Chapter 19
Questions: 1, 2, 3, 7, 10, 12, 14, 15, 19, 22
Problems: 1, 5, 12, 21, 22, 25, 35, 40, 51, 53, 55, 64
Questions
1. There are many ways that materials are in the same macroscopic state,
temperature, density, color, etc. and have di¤erent microscopic states, for
example di¤erent crystalline structure, di¤erent constituents, etc.
2. If the microscopic states are identical then the macroscopic states
(which are ensemble averages of these states) must be the same.
3. Since systems A and B are not in equilibrium, TA 6= TB : Additionally
systems B and C are not in equilibrium hence TB 6= TC : However there is
nothing we can state relating TA and TC :
7. Any standard thermometer would be subject to subtle changes in
pressure. The triple point of water is unique and can only be obtained at
one temperature and pressure.
10. If a thermometer is exposed to sunlight; then it is absorbing radiative
energy from the Sun and well as being in thermal contact with the air. Hence,
it will have a greater temperature than the air whose di¤erence will depend
on the thermometer’s emissivity as well as its heat transfer with the air.
12. Liquid water has a very large speci…c heat compared to the material
in its surroundings. So it requires more heat to change its temperature thus
moderating the temperature changes of the local surroundings.
14. The layers of Al or Cu provide better heat conduction properties so
that the temperature of the cooking surface is more uniform.
15. In baking, the air circulates inside the oven between the cooking
elements and the item being cooked. It is this circulation or “convection”
that dominates heat transfer during baking. When you broil, the item to be
cooked is close to the cooking element and it is radiation that dominates.
19. Double glazed windows have a layer of air between the two layers of
glass. Since air is poor heat conductor, this provide better insulation.
22. The amount of radiation that the Earth receives from the Sun is a
small percentage of the total radiation emitted by the Sun. This percentage
2
is the cross section of the Earth, RE
where RE is the radius of the Earth,
divided by 4 R2 where R is the distance from the Earth to the Sun. The
Earth also radiates but at a much lower temperature.
Problems
1. The …rst egg has 6 possible locations, the second egg 5, the third 4
etc. So the number of microscopic states is 6! = 720:
1
(b) For three identical eggs with 6 possible locations the number of microscopic states is 6 5 4= (3!) = 20:
5. When TC = TF then
5
5
(TF 32) ! T = (T
9
9
9T = 5T 160 ! 4T = 160
T =
40 either C or F:
TC =
32)
12. From equation (19.1) we know that
T = 273:16
P
;
P3
where P3 is the pressure at the triple point of water. Now the pressures are
given by P = gh and the quantity g cancels. Thus
57:8
h
= 273:16K
= 263:14K;
h3
60
T = (263:14 273:15) C = 10:01 C.
T = 273:16
21. (a) The speci…c heat of Cu is cCu = 386J=kg K: Hence
Q = mCu cCu T = :8
386
(90
15) = 23:2kJ
(b) If the pan contains 1kg of water then
Q = mCu cCu T + mw cw T = 23:2 + 4184
75 = 337kJ
(c) If the pan contains 4kg of mercury then
Q = mCu cCu T + mHg cHg T = 23:2 + 4
140
75 = 65:2kJ
22. (a) Since the heat transfer and masses are the same,
32
Tw
= 4184
Tx
76
cx Tx = cw Tw ! cx = cw
From table 19-1, the material is aluminium.
(b) The total heat transfer is
Q = :1
4184
2
12 = 5:02kJ:
20
= 896:6
20
Hence the rate of heat transfer is
Q
5:02kJ
=
= 83:7W
t
60
H=
25. From equation (19-4) the heat transfer is
Q = mc T:
Thus the rate of heat transfer is
Q
T
= mc
:
t
t
P =
In this problem the mass of water is given by
m=
1
c
1
P
500J=s
=
= 2:39kg:
T= t
4:184J=gmK 1=20K=s
35. Balancing the heat transfer yields
mCu cCu TCu =
mw cw Tw :
Using the speci…c heats from table 19-1 we …nd
mCu =
1
4184 5
kg ' 197g:
386 275
40. The rate of heat transfer via conduction is given by
H=
kA
T
:
x
The same insulating value implies that T and A are identical for the same
rate of heat transfer. Thus the ratio of thicknesses is found from
kcon
=
xcon
xcon
kf ib
;
xf ib
3:5
=
in ' 6:9f t:
:042
51. Since the insulation is such that H= T = 370W= C and heat production of 40 people is H = 4000W;
T = 4000=370 = 10:8 C.
3
Adding this to the outside temperature of 12 C, we …nd that the inside
temperature is approximately 23 C.
53. From the Stephan-Boltzmann law the radiated power is
P = e AT 4 = 5:67
10
8
325
10
(900)4 = 1:21kW
4
Since the input power is 1:5kW , the fraction of power loss due to radiation
is
1:21
= 80:6%
f=
1:5
55. The insulation for a thermal energy balance of 100W requires that
the heat transfer through a surface area of 1:5m2 in a down sleeping bag with
a loft of 4cm satis…es the expression
H =
kA
T
;
x
1:5m2
100W = :043 (W=m C)
37
Tc = 100
Tc = (37
(37
Tc ) C
;
:04m
:04= (:043 1:5) ;
62) C = 25 C:
64. From the Stefan-Boltzmann law
P = AT 4 = 5:57
For a star with T = 50
R=
(50
10
8
103 K and P = 4
4 1027
103 )4 5:57
W=m2 K 4 AT 4 :
1027 W , the radius of the star is
1=2
10
8
4
=6
107 m = 6
104 km: