17.1: From Eq.
       
F.1.134327.5659b) F.0.81328.6259a) ,1.17 
  
F.0.88321.3159c) 
17.2: From Eq.
       
C.7.413210795b) C.0.5320.4195a) ,2.17 
  
C.8.2732189/5c) 
17.3:
 F0.720.40so ,FC1
5
9

F140.2F0.70
12
 TT
17.4: a)
C.6.55)440.56(9)5(b) C.2.27))0.4((45.0)95( 
17.5:
  
F,4.104322.4059(17.1),Eq.Froma) 
which is cause for worry.

  
F54or F,6.53321259b) 
to two figures.
17.6:
  
 F2.218.1159

to arrive at the numerical results.)
 )85.126)(5/9(C,12715.273400a)
FC
TT
F.1079.232)1055.1)(5/9(C,1055.115.2731055.1c)
F.28932)15.178)(5/9(C,17815.27395b)F.26032
77
F
77
C
FC


TT
TT
17.11: From Eq.
K.23.2715.273)C92.245(),3.17(
K
T
17.12: From Eq.
C.1769273.15K2042.14K)3.16(7.476)(27),4.17( 
17.13: From Eq.
 
mm.444)mm0.325(),4.17(
K273.16
K15.373

17.14: On the Kelvin scale, the triple point is 273.16 K, so
R.491.69K5(9/5)273.1R 
One could also look at Figure 17.7 and note that the

P
17.16:
   
 
  
  
C,168m1.62C104.2m1025
1
52
0



αLLT
so the temperature is
C183
.
17.17:
.TL m39.0C)5.0)(Cm)(18.01410)()C(102.1(
15
0


17.18:
)1( Tαddd 

C)))78.0(C0.23)()C(10(2.4cm)(14500.0(
15



 
.C103.2
1
5



17.22: From Eq. (17.8),
C.4.49soC,4.29
15
3
0
K101.5
1050.1






TT
β
VV
17.23
 
 
  
L,11C0.9L1700C1075
1
5

 
 
 
 
.C107.1
0.55cm1000
cm95.8
K100.18
1
5
3
3
15
glass





C
β
17.26: a)
.222,
0
22
ALLLALA
L
L
L
L


ππD
A
b)
 
 
 
 
 
 
.cm437.1C150C1020.121cm431.121
252
0


TαAA

17.28: (a ) No, the brass expands more than the steel.
(b) call

D
the inside diameter of the steel cylinder at
BRST
:C150At C20 DD 
 
cm026.25
)C130)()C(102.1(1
)C130)()C(10(2.01cm)25(
1
)cm(125


17.29: The aluminum ruler expands to a new length of

cm048.20)]C100)()C(10(2.4cm)[10.20()1(
15
0


TαLL
The brass ruler expands to a new length of

cm040.20)]C100)()C(10(2.0cm)[10.20()1(
15
0


TαLL
The section of the aluminum ruler will be longer by 0.008 cm
17.30: From Eq. (17.12),

N.100.4
)m10C)(2.01110)()C(10Pa)(2.0109.0(
4
241511




TAYαF




b) There will be 1200 breaths per hour, so the heat lost is
J.104.6J)38)(1200(
4

17.34:
s,104.1
) W1200(
)C7)(KkgJ3480)(kg70(
3





P
Tmc
P
Q
t
about 24 min.
17.35: Using Q=mgh in Eq. (17.13) and solving for
Τ
gives
.C53.0
)kg.KJ4190(
)m225)(sm80.9(
2


23


T
17.37:
        
J.1003.3KkgJ470kg30.0KkgJ910kg60.1C20C210
5

17.38: Assuming
 
,1060.0 KQ 
 
    
 
 
.C1.45
KkgJ910kg1000.8
sm80.7kg80.16
61060.0
3
2
2
1
2
2
1




  
K.kgJ1051.2
C55.18C54.22kg780.0
W0.65s120
3





Tm
Q
c
b) An overstimate; the heat
Q
is in reality less than the power times the time
interval.
17.42: The temperature change is
soK,0.18T
  
  
K.kgJ240
K0.18N4.28
J1025.1sm80.9
42






melt

m
Q
L
F

(b)
Tm
Q
cTmcQLiquid


:

  
  



CkgJ000,1
C30kg50.0
min5.1minJ000,10
c
  
  





Q
should actually be larger than in
part (a), so the true
metal
c
is larger than we calculated; the value we calculated would be
smaller than the true value.
17.46: a) Let the man be designated by the subscript m and the “‘water” by w, and T is
the final equilibrium temperature.
wwwmmm
TCmTCm 

   
wwwmmm
TTCmTTCm 

   
wwwmmm
TTCmTTCm 
Or solving for T,
.
wwmm
wwwmmm
CmCm
TCmTCm
T



Inserting numbers, and realizing we can change K to








6.7or d,005.0
dayJ107
C)C)(0.15kg.Jkg)(3480355.70(
6



tt
minutes. This may acount for mothers
taking the temperature of a sick child several minutes after the child has something to
drink.
17.48:
)(
f
LTcmQ 

 
Btu.136kcal34.2J1043.1
kgJ10334K)0.18)(KkgJ(4190kg)350.0(
5
3








P
Tmc
P
Q
t
b)
min,230
)minJ(800
)kgJ10kg)(334550.0(
3
f


P
mL
so the time until the ice has melted is
min.252min230min 7.21 
17.51:
 
hr.Btu102.40kW01.7
s)(86,400
)kgJ10334()kglb2.205lb)4000(
4
3


K)K)(1.00kgJkg)(34800.70(
6
v
sweat






L
TMc
m
b) This much water has a volume of 101
,cm
3
about a third of a can of soda.
17.55: The mass of water that the camel saves is
kg,45.3
)kgJ10(2.42
K)K)(6.0kgJkg)(3480400(
6
v





L
TMc




T
17.57: The algebra reduces to

C5.27
)KkgJkg)(470(0.250
))KkgJkg)(4190(0.170)KkgJkg)(390500.0((
C))(85.0KkgJkg)(470(0.250
C))(20.0KkgJkg)(4190(0.170)KkgJkg)(390500.0((




















Q
m
ice




)C0.30)(KkgJ4190()kgJ10334()C0.20(K)kgJ(2100
J)10714.4(
3
4




g.94.0kg1040.9
2


17.60: The heat lost by the sample (and vial) melts a mass m, where
g.08.3
)kgJ10(334
K)))(19.5KkgJg)(2800(6.0)KkgJg)(22500.16((
3
f




L

3

























T
(The fact that a positive Celsius temperature was obtained indicates that all of the ice
does indeed melt.)
17.63: The steam both condenses and cools, and the ice melts and heats up along with

m0.450
K100

C.73.3m)1000.12)(mK(222Cc)100.0
2


17.66: Using the chain rule,
dt
dm
dt
dQ
LH
f

and solving Eq. (17.21) for k,

K.mW227
K)100)(m10250.1(
s)100.60(
s)600(
kg)1050.8(
)kgJ10334(
24
23
3
f




cm0.3Km W080.0cm2.2Km W010.0
C0.10cm0.3Km W080.0C0.19cm2.2Km W010.0



C.8.5 
Note that the conversion of the thickness to meters was not necessary. b) Keeping extra
figures for the result of part, (a), and using that result in the temperature difference across
either the wood or the foam gives
 
  
m102.2
C767.5C0.19
m W010.0
2
woodfoam



 K
A
H
A
H

 
  
m100.3
C0.10C767.5
Km W080.0

 
,23.17
the energy that flows in time
t
is
 
 
 
 
J.107.5Btu708h0.5
BtuhFft30
F34ft125
5
2
2






t
R
TA
tH
17.70: a) The heat current will be the same in both metals; since the length of the
copper rod is known,
 
 
 

2







T
T
k
k
LL
17.71: Using
(17.21),Eq.in 17.66)Problem(see
dt
dm
v
LH 

kA
L
dt
dm
LT
v


,C5.5
)m150.0)(Km(50.2

2
23



























