CHAPTER 19: Heat and the First Law of Thermodynamics
Responses to Questions
26. At night, the Earth cools primarily through radiation of heat back into space. Clouds reflect energy
back to the Earth and so the surface cools less on a cloudy night than on a clear one.
30. A thermos bottle is designed to minimize heat transfer between the liquid contents and the outside
air, even when the temperature difference is large. Heat transfer by radiation is minimized by the
silvered lining. Shiny surfaces have very low emissivity, e, and thus the net rate of energy flow by
radiation between the contents of the thermos and the outside air will be small. Heat transfer by
conduction and convection will be minimized by the vacuum between the inner and outer walls of
the thermos, since both these methods require a medium to transport heat.
Solutions to Problems
58. The heat conduction rate is given by Eq. 19-16a.
Q
t
 kA
T1  T2
l
  380 J sgmgC    0.010 m 
2
 460C  22C 
0.45m
 116 W  120 W
61.
(a)
The rate of heat transfer due to radiation is given by Eq. 19-17. We assume that each
teapot is a sphere that holds 0.55 L. The radius and then the surface area can be found from that.
1/ 3
V  r
4
3
Q
t
3
 3V 
 r

 4 
 3V 
 S . A.  4 r  4 

 4 

 3V 

 4 

2/3
2
  A T14  T24  4 
2/3
T
4
1
 T24

3
2
 Q 
 5.67  10 8 W   3  0.55  10 m  

4

0.70







m 2 gK 4  
4
 t ceramic


 14.13W  14 W
 Q    Q 
 0.10   2.019 W  2.0 W






 t shiny  t ceramic  0.70 
2/3
 368 K  4   293 K  4 


(b)
We assume that the heat capacity comes primarily from the water in the teapots, and
ignore the heat capacity of the teapots themselves. We apply Eq. 19-2, along with the results from part
(a). The mass is that of 0.55 L of water, which would be 0.55 kg.
Q  mcT  T 
 T ceramic 
1  Q 
telapsed


mc  t  radiation
14.13W
J 

 0.55 kg   4186

kggC 

1
 T shiny   T ceramic 
7
1800s  
11C
1.6 C
62. For the temperature at the joint to remain constant, the heat flow in both rods must be the same.
Note that the cross-sectional areas and lengths are the same. Use Eq. 19-16a for heat conduction.
 Q    Q   k A Thot  Tmiddle  k A Tmiddle  Tcool
 
 
Cu
Al
l
l
 t Cu  t  Al
Tmiddle 
kCuThot  kAlTcool
kCu  kAl


 380 J sgmgC 225C    200 J sgmgC0.0C 
380 J sgmgC  200 J sgmgC
 147C
63. (a) The cross-sectional area of the Earth, perpendicular to the Sun, is a circle of radius REarth , and
2
so has an area of  REarth
. Multiply this area times the solar constant to get the rate at which
the Earth is receiving solar energy.
Q
t

2
  REarth
 solar constant    6.38 106 m
 1350 W m   1.73 10
2
2
17
W
(b) Use Eq. 19-18 to calculate the rate of heat output by radiation, and assume that the
temperature of space is 0 K. The whole sphere is radiating heat back into space, and so we use the
2
full surface area of the Earth, 4 REarth
.
Q
t
1/ 4
 e AT
4
Q 1 
 T 

 t  A 
1/ 4


1

 1.73  1017 J s 
2
8
2
4
6

1.0
5.67

10
W
m
K
4

6.38

10
m
g
 
 
 
 278 K  5C
64. This is an example of heat conduction. The temperature difference can be calculated by Eq. 19-16a.
 95 W  5.0  104 m
Q
T1  T2
Pl
 P  kA
 T 

 5.0C
2
t
l
kA
0.84 J sgmgCo 4 3.0  102 m


 


CHAPTER 20: Second Law of Thermodynamics
Responses to Questions
1.
Yes, mechanical energy can be transformed completely into heat or internal energy, as when an
object moving over a surface is brought to rest by friction. All of the original mechanical energy is
converted into heat. No, the reverse cannot happen (second law of thermodynamics) except in
very special cases (reversible adiabatic expansion of an ideal gas). For example, in an explosion, a
large amount of internal energy is converted into mechanical energy, but some internal energy is
lost to heat or remains as internal energy of the explosion fragments.
2.
Yes, you can warm a kitchen in winter by leaving the oven door open. The oven converts electrical
energy to heat and leaving the door open will allow this heat to enter the kitchen. However, you
cannot cool a kitchen in the summer by leaving the refrigerator door open. The refrigerator is a
heat engine which (with an input of work) takes heat from the low-temperature reservoir (inside
the refrigerator) and exhausts heat to the high-temperature reservoir (the room). As shown by the
second law of thermodynamics, there is no “perfect refrigerator,” so more heat will be exhausted
into the room than removed from the inside of the refrigerator. Thus, leaving the refrigerator door
open will actually warm the kitchen.
5.
A 10ºC decrease in the low-temperature reservoir will give a greater improvement in the efficiency
of a Carnot engine. By definition, TL is less than TH, so a 10ºC change will be a larger percentage
change in TL than in TH, yielding a greater improvement in efficiency.
11. The isothermal process will result in a greater change in entropy. The entropy change for a
reversible process is the integral of dQ/T. Q = 0 for an adiabatic process, so the change in entropy is
also 0.
Solutions to Problems
1.
The efficiency of a heat engine is given by Eq. 20-1a. We also invoke energy conservation.
e
W
QH

W
W  QL

2600 J
2600 J  7800 J
 0.25  25%
6. (a) For the net work done by the engine to be positive, the path must be carried out clockwise. Then
the work done by process bc is positive, the work done by process ca is negative, and the work done by
process ab is 0. From the shape of the graph, we see that Wbc  Wca .
(b) The efficiency of the engine is given by Eq. 20-1a. So we need to find the work done and the
heat input. At first glance we might assume that we need to find the pressure, volume, and
temperature at the three points on the graph. But as shown here, only the temperatures and
the first law of thermodynamics are needed, along with ratios that are obtained from the ideal
gas law.
ab:
Wab  PV  0 ; Qab  Eint  nCV Tb  Ta   23 nR Tb  Ta   0
ab
bc:
Eint  nCV Tc  Tb   0 ; Qbc  Wbc  nRTb ln
bc
ca:
Wca  Pa Va  Vc  
nRTa
Va
Vc
Vb
 nRTb ln
Vc
Va
 nRTb ln

V 

T 

a


a
Tc
Ta
0
Va  Vc   nRTa  1  c   nRTa  1  c   nR Ta  Tc 
V
T

Qca  nCP Ta  Tc   25 nR Ta  Tc   0
Tc
T
T
 nR  Ta  Tc 
Tb ln c   Ta  Tc 
Tb ln b  Ta  Tb 
W
W  Wca
Ta
Ta
Ta
e
 bc



T
T
T
Qinput Qab  Qbc 3 nR T  T  nRT ln c
3
3
T  Ta   Tb ln c
T  Ta   Tb ln b
 b a
2
2 b
2 b
b
Ta
Ta
Ta
nRTb ln
423 K
  273 K  423 K 
273 K

 0.0859  8.59%
423 K
3
423 K  273 K    423 K  ln
2
273 K
 423 K  ln
Of course, individual values could have been found for the work and heat on each process, and
used in the efficiency equation instead of referring everything to the temperatures.
8.
The maximum efficiency is the Carnot efficiency, given in Eq. 20-3.
e  1
TL
TH
 1
 365  273 K

 550  273 K
0.225 or 22.5%
We assume that both temperatures are measured to the same precision – the nearest degree.
12. This is a perfect Carnot engine, and so its efficiency is given by Eq. 20-1a and Eq. 20-3. Use these
two expressions to solve for the rate of heat output.
e  1
TL
TH
QL t  W
W
W
 45  273 K
 0.3416 e 


QH W  QL
 210  273 K
t 1 e  1   950 W 1 0.3416  1  1831W  1800 W
 1
18. The heat input must come during the isothermal expansion.
From section 20-3, page 534, we have
V
QH  nRTH ln b  nRTH ln 2. Since this is a Carnot cycle, we may
Va
QL  W 1 e  1
p
a
TH
use Eq. 20-3 combined with Eq. 20-1.
b
e  1
TL
TH

W
QH
d

T T 
T T 
W  QH  H L    nRTH ln 2   H L   nR  TH  TL  ln 2
 TH 
 TH 
The adiabatic relationship between points b and c and the ideal
gas law are used to express the temperature ratio in terms of the volume ratio.
TL
c
V

b b

c c
PV  PV
nRTH

Vb

b
V 

nRTL
Vc

c
V

V 
 TH  TL  c 
 Vb 
 1

 TL 5.7 2 / 3

W  nR  TH  TL  ln 2  nRTL 5.7 2 / 3  1 ln 2 
TL 
W
nR ln 2 5.7
2/3

 1

920 J
1.00 mol 8.314 J

molgK  ln 2 5.7 2 / 3  1
 72.87 K  73 K
TH   72.87 K  5.7 2 / 3  232.52 K  233 K
20.
(a)
We use the ideal gas law and the adiabatic process
relationship to find the values of the pressure and volume at each of the four points.
Pa  8.8atm ; Ta  623 K ;
Va 
nRTa 1.00 mol  0.0821Lgatm molgK  623 K 

Pa
8.8atm
 5.81L  5.8 L
Tb  623K ; Vb  2Va  2  5.81L   11.62 L  11.6 L
Pb  Pa
Va
Vb
 12 Pa  4.4 atm


 PV

Tc  483 K ; PV
b b
c c
1
nRTb
Vb
 T   1
 623 K 
Vc  Vb  b   11.62 L  

 483 K 
 Tc 
Pc 
nRTc
Vc

Vb 
nRTc
Vc
Vc 
3/ 2
1.00 mol  0.0821Lgatm
 17.02 L  17.0 L
molgK  483 K 
17.02 L
1
 T   1
 623 K 
Td  483 K ; Vd  Va  a    5.81L  

 483 K 
 Td 
Pd 
nRTd
Vd
To summarize:

3/ 2
 8.51L  8.5 L
1.00 mol  0.0821Lgatm molgK  483 K 
8.51L
 2.33atm  2.3atm
 4.66 atm  4.7 atm
Pa  8.8atm ; Va  5.8 L ; Pb  4.4 atm ; Vb  11.6 L
Pc  2.3atm ; Vc  17.0 L ; Pd  4.7 atm ; Vd  8.5 L
(b) Isotherm ab:
Eint  0 ;
ab
Qab  Wab  nRTa ln
Vb
Va
 1.00 mol 8.314 J molgK  623K  ln 2  3590 J  3600 J
Qbc  0 ;
Adiabat bc:
Eint  nCV Tc  Tb   23 nR Tc  Tb  
3
2
1.00 mol 8.314 J
molgK  140 K 
bc
 1746 J  1700 J ; Wbc  Qbc  Eint  1746 J  1700 J
bc
Isotherm cd:
Eint  0 ;
cd
Qcd  Wcd  nRTc ln
Vd
Vc
 1.00 mol 8.314 J molgK  483K  ln 12  2783J  2800 J
Qda  0 ;
Adiabat da:
Eint  nCV Tc  Tb   23 nR Tc  Tb  
3
2
1.00 mol 8.314 J
molgK 140 K 
bc
 1746 J  1700 J ; Wbc  Qbc  Eint  1746 J  1700 J
bc
To summarize: ab : Eint  0 ;
Q  3600J ; W  3600J
bc: Eint  1700 J ; Q  0 ;
cd: Eint  0 ;
W  1700 J
Q  2800 J ; W  2800 J
da: Eint  1700 J ; Q  0 ;
(c) Using Eq. 20-1: e 
W
Qinput
Using Eq. 20-3: e  1 
TL
TH

3590 J  1746 J  2783J  1746 J
 1
3590 J
 273  210 K
 0.2247 
 273  350 K

W  1700 J
807 J
3590 J
 0.2248  0.22
0.22
The slight disagreement is due to rounding of various calculations.
26. The coefficient of performance for an ideal refrigerator is given by Eq. 20-4c, with temperatures in
Kelvins. Use that expression to find the temperature inside the refrigerator.
COP 
TL
 TL  TH
TH  TL
COP
1  COP
  32  273 K 
5.0
6.0
 254 K  19C
27. The efficiency of a perfect Carnot engine is given by Eq. 20-1a and Eq. 20-3. Equate these two
expressions to solve for the work required.
e  1
TL
TH
 ; e

W
QH
 1
TL
TH

W
QH

TL 

TH 
 W  QH 1 
TL 

0  273 

  3100 J   1 
  231.2 J  230 J

TH 
 22  273 

TL 
(a) W  QH  1 
(b) W  QH  1 


 15  273   388.8J  390 J
  3100 J   1 


TH 
22  273 

31. The coefficient of performance is the heat removed from the low-temperature area divided by the
work done to remove the heat. In this case, the heat removed is the latent heat released by the
freezing ice, and the work done is 1.2 kW times the elapsed time. The mass of water frozen is its
density times its volume.
VLf

W
W
Pt
 COP  Pt
 7.01200 W  3600s 
V

 0.0908m 3  91L
3
3
5
 Lf
1.0  10 kg m  3.33  10 J kg 
COP 
QL

mLf
