠
nRT
⎝
V =
=
= 1.705 m3
p
1.01 × 105 N/m 2
In the expansion of the water vapor, work is done by the gas against the atmosphere equal to
W = p∆V = (1.01 × 105 N/m 2 ) ( (1.705 − 0.00100)m3 ) = 1.72 × 105 J
The change in the internal energy during the expansion is
∆E = Q − W = mLV − W = (1.00 kg)(2.26 × 106 J/kg) − (1.72 × 105 J) = 2.09 × 106 J
During the process the internal energy of the system increases due to the heat added to the
system, but that energy is reduced by the work done by the vapor. About 8% of the heat input to
the system goes into doing work.
(b) Because liquid water is incompressible, there is no volume change as the pressure is
increased from 1 atm to 2 atm. No work is done, therefore, on the liquid water. Because there is
also no temperature change, there is no change in the internal energy of the water. As for water
vapor at 1 atm, the heat of vaporization is LV, 1atm = 2.26 × 106 J. Because the temperature is
constant at 373 K, ∆E = 0. To compress water vapor from 1 atm to 2 atm, the work done,
applying the Ideal-Gas Law, p = nRT/V and p1V1 = p2V2, is
V2
V2
dV
V
W = ∫ pdV = ∫ nRT
= nRT ln 2
V1
V1
V
V1
21-9.
⎛
⎞
1.00 kg
5
=⎜
⎟ ( 8.31 J/(mole i K) )( 373 K ) ln ( 0.50 ) = −1.19 × 10 J
0.018
kg/mole
⎝
⎠
The heat of vaporization at 2 atm is
LV, 2 atm = LV, 1atm + W = 2.26 × 106 J − 1.19 × 105 J = 2.14 × 106 J
(a) W = p∆V = (1.01 × 105 N/m 2 ) (1.000 × 10−3 m3 − 1.091 × 10−3 m3 ) = −9.19 J
∆E = Q − W = mLF − W = (1.00 kg)(3.34 × 105 J/kg) − ( −9.19 J) = 3.34 × 105 J
The work done by the atmosphere on the ice is negligible, so the change in internal energy is that
of the heat added to the ice to melt it.
(b) Since the densities, and hence the volumes, of the water and ice are assumed to remain the
same at 2.00 atm as at 1.00 atm, the work done by the atmosphere will be
W = p∆V = (2.02 × 105 N/m 2 ) (1.000 × 10−3 m3 − 1.091 × 10−3 m3 ) = −18.4 J
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CHAPTER
21-10.
Given the assumptions that the volumes of water and ice do not change between 1.00 and 2.00
atm and the negligible amount of work done by the atmosphere, one concludes that the heat of
vaporization remains unchanged.
W
V′
dV
(a) ∫ dW = ∫ nRT
V
0
V
V′
V ′ dV
V′
= nRT ln V = nRT (ln V ′ − ln V ) = nRT ln
W = nRT ∫
V V
V
V
(b) In an isothermal expansion, the change in internal energy ∆E = Q − W = 0 J. Therefore, W =
Q. Using the result from part (a),
V′
W = nRT ln
V
eW / nRT = e
eW / nRT =
⎛V′⎞
ln ⎜ ⎟
⎝V ⎠
V′
V
V ′ = VeW / nRT = ( 0.0020 m3 ) e
†21-11.
21-12.
21
500 J/ ( (0.10 mole)(8.31 J/mole i K)(700 K) )
= 0.0047 m3
The work done during this process is
W = p∆V
= 0.25 atm (1.01 × 105 N/m2/1.0 atm)(0.5 × 10−3 m3 − 2.0 × 10−3 m3) = −37.9 J
During the process 75 J flows out of the system, therefore Q = −75 J. The change in internal
energy is ∆E = Q − W = −75 J − (−37.9 J) = −37.1 J
The change in the volume of the cylinder in which the gas is contained decreases by
∆V = Ah = ( 90 cm 2 ) ( 1001 mcm ) ( 0.12 m ) = 1.08 × 10−3 m3
2
where A is the area of the cylinder and h is the displacement of the cylinder. The work done
during this process is
W = p∆V = 15 atm (1.01 × 105 N/m2/1.0 atm)(−1.08 × 10−3 m3) = −1636 J
Because work is done on the gas to compress it and the internal energy decreases, heat must be
removed from the system. The quantity of this heat is found by applying the First Law of
Thermodynamics,
∆E = Q − W
Q = ∆E + W = −15 J + 15 atm
†21-13.
(
) ( − 1.08 × 10
1.01 × 105 N/m 2
1.0 atm
−3
m3 ) = −1651 J
In the temperature range of 5°C to 30°C, ethanol is a liquid. The amount of heat added to the
ethanol is
Q = mc∆T
According to Table 20.1, the specific heat of ethanol at 20°C is 2430 J/(kg • °C). To determine
the mass of the liquid, we need to know its density, which is given for 20°C. Therefore, we need
to know the volume of the ethanol at 20°C. At 5.0°C, the volume is 1.00 liter.
As the alcohol is heated, it undergoes volumetric thermal expansion
∆V = β V ∆T = (1.01 × 10−3 / °C)(1.00 liter)
(
1.00 × 10−3 m3
1.00 liter
) (20°C − 5°C) = 1.51 × 10
−5
m3
where β is the volumetric thermal expansion coefficient listed in Table 20.2. The volume at 20°C
is
V + ∆V = (1.00 × 10−3 m3) + (1.51 × 10−5 m3) = (1.015 × 10−3 m3)
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CHAPTER
21
The mass of the ethanol is therefore,
m = ρV = (789 kg/m3 )(1.015 × 10−3 m3 ) = 0.801 kg
Then,
Q = mc∆T = (0.801 kg) ( 2430 J/(kg i °C) ) (30°C − 5°C) = 4.87 × 104 J
The volume expansion between 5°C and 30°C is
∆V = β V ∆T = (1.01 × 10−3 / °C)(1.00 liter)
21-14.
(
1.00 × 10−3 m3
1.00 liter
) (30°C − 5°C) = 2.53 × 10
−5
m3
The work done by the alcohol in its expansion on the surrounding air is
W = p∆V = 1.00 atm (1.01 × 105 N/m2/1.00 atm)(2.53 × 10−5 m3) = 2.56 J
In an adiabatic compression, Q = 0 J and ∆E = W. Assuming air behaves as an ideal gas,
3
∆E = W = p∆V = (1.01 × 105 N/m 2 )( 20 cm3 − 80 cm3 ) ( 1001 mcm ) = −6.06 J
To determine the new pressure when the gas is compressed, we use the fact that pV γ is constant
for an ideal gas, where γ = 1.40 for a diatomic gas.
p1V11.40 = p2V21.40
1.40
1.4
⎛V ⎞
⎛ 80 cm3 ⎞
p2 = p1 ⎜ 1 ⎟ = (1.01 × 105 N/m 2 ) ⎜
= 7.03 × 105 N/m 2
3 ⎟
20
cm
V
⎝
⎠
⎝ 2⎠
Using the Ideal-Gas Law, the number of moles of air present and the temperature of the
compressed gas may be determined.
pV = nRT
5
2
3
1m
p1V1 (1.01 × 10 N/m )( 80 cm ) ( 100 cm )
n=
=
= 0.0032 moles
RT1
( 8.31 J/(mole i K) )( 300 K )
3
(1.01 × 105 N/m 2 )( 20 cm3 ) ( 1001 mcm ) = 75 K
pV
T2 = 2 2 =
nR
( 8.31 J/(mole i K) )( 0.0032 moles )
3
The gas then loses heat to its surroundings until the temperature reaches 300 K. This occurs at
constant volume, or isovolumetrically. The work done is equal to 0 J, since there is no volume
change. The mass of 1 mole of air is 29.0 g. Therefore, the change in internal energy is
kg
∆E = Q = mc∆T = 0.0032 moles 0.0029
( 20.8 J/(mole i K) ) (300 K − 75 K) = 0.043 J
1 mole
(
)
The initial internal energy of the air is found by integrating equation (20.21)
dE = nCV dT
∫ dE =nC ∫ dT
V
E1 = nCVT1 = (0.0032 moles) ( 20.8 J/(mole i K) ) (300 K) = 19.97 J
As a result of the two steps taken to compress and cool the air, the final internal energy is
E2 = E1 + ∆E1 + ∆E2 = 19.97 J − 6.06 J + 0.043 J = 13.95 J
†21-15.
This is an example of an isothermal process, ∆T = 0 K. Since the temperature does not change,
there is no change in the internal energy of the ideal gas; and W = Q. The work done by the gas in
its expansion, as determined in problem 21-10(a), is
V
⎛ 10 liters ⎞
4
W = nRT ln 2 = (10 moles) ( 8.31 J/(mole i K) ) (323 K) ln ⎜
⎟ = 4.3 × 10 J
V1
⎝ 2 liters ⎠
442
CHAPTER
21-16.
†21-17.
21
This is an example of an isothermal process, ∆T = 0 K. Since the temperature does not change,
there is no change in the internal energy of the ideal gas; and W = Q.
V
W = Q = nRT ln 2 = (2.0 moles) ( 8.31 J/(mole i K) ) (373 K) ln(10) = 1.4 × 104 J
V1
The temperature may be determined from the Ideal-Gas Law,
pV = nRT
T =
(1.01 × 105 N/m2 )( 0.10 m3 ) = 243 K
p2V2
=
nR
( 8.31 J/(mole i K) )( 5.00 moles )
Because this is an isothermal process, ∆T = 0 K; and there is no change in the internal energy of
the ideal gas; and the work done as the gas expands is
V
W = Q = nRT ln 2
V1
from which the initial volume may be determined since the quantity of work done by the gas is
given.
W
V
= ln 2
nRT
V1
W
e nRT =
V2
V1
V1 = V2 e
−
W
nRT
= ( 0.10 m3 ) e
− (2.0 × 104 J)/ ⎡⎣ (5.00 moles) ( 8.31 J/(mole i K) ) (243 K) ⎤⎦
= 0.014 m3
The initial pressure was
nRT ( 5.00 moles )( 8.31 J/(mole i K) ) (243 K)
p1 =
=
= 7.2 × 105 N/m 2
V1
0.014 m3
21-18.
†21-19.
At 10 m below the surface of the water, the hydrostatic pressure is (see example 6, chapter 18),
p1 = p0 − ρ gy = 1.01 × 105 N/m 2 − (1000 kg/m3 )(9.81 m/s 2 )(−10 m) = 1.99 × 105 N/m 2
The bubble forms from the diver’s airline from an initial volume of zero m3. So, the work done in
forming the bubble is done against the surrounding water is
3
W = p∆V = (1.99 × 105 N/m 2 ) ⎡ ( 4.0 cm3 ) ( 1001 mcm ) − 0 m3 ⎤ = 0.80 J
⎣
⎦
As this is an isothermal process, ∆T = 0 K; and there is no change in the internal energy of the
ideal gas; and the heat absorbed by the bubble as the gas expands is equal to the work,
W = Q = 0.80 J
The diagram shows the indirect process
Water at
Water vapor
described in the problem. The net work done
Boil
100°C
at 100°C
against the atmosphere is the same in the
indirect process as in the direct process since
Cool
the changes in the volume are the same. The
Heat
net changes in the internal energies is also the
same. Because the work done and the internal
Water at
Water vapor
energy are the same, then the heat must also be
20°C
at 20°C
the same. For 1.0 kg of water, the latent heat
per kilogram is
443
CHAPTER
21
LV(20°C) = c∆T + LV(100°C) + cp ∆T
= ( 4187 J/(kg i °C) ) (80°C) + 2.26 × 106 J/kg − ( 2010 J/(kg i °C) ) (80°C)
= 2.43 × 106 J/kg
21-20.
LV(140° C) = c∆T + LV(100° C) + cp ∆T
= ( 4187 J/(kg i °C) ) (−40°C) + 2.26 × 106 J/kg + ( 2010 J/(kg i °C) ) (40°C)
= 2.17 × 106 J/kg
21-21.
e=
21-22.
e=
†21-23.
120 hp
(
746 W
1 hp
5
) = 0.203 or 20.3%
4.40 × 10 W
The work done by the runner is W = ∆U = mgh. So, the rate that work is being done is
∆W
∆h
= mg
= mgv
∆t
∆t
The efficiency is determined by the ratio of this work to the total energy generated by the body
that includes the work done plus the waste heat generated.
(∆W/∆t )
mgv
e=
=
(∆W/∆t ) + (∆Q/∆t ) mgv + (∆Q/∆t )
=
21-24.
W (W/t ) 300 MW
=
=
= 0.35 or 35%
Q (Q/t ) 850 MW
e=
( 70 kg ) ( 9.81 m/s 2 ) ( 0.30 m/s )
( 70 kg ) ( 9.81 m/s 2 ) ( 0.30 m/s ) + (1300 W)
= 0.14 or 14%
W
2.0 × 104 J
=
= 0.67 or 67%
Q1 3.0 × 104 J
The waste heat produced is Q2 = Q1 − W = 1.0 × 104 J
†21-25.
According to equation 21.5, e =
Q
W
= 1 − 2 , the efficiency may be determined in more than
Q1
Q1
one way. In this case, we are given Q1 and Q2 and asked to find the efficiency and the work done
by the engine.
Q
8.0 × 106 J
e =1− 2 =1−
= 0.60 or 60%
2.0 × 107 J
Q1
The work done is W = Q1 − Q2 = 1.2 × 107 J
21-26.
21-27.
21-28.
Q
W
= e 1 = 0.95(3.0 × 103 W) = 2.85 × 103 W or 2.85 kW
t
t
(Q2/t) = (Q1/t) − (W/t) = 0.15 kW
The overall efficiency is the product of the individual efficiencies.
e = (0.90)(0.50)(0.99) = 0.445 or 44.5%
Summing the individual expenditures of mechanical power gives 2.77 hp. The efficiency of the
runner is
2.77 hp
e=
= 0.21 or 21%
13 hp
444
CHAPTER
21-29.
21
The maximum efficiency of the body as a heat engine would be
T
293 K
e =1− 2 =1−
= 0.055 or 5.5%
T1
310 K
The problem asks the student to calculate the work that he or she does to climb to a height of 3.0
m, the work done by the body would be W = ∆U = mgh. The student would insert his/her mass in
this calculation. Therefore, the general result for the work done is
W = myou ( 9.81 m/s 2 ) ( 3.0 m )
21-30.
With the assumption that the body is an inefficient heat engine, the amount of kcal one would
have to consume to climb 3.0 m is
2
W myou ( 9.81 m/s ) ( 3.0 m ) ⎛ 1 kcal ⎞
Q=
=
⎜
⎟
e
0.055
⎝ 4187 J ⎠
Q
1W
1
The rate at which chemical energy is consumed is 1 =
(50 W) = 135 W
=
0.37
t
e t
Q
Because, e = 1 − 2 , the rate at which waste heat is produced is
Q1
Q2
t
=
Q1
t
(1 − e ) = (135 W)(1 − 0.37) =
85 W
The rate at which glucose is consumed is
Q1
135 J/s
kg
=
= 8.7 × 10−6
= 8.7 mg/s
3 kcal 4187 J
glucose yield 3.7 × 10 kg ( 1 kcal )
s
†21-31.
The engines each do 1100 hp of work per second and the efficiency is 20%, so the heat input to
the system to deliver this work is somewhat greater. To find that rate of heat input, remember that
efficiency is defined as e = W/Q1. Dividing both factors on the right side of the equation by t
gives the ratio of the rate of work to the rate of heat input.
W/t
e=
Q1/t
(
)
745.7 W
Q1 W/t 2200 hp 1 hp
=
=
= 8.2 × 106 W = 8.2 × 106 J/s
t
e
0.20
The rate of fuel input needed to supply this heat input is
m
Q/t
8.2 × 106 J/s
=
=
= 0.19 kg/s
t
E/m 4.4 × 107 J/kg
21-32.
(a)
W2
W1
Q 1 , T1
Steam
Q2,T2
High pressure
stage
Steam
Low pressure
stage
Q3, T3
Water
(b) Because some heat is put back into the system in the low-pressure stage, there is a gain in
efficiency because additional work is the result. Therefore, the overall efficiency e is calculated as
follows. Given that,
T
T
e1 = 1 − 2
e2 = 1 − 3
T1
T2
and
e1T1 = T1 − T2
e2T2 = T2 − T3
The overall efficiency is
445
CHAPTER
21
e =1−
T3
T1
eT1 = T1 − T3 = ( T1 − T2 ) + ( T2 − T3 ) = e1T1 + e2T2
†21-33.
21-34.
⎛T ⎞
e = e1 + e2 ⎜ 2 ⎟ = e1 + e2 − e1e2 = 0.4 + 0.2 − (0.4)(0.2) = 0.52 or 52%
⎝ T1 ⎠
The efficiency is determined by the temperatures of the high temperature (T1) and low
temperature (T2) reservoirs.
T
300 K
e =1− 2 =1−
= 1 − (3 × 109 ) = 0.999 999 997
11
T1
1 × 10 K
This is very close to 100% efficiency. However, working with such high temperatures can be a
difficult engineering problem in practice.
If the efficiency is 33% and the amount of useful power produced is 1 × 109 W, then the rate of
waste heat produced is ∆Q/∆t = 2 × 109 W. This the heat that raises the temperature of the water,
∆Q ∆m
=
c∆T
∆t
∆t
∆m ( ∆Q / ∆t )
2 × 109 J/s
3
1 m3
=
=
= 6 × 104 kgs 1000
kg = 60 m /s
∆t
c∆T
( 4187 J/(kg i K) )( 8 K )
(
†21-35.
)
The efficiency is determined by the temperatures of the high temperature (T1) and low
temperature (T2) reservoirs. The efficiency e is the ratio of the work done W by the engine to the
amount of heat absorbed Q1. Therefore,
W
T
e=
=1− 2
Q1
T1
5.0 × 104
W
=
= 1.9 × 105 J
T2
273 K ⎞
⎛
1−
1− ⎜
⎟
T1
⎝ 373 K ⎠
The amount of waste heat produced is the difference between the heat absorbed and the work
done,
Q2 = Q1 − W = (1.9 × 105 J) − (5.0 × 104 J) = 1.4 × 105 J
T
300 K
W
e=
=1− 2 =1−
= 0.25 or 25%
400 K
Q1
T1
Q1 =
21-36.
Increasing the efficiency by 10%, gives
T
1 − 2 = 0.35
T1
†21-37.
T2 = 0.65T1
Therefore, if T2 = 300 K, then T1 = 462 K. Alternatively, if T1 = 400 K, then T2 = 260 K.
Q
W
According to equation 21.5, e =
= 1 − 2 , the efficiency may be determined in more than
Q1
Q1
one way. In this case, we are given Q1 and Q2 and asked to find the efficiency and the mechanical
power output (the work done per second) the engine. Therefore,
Q
Q /t
1400 J/s
e =1− 2 =1− 2 =1−
= 0.44 or 44%
Q1
Q1/t
2500 J/s
446
CHAPTER
21
The power output is
W W/t
e=
=
Q Q1 /t
W
= e(Q1/t ) = (0.44)(2500 J/s) = 1100 J/s = 1100 W
t
The efficiency may be calculated in several ways. Of relevance here is
Q
W
e=
=1− 2
Q1
Q1
P=
21-38.
Q2
=1− e
Q1
5.0 × 103 J
Q2
=
= 5880 J = 5.9 × 103 J
1− e
1 − 0.15
Given the heat absorbed for one cycle, we can determine the amount of work done during the
cycle.
W = eQ1 = (0.15)(5880 J) = 882 J = 8.8 × 102 J
Q1 =
Since the rate of work done is P = 12 kW = 1.2 × 103 J/s, the time for one cycle is
W
882 J
=
= 0.074 s
P 1.2 × 104 J/s
†21-39.
The flowchart for the operation of a heat pump is shown in Figure 21.16. Heat is taken from the
inside of the house at T2 and sent to the high temperature reservoir at T1 outside through the work
done by heat pump unit. The efficiency of the heat pump is
Q
T
W
e=
=1− 2 =1− 2
Q1
Q1
T1
To apply the thermodynamic relationships, temperatures must be converted into degrees Kelvin.
Each °F = (5/9)°C. Since the freezing point of water is at 32°F, the temperatures given are:
⎛ 5°C ⎞
T2 = ( 75°F − 32°F ) ⎜
⎟ + 273°C = 297 K
⎝ 9°F ⎠
and
⎛ 5°C ⎞
T1 = (104°F − 32°F ) ⎜
⎟ + 273°C = 313 K
⎝ 9°F ⎠
Therefore, the heat transferred to the hot reservoir by the heat pump operating in cooling mode
each hour will be
Q2 T2
=
Q1 T1
⎛T ⎞
⎛ 313 K ⎞
6
Q1 = Q2 ⎜ 1 ⎟ = ( 5.0 × 106 J ) ⎜
⎟ = 5.27 × 10 J
⎝ 297 K ⎠
⎝ T2 ⎠
Therefore, the work input to the heat pump is
W = Q1 − Q2 = 5.27 × 106 J − 5.0 × 106 J = 2.7 × 105 J
The rate of work done is
W 2.7 × 105 J ⎛ 1 h ⎞
P=
=
⎜
⎟ = 75 W
t
1h
⎝ 3600 s ⎠
447
CHAPTER
21-40.
21
From the efficiency relationship, we may find the temperature of the cold reservoir.
T
e =1− 2
T1
T2 = T1 (1 − e) = (350 +273 K)(1 − 0.15) = 530 K
Then, to increase the efficiency while maintaining the temperature of the cold reservoir, the
temperature of the hot reservoir should be increased to
T2
530 K
T1 =
=
= 707 K or 434°C
(1 − e) 1 − 0.25
†21-41.
When a heat pump is used to provide heat to a home, the cofficient of performance (CP) is
defined as
energy transferred to the high temperature reservoir Q1
CP =
=
work done by the heat pump
W
The efficiency of a carnot engine is
Q
T
T − T2
W
=1− 2 =1− 2 = 1
e=
Q1
Q1
T1
T1
Therefore,
Q
T1
293 K
CP = 1 =
=
= 19.5
W T1 − T2 293 − 278 K
21-42.
This value represents the maximum CP for a system operating between these two temperatures.
In practice, the CP values for real heat pumps is somewhat smaller than this value.
T
e =1− 2
T1
T1 =
21-43.
21-44.
T2
273 K
=
= 403 K
(1 − e) 1 − 0.322
W = Q1 − Q2
Q1 = W + Q2 = 2.5 × 103 J + 6.0 × 103 J = 8.5 × 103 J
Q
8.5 × 103 J
CP = 1 =
= 3.4
W
2.5 × 103 J
T
T − T2 373 K − 273 K
W
e=
=1− 2 = 1
=
= 0.268
Q1
T1
T1
373 K
W = eQ1 = e(W − Q2 )
†21-45.
⎛ e ⎞
⎛ 0.268 ⎞
W = Q2 ⎜
⎟ = Q2 ⎜
⎟ = 0.366Q2
⎝1− e ⎠
⎝ 1 − 0.268 ⎠
For every joule of heat removed from the low temperature reservoir, 0.366 J of work must be
done.
(a) The work done during each step for the ideal monatomic gas is W = p∆V. Therefore, the work
done during step 1 is zero J, since there is no change in volume. During step 2, the work done is
W2 = p2 (V2 − V1 ) = (1 × 105 N/m 2 )(0.03 m3 − 0.01 m3 ) = 2 × 103 J
In step 3, both the pressure and the volume change, such processes may be either isothermal or
adiabatic. To determine which type of process occurs in this system, consider the beginning and
ending points of step 1, applying the Ideal-Gas Law
448
CHAPTER
21
PV
1 1 = nRT1
T1 =
PV
(3 × 105 N/m 2 )(0.01 m3 )
1 1
=
= 361 K
nR
(1.00 mole) ( 8.31 J/(mole i K) )
P1 P2
=
T1 T2
P2
T1 = 13 T1 = 13 (361 K) = 120 K
P1
Then, in step 2, there is an isobaric expansion, the temperature at the beginning of step 3 is
V2 V3
=
T2 T3
T2 =
V3
T2 = 3T2 = 3 ( 13 T1 ) = T1
V2
The temperature in step 3 does not change, so the process is isothermal. The work done in an
isothermal process is
⎛ 0.01 m3 ⎞
V
= −3.3 × 103 J
W3 = p2V2 ln 3 = (1 × 105 N/m 2 )(0.03 m3 ) ln ⎜
3 ⎟
0.03
m
V2
⎝
⎠
T3 =
The total work done during one cycle is
W = W1 + W2 + W3 = 0 + 2.0 × 103 J − 3.3 × 103 J = −1.3 × 103 J
The negative value of the net work indicates that work is done on the gas during the cycle.
(b) During the isobaric step 2, the heat flow Q = mcp∆T. Since the change in temperature is
positive, there is a net flow of heat into the system. Heat is absorbed during step 2. During
isothermal step 3, the heat flow is equal to the amount of work, since the work is negative, the
heat flow is again negative and there is a net heat flow out of the system. Heat is rejected in step
3.
(c) During isovolumetric step 1, heat is rejected by the system, since Q = mcp∆T and T1 is greater
than T2.
(d) Just as in single step processes that occur between high and low temperature reservoirs, the
efficiency is still defined in a cyclical processes as the ratio of the work done by the system to the
quantity of heat rejected to the high temperature reservoir. Heat is rejected to the high
temperature reservoir in step 3.
W
W
W
−1.3 × 103 J
e=
=
=
=
= 0.39
Q3 W3 W3 −3.3 × 103 J
21-46.
(a) In steps 2 and 4, no work is done since these are isovolumetric steps. During steps 1 and 3,
the work done depends on whether these are isothermal or adiabatic processes. At the beginning
of step 1, the temperature is given by the Ideal-Gas Law,
PV
1 1 = nRT1
T1 =
PV
(2 × 105 N/m 2 )(0.01 m3 )
1 1
=
= 240 K
nR
(1.00 mole) ( 8.31 J/(mole i K) )
At the end of step 1, the temperature is
PV
(1 × 105 N/m 2 )(0.03 m3 )
T2 = 2 2 =
= 360 K
nR
(1.00 mole) ( 8.31 J/(mole i K) )
Step 1 must be an adiabatic process, therefore the work done during step 1 is
449
CHAPTER
21
W1 =
p1V1 − p2V2
γ
=
(2 × 105 N/m 2 )(0.01 m3 ) − (1 × 105 N/m 2 )(0.03 m3 )
= −600 J
1.66
At the beginning of step 3, the temperature is
PV
(3 × 105 N/m 2 )(0.03 m3 )
T3 = 3 3 =
= 1100 K
nR
(1.00 mole) ( 8.31 J/(mole i K) )
At the end of step 3, the temperature is
PV
(4 × 105 N/m 2 )(0.01 m3 )
T4 = 4 4 =
= 480 K
nR
(1.00 mole) ( 8.31 J/(mole i K) )
The work done during step 3 is
p V − p4V4 (3 × 105 N/m 2 )(0.03 m3 ) − (4 × 105 N/m 2 )(0.01 m3 )
W3 = 3 3
=
= 3000 J
γ
1.66
Therefore, the net work done during one cycle is W = −600 J + 0 + 3000 J + 0 = 2400 J
(b) Steps 2 and 4 are isovolumetric, so the heat transferred during these steps is Q = nCV∆T.
Q2 = nCv ∆T = (1.00 mole) (12.5 J/(mole i K) ) (1100 − 360 K) = 9250 J
Q4 = nCv ∆T = (1.00 mole) (12.5 J/(mole i K) ) (240 − 480 K) = −3000 J
21-47.
(c) Since steps 1 and 3 are adiabatic, Q1 = Q3 = 0 J and ∆E = −W.
2.4 × 103 J
W
(d) e =
=
= 0.26
Q2 9.25 × 103 J
T
T − T2 298 K − 278 K
W
=1− 2 = 1
=
= 0.067
(a) e =
Q1
T1
T1
298 K
(b) W = eQ1
W
e
Q1 1 ⎛ W ⎞
1
= ⎜ ⎟=
(1.00 × 106 W) = 1.49 × 107 W
t
e ⎝ t ⎠ 0.067
Q W
= 1.49 × 107 W − 1.00 × 106 W = 1.39 × 107 W
Pwaste = 1 −
t
t
Q1 m
(c)
= c∆T
t
t
1.49 × 107 W
m Q1/t
=
=
= 180 kg/s
t
c∆T ( 4187 J/(kg i °C) )( 20°C )
Q1 =
21-48.
(a)
e=
Q
T
T − T2
W
=1− 2 =1− 2 = 1
Q1
Q1
T1
T1
and
Q2 = Q1 − W
W
⎛1
⎞
− W = W ⎜ − 1⎟
e
⎝e
⎠
⎞
Q2 W ⎛ T1
⎛ 520 + 273 K
⎞
= ⎜
− 1 ⎟ = 5 × 108 W ⎜
− 1 ⎟ = 3 × 108 W
t
t ⎝ T1 − T2
⎝ 520 − 30 K
⎠
⎠
Q2 =
⎛1
⎞
⎛ 1
⎞
(b) Q2 = W ⎜ − 1 ⎟ = 5 × 108 W ⎜
− 1 ⎟ = 1 × 109 W
⎝e
⎠
⎝ 0.33
⎠
450
CHAPTER
†21-49.
21
(a) For the carnot heat pump operating in cooling mode, the heat rejected each hour to the high
temperature reservoir is found from
Q2 T2
=
Q1 T1
⎛T ⎞
⎛ 300 K ⎞
6
Q1 = Q2 ⎜ 1 ⎟ = ( 8.4 × 106 J ) ⎜
⎟ = 8.57 × 10 J
⎝ 294 K ⎠
⎝ T2 ⎠
Therefore, the work input to the heat pump each hour is
W = Q1 − Q2 = 8.57 × 106 J − 8.4 × 106 J = 1.71 × 105 J
The rate of work done is
W 1.71 × 105 J ⎛ 1 h ⎞
P=
=
⎜
⎟ = 48 W
1h
t
⎝ 3600 s ⎠
(b) The air conditioner unit requires 950 W of work input, which is somewhat more than the heat
pump from part (a).
P
950 W
factor = ac =
= 20
Php
48 W
21-50.
The heat removed from ice of mass m in cooling it from 30°C to −5°C is
Q2 = mcwater ∆Twater + mLF + mcice ∆Tice
Q2 m
= ( cwater ∆Twater + LF + cice ∆Tice )
t
t
1 day
⎡ 4187 kg Ji ° C ( 30°C ) + 3.34 × 105
= 1 × 104 1 kg
day 86 400 s ⎣
= 5.45 × 104 W
The maximum efficiency of this ice making process is
Q
T
T − T2
W
e=
=1− 2 =1− 2 = 1
Q1
Q1
T1
T1
(
e =1−
†21-51.
)(
)
(
J
kg
) + ( 2230
J
kg i ° C
) ( 5°C ) ⎤⎦
T2
268 K
=1−
= 0.116
303 K
T1
Combining the efficiency relations and the fact that W = Q1 − Q2, then dividing by the time, we
find the power required.
W
Q2 / t
5.45 × 104 W
=
=
= 7.2 × 103 W
t
⎛1
⎞
⎛ 1
⎞
− 1⎟
⎜ − 1⎟
⎜
⎝e
⎠
⎝ 0.116
⎠
T2
260 + 273 K
=1−
= 0.34
(a) eturbine = 1 −
T1
540 + 273 K
eengine = 1 −
T3
38 + 273 K
=1−
= 0.42
T2
260 + 273 K
(b) The maximum net efficiency for the two components would occur if all of the heat that is
rejected from the turbine goes into the stream engine. In practice, heat would be lost to the
surroundings. Here, we’ll calculate the maximum efficiency,
451
CHAPTER
e=
21
Wnet Wturbine + Wengine eturbine Q 1 +eengine Q 2
=
=
Q1
Q1
Q1
⎛Q ⎞
⎛T ⎞
⎛ 260 + 273 K ⎞
= eturbine + eengine ⎜ 2 ⎟ = eturbine + eengine ⎜ 2 ⎟ = 0.34 + 0.42 ⎜
⎟ = 0.62
⎝ 540 + 273 K ⎠
⎝ Q1 ⎠
⎝T1 ⎠
If a single engine were utilized between the high and low temperature reservoirs,
T
38 + 273 K
e =1− 3 =1−
= 0.62
T1
540 + 273 K
21-52.
The two efficiencies are the same.
From the carnot efficiency relationships, the minimum amount of work needed to cool the water
from 30°C to 5°C can be determined.
Q
T
T − T2
W
and
e=
W = Q1 − Q2
=1− 2 =1− 2 = 1
Q1
Q1
T1
T1
⎛T ⎞
⎛T
⎞
⎛T
⎞
∆W = Q2 ⎜ 1 ⎟ − Q2 = Q2 ⎜ 1 − 1 ⎟ = −mc∆T ⎜ 1 − 1 ⎟
⎝ T2 ⎠
⎝ T2
⎠
⎝ T2
⎠
This assumes that the heat rejected from the low temperature reservoir has been used to cool the
water. Assuming that a sequence of Carnot engines is utilized, the above equation may be written
in infinitesimal form. The mass of one liter of water is one kilogram.
⎛T
⎞
dW = −mc ⎜ 1 − 1 ⎟ dT
⎝T
⎠
T1 ⎛ T
303 K
⎞
∫ dW = − mc ∫T2 ⎜⎝ T1 − 1⎟⎠ dT = mc ( T1 ln T − T ) 278 K
⎡
⎤
⎛ 303 K ⎞
= (1 kg) ( 4187 J/(kg i °C) ) ⎢ (303 K) ln ⎜
⎟ − 303 K + 278 K ⎥
⎝ 278 K ⎠
⎣
⎦
†21-53.
= 4570 J
The change in entropy as the water vaporizes is
6
dQ ∆Q mLV (1.0 kg ) ( 2.26 × 10 J/kg )
∆S1 = ∫
=
=
=
= 6.06 × 103 J/K
T
T
T
373 K
In addition, the water vapor expands as the vaporization occurs and does work against the
surrounding air. For the expanding gas, dQ = dW = pdV, so the change in entropy due to the
expansion is
V
dQ
pdV
dV
∆S 2 = ∫
=∫
= ∫ nR
= nR ln 2
T
T
V
V1
The initial volume of the water, assuming thermal expansion is negligible, is
V1 = m/ρ = 1.0 kg/(1000 kg/m3) = 0.0010 m3
The volume of 1.0 kg of water vapor at 1.0 atm is
1 mole
nRT (1.0 kg) 0.018 kg ( 8.31 J/(mole i K) ) (373 K)
V2 =
=
= 1.705 m3
p
1.01 × 105 N/m 2
(
)
Therefore,
∆S 2 = nR ln
V2
= (1.0 kg)
V1
(
3
1 mole
0.018 kg
m
) ( 8.31 J/(mole i K) ) ln 1.705
0.001 m
3
= 3.44 × 103 J/K
The net change in entropy is ∆S = ∆S1 + ∆S2 = 6.06 × 103 J/K + 3.44 × 103 J/K = 9.5 × 103 J/K
452
CHAPTER
21-54.
The net change in entropy is
⎛1
∆Q1 ∆Q2
1⎞
∆S =
+
= ∆Q1 ⎜ + ⎟
T1
T2
⎝ T1 T2 ⎠
The rate of change of entropy is then,
∆S ∆Q1 ⎛ 1
1⎞
=
⎜− + ⎟
∆t
∆t ⎝ T1 T2 ⎠
= ( 2.5 × 104
†21-55.
kcal
h
)(
4187J J
1 kcal
⎛
⎞
1h
+
)( 3600
⎟ = 9.6 W/K
s )⎜ −
21 + 273 K −5 + 273 K
1
1
⎝
⎠
From problem 49, the rate of heat removed from the inside is
∆Q2
J⎛ 1h ⎞
3
= 8.4 × 106 ⎜
⎟ = 2.33 × 10 W
h ⎝ 3600 s ⎠
∆t
The total entropy change is the sum of the entropy change on the inside and outside. The change
in heat on the inside is ∆Q2 where the temperature is T2; and ∆Q1 on the outside where the
temperature is T1.
∆Q1 ∆Q2
∆S =
−
T1
T2
∆S ∆Q1/∆t ∆Q2 /∆t 2.33 × 103 W + 950 W 2.33 × 103 W
=
−
=
−
= 3.0 W/K
∆t
T1
T2
27 + 273 K
21 + 273 K
21-56.
Typical body temperature is 37°C = 310K; and typical room temperature is around
20°C = 293 K.
⎛ 1
∆Q1 ∆Q2
1⎞
∆S =
+
= ∆Q1 ⎜ − + ⎟
T1
T2
⎝ T1 T2 ⎠
∆S ∆Q1 ⎛ 1
1⎞
=
⎜ − ⎟
∆t
∆t ⎝ T2 T1 ⎠
1
1 ⎞
−
⎟ = 0.44 W/K
⎝ 293 K 310 K ⎠
Using the Carnot efficiency relationships
Q
T
T − T2
W
e=
=1− 2 =1− 2 = 1
and
W = Q1 − Q2
Q1
Q1
T1
T1
= ( 2.0 × 103
21-57.
21
kcal
h
)(
4187J J
1 kcal
⎛
1h
)( 3600
s )⎜
we can find the rate of change of entropy for this system. The absolute temperatures are
T1 = 480 + 273 K = 753 K and T2 = 27 + 273 K = 300 K
∆Q1 ∆Q2
∆S =
+
T1
T2
The rate of change of the entropy for this system is
∆S ∆Q1/∆t ∆Q2 /∆t
=
+
∆t
T1
T2
=
( ∆W/∆t ) /e + (∆Q1 − ∆W )/∆t
T1
⎛ 1 ⎞
⎜
⎟ (2000 hp)
0.40 ⎠
=⎝
753 K
T2
(
745.7 W
1 hp
)
=
( ∆W/∆t ) /e + ( (∆W/e) − ∆W ) /∆t
T1
(2000 hp)
+
(
T2
745.7 W
1 hp
1
⎞
− 1⎟
) ⎛⎜⎝ 0.40
⎠
300 K
453
= 12400 W/K
CHAPTER
21-58.
21-59.
21-60.
21
5
∆Q mLF (1.0 kg ) ( 3.34 × 10 J/kg )
=
=
= 1.22 × 103 J/K
273 K
T
T
5
∆Q mLV (1.0 kg ) ( 5.8 × 10 J/kg )
∆S =
=
=
= 3.0 × 103 J/K
T
T
−79 + 273 K
5
∆Q mLF (1.00 kg ) ( 2.05 × 10 J/kg )
∆S melt =
=
=
= 150 J/K
Tmelt Tmelt
1083 + 273 K
∆S =
∆S vapor
†21-61.
The melting temperatures and heats of fusion are given in Table 20.4.
5
∆Q mLF (1.00 kg ) ( 3.99 × 10 J/kg )
∆S Al =
=
=
= 430 J/K
Tmelt Tmelt
660 + 273 K
∆S Fe
∆S Ag
5
mLF (1.00 kg ) ( 2.7 × 10 J/kg )
=
=
= 150 J/K
Tmelt
1535 + 273 K
4
mLF (1.00 kg ) ( 9.9 × 10 J/kg )
=
=
= 80 J/K
Tmelt
962 + 273 K
∆S Hg =
21-62.
†21-63.
21-64.
(1.00 kg ) ( 5.2 × 106 J/kg )
mLV
∆Q
=
=
=
= 1.8 × 103 J/K
Tvapor Tvapor
2567 + 273 K
4
mLF (1.00 kg ) (1.1 × 10 J/kg )
= 47 J/K
=
Tmelt
−39 + 273 K
The change in entropy seems to decrease with increasing atomic number. The largest value
occurs for aluminum (Z = 13) and the smallest value occurs for mercury (Z = 80).
As the parachute descends, it does work against air friction and absorbs heat.
W = fs = ∆Q
Because the parachutist is failing at a constant rate, the upward force due to air friction f is equal
in magnitude to the downward force due to gravity, mg. The rate of work done is thus,
W
∆y
∆Q
= mg
= mgv =
t
∆t
∆t
The rate of change of entropy is then
2
∆S ∆Q / ∆t mgv ( 80 kg ) ( 9.81 m/s ) (5.0 m/s)
=
=
=
= 13 W/K
∆t
T
T
20 + 273 K
As the automobile moves, it does work against the various sources of friction and absorbs heat.
W = fs = ∆Q
The rate of work done is thus,
W
∆x ∆Q
= f
=
t
∆t
∆t
The rate of change of entropy is then
∆S ∆Q / ∆t 12 × 103 W
=
=
= 41 W/K
∆t
T
20 + 273 K
(a) The work done by the brakes is equal to the change in kinetic energy. This is also equal to the
heat absorbed by the brakes.
454
CHAPTER
W = ∆K = 12 mv1 = ∆Q
dQ
=
T
∫ dS = ∫
=
=
1
2
∫
mcdT
T
∆Q T2
ln
=mc ln 2 =
T
T1
∆T T1
mv12 T2
ln
∆T
T1
1
2
(2100 kg) ⎣⎡ ( 80
(b) ∆S = ∆Sair
km
h
1000 m
1h
⎤
) ( 3600
s )( 1 km ) ⎦
2
40 K
∆Q ∆Q T2
+ ∆S brake =
+
ln
T1
∆T T1
⎛ 60 + 273 K ⎞
3
ln ⎜
⎟ = 1.66 × 10 J/K
⎝ 20 + 273 K ⎠
So, the additional entropy is
21-66.
21-67.
21-68.
†21-69.
(2100 kg) ⎡⎣ ( 80
km
h
1000 m
1h
⎤
) ( 3600
s )( 1 km ) ⎦
2
= 1.8 × 103 J/K
20 + 273 K
The potential energy of the water is dissipated as heat at a rate
∆m
∆Q
gy =
∆t
∆t
The rate of change of entropy is then
⎛ ∆m ⎞
gy
( 5700 m3 /s ) 10001 mkg3 ( 9.81 m/s2 ) (50 m) = 9.5 × 106 W/K
∆S ∆Q / ∆t ⎜⎝ ∆t ⎟⎠
=
=
=
∆t
T
T
20 + 273 K
Entropy of the water increases as it does work on the diver to change his kinetic energy. That
energy is then dissipated as heat in the water.
∆Q mgy (70 kg)(9.81 m/s 2 )(36 m)
∆S =
=
=
= 84 J/K
295 K
T
T
−∆Q1 ∆Q2
∆S =
+
T1
T2
∆Sair =
†21-65.
1
2
∆S −∆Q / ∆t ∆Q / ∆t ∆Q ⎛ 1
1⎞
=
+
=
⎜− + ⎟
∆t
T1
T2
∆t ⎝ T1 T2 ⎠
1
1
⎛
⎞
= 1.8 × 103 W ⎜ −
+
⎟ = 0.94 W/K
⎝ 21 + 273 K −18 + 273 K ⎠
The melting represents an increase in entropy. The change in entropy is
∆Q mLF (20 kg)2.9 × 104 J/kg
∆S =
=
=
= 965 J/K
328 + 273 K
T
T
In the free expansion of the gas, we remember from the First Law of Thermodynamics that
∆E = ∆Q − W = ∆Q − p∆V
Since the process is an isothermal one, ∆E = 0 J and ∆Q = p∆V
If we consider the system by adding and removing infinitesimally small amounts of heat, the
change in entropy is
dQ pdV
dV
dS =
=
= nR
T
T
V
V′
dV
V′
2.0 liters
∆S = ∫ dS = ∫ nR
= nR ln
= (1.00 mole )( 8.31 J/(mole i K) ) ln
= 5.8 J/K
V
V
V
1.0 liter
455
21
CHAPTER
21-70.
21
At its initial temperature, the silver is solid since it is below its melting temperature. Heat will be
transferred from the silver to the water until the equilibrium temperature is reached.
mW cW ∆TW = mAg cAg ∆TAg
heat gained by the water
heat lost by the silver
mW cW (T − TW ) = mAg cAg (TAg − T )
mW cW T − mW cW TW = mAg cAgTAg − mAg cAgT
T =
=
21-71.
mW cW TW + mAg cAgTAg
mW cW + mAg cAg
5.0 kg ( 4187 J/(kg i °C) ) (20°C) + (0.50 kg) ( 240 J/(kg i °C) ) (950°C)
5.0 kg ( 4187 J/(kg i °C) ) + (0.50 kg) ( 240 J/(kg i °C) )
= 25.3°C = 298 K
The heat lost by the silver has been added to the water, so there is a change in entropy of the
water.
∆Q mW cW ∆TW 5.0 kg ( 4187 J/(kg i °C) ) (25.3 − 20°C)
∆S =
=
=
= 370 J/K
298 K
T
T
(a) In the limit of adding or removing infinitesimally small quantities of heat from a system, we
have
dQ
mcdT
T
∫ dS = ∫ T = ∫ T = mc ln T12
S 2 − S1 = ∆S = mc ln
(b) ∆S = mc ln
21-72.
†21-73.
T2
T1
T2
⎛ 80 + 273 K ⎞
= (1.0 kg) ( 4187 J/(kg i °C) ) ln ⎜
⎟ = 780 J/K
T1
⎝ 20 + 273 K ⎠
The total entropy change is ∆S = ∆S1 + ∆S2 where ∆S1 is the change in entropy of the boiling
water and ∆S2 is the change in entropy of the ice cubes.
T
⎛ 40 + 273 K ⎞
∆S1 = mc ln 2 = (1.0 kg) ( 4187 J/(kg i °C) ) ln ⎜
⎟ = −734 J/K
T1
⎝ 100 + 273 K ⎠
mLF
T
(0.5 kg)(3.34 × 105 J/kg)
⎛ 313 K ⎞
∆S 2 =
+ mc ln 2 =
+ (0.5 kg) ( 4187 J/(kg i °C) ) ln ⎜
⎟
273 K
T0
T0
⎝ 273 K ⎠
= 900 J/K
∆S = ∆S1 + ∆S2 = −734 J/K + 900 J/K = 164 J/K
Because we are mixing equal amounts of warm and cool water, the final temperature will be
50°C, the mid-point between the two temperatures. The total entropy change is ∆S = ∆S1 + ∆S2
where ∆S1 is the change in entropy of the initially cooler water and ∆S2 is the change in entropy
of the initially warmer water.
T
50 + 273 K ⎞
kg
∆S1 = mc ln 2 = (1.0 liter) 11liter
( 4187 J/(kg i °C) ) ln ⎛⎜
⎟ = 408 J/K
T1
⎝ 20 + 273 K ⎠
T
50 + 273 K ⎞
kg
∆S 2 = mc ln 2 = (1.0 liter) 11liter
( 4187 J/(kg i °C) ) ln ⎛⎜
⎟ = −371 J/K
T1
⎝ 80 + 273 K ⎠
(
)
(
)
∆S = ∆S1 + ∆S2 = 408 J/K − 371 J/K = 37 J/K
456
CHAPTER
21-74.
(a) In an isobaric process, the heat transferred into or out of a system is given by ∆Q = nCp∆T. If
the heat is transferred in infinitesimally small amounts, the the change in entropy is
nCp dT
T
dQ
∆S = ∫ dS = ∫
=∫
= nCp ln 2
T
T
T1
(b) In an isovolumetric process, the heat transferred into or out of a system is given by ∆Q =
nCV∆T.
If the heat is transferred in infinitesimally small amounts, the the change in entropy is
nC dT
T
dQ
∆S = ∫ dS = ∫
=∫ V
= nCV ln 2
T
T
T1
21-75.
21-76.
†21-77.
In the free expansion of the gas, we remember from the First Law of Thermodynamics that
∆E = ∆Q − W = ∆Q − p∆V
Because the process is an isothermal one, ∆E = 0 J and ∆Q = p∆V
If we consider the system by adding and removing infinitesimally small amounts of heat, the
change in entropy is
dQ pdV
dV
dS =
=
= nR
T
T
V
′
V
dV
V′
∆S = ∫ dS = ∫ nR
= nR ln
V
V
V
∆E = ∆Q − W = 3000 J − 2000 J = 1000 J
The work done is W = p∆V = nR∆T, therefore
p∆V
W
2000 J
∆T =
=
=
= 240 K
nR
nR (1 mole) ( 8.31 J/(mole i K) )
For n moles of a diatomic gas at temperature T, the internal energy is E =
5
nRT
2
The change in internal energy is ∆E = ∆Q − W = −W
Because work is done on the system, the sign for the work is negative, W = −2500 J
Combining the above equations, gives
5
∆E = nR∆T = −W
2
−2 ( −2500 J )
−2W
∆T =
=
= 120 K
5nR
5 (1 mole )( 8.31 J/(mole i K) )
21-78.
(a) W = 12 ( p1 − p2 )(V2 − V1 ) + p2 (V2 − V1 ) = 12 ( p1 + p2 )(V2 − V1 )
CV
( p2V2 − p1V1 ) = 32 ( p2V2 − p1V1 )
R
(c) Q = ∆E + W = 32 ( p2V2 − p1V1 ) + 12 ( p1 + p2 )(V2 − V1 )
(b) ∆E = CV (T2 − T1 ) =
= 2( p2V2 − p1V1 ) + 12 ( p1V2 − p2V1 )
21-79.
21
(a) p1 =
nRT (0.20 mole) ( 8.31 J/(mole i K) )( 300 K )
=
= 4.16 × 105 N/m 2
3
V1
0.0012 m
p2 =
nRT (0.20 mole) ( 8.31 J/(mole i K) )( 300 K )
=
= 2.27 × 105 N/m 2
V2
(0.0012 + 0.0010) m3
457
CHAPTER
21
(b) In an isothermal compression, the work done on the gas is
V
W = p2V2 ln 3
V2
21-80.
⎛ 0.0012 m3 ⎞
= ( 2.27 × 105 N/m 2 )( 0.0022 m3 ) ln ⎜
= −3.0 × 102 J
3 ⎟
0.0022
m
⎝
⎠
Also, since ∆E is equal to 0 J in such a process, ∆Q = W = − 3.0 × 102 J
(a) At 0°C, one mole of an ideal gas occupies 22.4 liters.
p2V2 = nRT2
1 mole
nRT 4.0 liters ( 22.4
liters ) ( 8.31 J/(mole i K) )( 60 + 273 K )
=
= 0.0049 m3 = 4.9 liters
V2 =
p
1.01 × 105 N/m 2
(b) W = p∆V = (1.01 × 105 N/m2)(0.9 × 10−3 m3) = 91 J
∆E = Q − W = nCp ∆T − W
1 mole
= (4.0 liters) ( 22.4
29.1 J/(mole i K) ) (60 K) − 91 J = 220 J
liters ) (
1 mole
Q = (4.0 liters) ( 22.4
29.1 J/(mole i K) ) (60 K) = 310 J
liters ) (
21-81.
(
)
745.7 W
W W/∆t 20 hp 1 hp
=
=
= 0.24 = 24%
Q Q/∆t
6.3 × 104 W
Q
e =1− 2
Q1
e=
Q2
=1− e
Q1
21-82.
21-83.
Q2
Q
= (1 − e ) 1 = (1 − 0.24)(6.3 × 104 W) = 4.8 × 104 W
∆t
∆t
869 W
W W / ∆t
e=
=
=
= 0.23 = 23 %
Q
Q / ∆t
⎛ 10.8 liters ⎞ ⎛ 1min ⎞
4
( 2.1 × 10 J/liter ) ⎜⎝ min ⎟⎠ ⎜⎝ 60 s ⎟⎠
Q2 T2
=
Q1 T1
∆Q2 /∆t T2
=
∆Q1 /∆t T1
∆Q1 ∆Q2 ⎛ T1 ⎞
7
=
⎜ ⎟ = ( 3.8 × 10
∆t
∆t ⎝ T2 ⎠
W = Q1 − Q2
21-84.
J
h
)(
1h
3600 s
32 + 273 K
) ⎛⎜ 21 + 273 K ⎞⎟ = 1.095 × 104 W
⎝
⎠
∆W ∆Q1 − ∆Q2
1h
=
= 1.0950 × 104 W − ( 3.8 × 107 hJ ) ( 3600
= 4.0 × 102 W
s)
∆t
∆t
Q
T
30 + 273 K
W
(a) e =
=1− 2 =1− 2 =1−
= 0.61
500 + 273 K
Q1
Q1
T1
W
(b) e =
Q1
Q1 =
W 1.5 × 103 J
=
= 2460 J
e
0.61
458
CHAPTER
†21-85.
(a) Beginning with the point at the upper left, the gas undergoes an isobaric expansion in step 1
as the volume is increased, followed by a isovolumetric reduction of pressure in step 2 as the
temperature is reduced. The gas is then compressed isobarically in step 3 by reducing the volume,
before an isovolumetric increase in pressure in step 4 by increasing the temperature.
(b) Step 1: W1 = p∆V = (1.5 × 105 N/m2)(0.021 − 0.007 m3) = 2100 J
Step 2: W2 = 0 J (no work is done in an isovolumetric process)
Step 3: W3 = p∆V = (0.5 × 105 N/m2)(0.007 − 0.021 m3) = −700 J
Step 4: W4 = 0 J (no work is done in an isovolumetric process)
(c) Step 1: At the beginning of step 1, the temperature is
PV
1 1 = nRT1
T1 =
PV
(1.5 × 105 N/m 2 )(0.007 m3 )
1 1
=
= 126 K
nR
(1.00 mole) ( 8.31 J/(mole i K) )
At the end of step 1, the temperature is
PV
(1.5 × 105 N/m 2 )(0.021 m3 )
T2 = 2 2 =
= 379 K
nR
(1.00 mole) ( 8.31 J/(mole i K) )
Q1 = nCp∆T1 = (1 mole)(20.8 J/(mole • K))(379 K − 126 K) = 5260 J
Step 2: At the end of step 2, the temperature is
PV
(0.5 × 105 N/m 2 )(0.021 m3 )
T3 = 3 3 =
= 126 K
nR
(1.00 mole) ( 8.31 J/(mole i K) )
Q2 = nCV∆T2 = (1 mole)(12.5 J/(mole • K))(126 K − 379 K) = −3160 J
Step 3: At the end of step 3, the temperature is
PV
(0.5 × 105 N/m 2 )(0.007 m3 )
T4 = 4 4 =
= 42 K
nR
(1.00 mole) ( 8.31 J/(mole i K) )
21-86.
Q3 = nCp∆T3 = (1 mole)(20.8 J/(mole • K))(42 K − 126 K) = −1750 J
Step 4: Q4 = nCV∆T4 = (1 mole)(12.5 J/(mole • K))(126 K − 42 K) = 1050 J
In part (b), the net work done is 2100 − 700 J = 1400 J. From part (c), the sum of the four
quantities of heat also equals 1400 J.
(d) The efficiency of this four-step cycle is
W
1400 J
e=
=
= 0.44 or 44%
Q2 3160 J
Q2 T2
=
Q1 T1
Q2 = Q1
21-87.
21
T2
200 K
= (1200 J)
= 400 J
T1
600 K
W = Q1 − Q2 = 1200 J − 400 J = 800 J
Q
T
W
25 + 273 K
e=
=1− 2 =1− 2 =1−
= 0.48 or 48%
Q1
Q1
T1
300 + 273 K
Q1 W / t 1 × 107 W
=
=
= 2 × 107 W
0.48
t
e
The rate of waste heat is then,
Q2 Q1 W
=
−
= 2 × 107 W − 1 × 107 W = 1 × 107 W
t
t
t
459
CHAPTER
21-88.
21
Q2 T2
=
Q1 T1
Q2 = Q1
( 4 + 273 ) K = 4727 J
T2
= (5.0 × 103 J)
T1
( 20 + 273 ) K
W = Q1 − Q2 = 5.0 × 103 J − 4727 J = 270 J
†21-89.
The boiling point temperatures and heats of vaporization are given in Table 20.4.
5
∆Q mLV (1.00 kg ) ( 2.00 × 10 J/kg )
∆S N =
=
=
= 2600 J/K
Tboil
Tboil
−196 + 273 K
∆SO =
5
∆Q mLV (1.00 kg ) ( 2.1 × 10 J/kg )
=
=
= 2300 J/K
Tboil
Tboil
−183 + 273 K
5
∆Q mLV (1.00 kg ) ( 4.5 × 10 J/kg )
∆S H =
=
=
= 2.25 × 104 J/K
Tboil
Tboil
−253 + 273 K
21-90.
Hydrogen is largest and oxygen the smallest.
(a) Because both the lead and the ice are at their respective melting temperatures, some of the ice
will melt into water and the lead will solidify. Because the amount of heat removed from the lead
in solidifying is much less than the heat required to melt all of the ice, the equilibrium
temperature will be 273 K. As a check, let’s calculate how much of the ice melts.
mlead clead (Tlead − T ) + mlead LF, lead = mice LF
heat lost from the lead
mice =
(b)
∆S
(c)
21-91.
(a)
(b)
heat gained by ice
mlead clead (Tlead − T ) + mlead LF, lead
LF
(1.0 kg) ⎡⎣ (130 J/(kg i °C) )( 328 °C ) + 2.32 × 104 J/kg ⎤⎦
=
= 0.20 kg
3.34 × 105 J/kg
The entropy of the lead decreases by
T
273 K ⎞
⎛
= mc ln 2 = (1.0 kg) (130 J/(kg i K) ) ln ⎜
⎟ = −100 J/K
T1
⎝ 328 + 273 K ⎠
∆Q mLF (0.20 kg)(3.34 × 105 J/kg)
∆S water =
=
=
= 245 J/K
273 K
T
T
Q
800 J
S = 1 =
= 1.1 J/K
T1
700 K
Q2 T2
=
Q1 T1
Q2 = Q1
300 K
T2
= (800 J)
= 340 J
700 K
T1
Q2
340 J
=
= 1.1 J/K
300 K
T2
(c) The entropy remains constant during adiabatic processes. ∆S = 0 J/K
S =
460