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Heat
Chapter 10
I
Practice 10A, p. 363
Givens
1. TF = −128.6°F
Solutions
5
5
5
TC = 9(TF − 32.0) = 9(−128.6 − 32.0)°C = 9(−160.6)°C
TC = −89.22°C
T = TC + 273.15 = (−89.22 + 273.15) K = 183.93 K
2. TF,1 = 105°F
TF,2 = −25°F
5
5
5
TC,1 = 9(TF,1 − 32.0) = 9(105 − 32.0)°C = 9(73)°C
TC,1 = 41°C
T = (TC + 273)K
T1 = (41 + 273)K = 314 K
5
5
5
TC,2 = 9(TF,2 − 32.0) = 9(−25 − 32.0)°C = 9(−57)°C
TC,2 = −32°C
T2 = (−32 + 273)K = 241 K
3. TF,1 = 98.6°F
TF,2 = 102°F
5
5
5
TC,1 = 9(TF,1 − 32.0) = 9(98.6 − 32.0)°C = 9(66.6)°C
TC,1 = 37.0°C
5
5
5
TC,2 = 9(TF,2 − 32.0) = 9(102° − 32.0)°C = 9(7.0 × 101)°C
Copyright © by Holt, Rinehart and Winston. All rights reserved.
TC,2 = 39°C
4. TC,i = 23°C
TC,f = 78°C
T = TC + 273.15
Ti = TC,i + 273.15 = (23 + 273.15) K = 296 K
Tf = TC,f + 273.15 = (78 + 273.15) K = 351 K
∆T = Tf − Ti = 351 K − 296 K = 55 K
Alternatively, because a degree Celsius equals a kelvin,
∆T = ∆T C = TC,f − TC,i = 78°C − 23°C
∆T = 55°C = 55 K
9
TF = 5TC + 32.0
9
9
TF,i = 5TC,i + 32.0 = 5(23)°F + 32.0° F = (41 + 32.0)°F = 73°F
9
9
TF,f = 5TC,f + 32.0 = 5(78)°F + 32.0° F = (140 + 32.0)°F = 172°F
9
9
∆TF = TF,f − TF,i = 5 (78 − 23)°F = 5 (55)°F = 99°F
5. T = 77.34 K
TC = T − 273.15 = (77.34 − 273.15)°C = −195.81°C
9
9
TF = 5TC + 32.0 = 5(−195.81)°F + 32.0°F = (−352.46 + 32.0)°F
TF = −320.5°F
Section One—Pupil’s Edition Solutions
I Ch. 10–1
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Section Review, p. 364
Givens
Solutions
3. T = 90.2 K
TC = T − 273.15 = (90.2 − 273.15)°C = −183.0°C
9
9
TF = 5(TC) + 32.0 = 5(−183.0)°F + 32.0°F = (−329.4 + 32.0)°F
I
TF = −297.4°F
4. boiling point = 444.6°C
5
5
a. ∆TC = 9(∆TF) = 9(586.1)°C = 325.6°C
melting point = 586.1°F
below boiling point
melting point = 444.6°C − 325.6°C = 119.0°C
9
9
b. TF,1 = 5(TC ,1) + 32.0 = 5(119.0)°F + 32.0°F = (214.2 + 32.0)°F
TF,1 = 246.2°F
9
9
TF,2 = 5(TC ,2 ) + 32.0 = 5(444.6)°F + 32.0°F = (800.3 + 32.0)°F
TF,2 = 832.3°F
c. T1 = TC ,1 + 273.15 = (119.0 + 273.15) K = 392.2 K
T2 = TC ,2 + 273.15 = (444.6 + 273.15) K = 717.8 K
Practice 10B, p. 370
1. m = 11.5 kg
∆U = mgh = (11.5 kg)(9.81 m/s2)(6.69 m) = 755 J
g = 9.81 m/s2
h = 6.69 m
mh = 2.50 kg
vh = 65.0 m/s
3. m = 3.0 × 10−3 kg
h = 50.0 m
4. m = 2.5 kg
vi = 5.7 m/s
5
3.3 × 10 J melts 1.0 kg of ice
5. ∆U = 209.3 J
∆U = 3(KEh ) = 3 2mh vh2 = 6 (2.50 kg)(65.0 m/s)2
1
1
∆U = 1.76 × 103 J
∆U = (0.65)(PEi) = (0.65)(mgh) = (0.65)(3.0 × 10−3 kg)(9.81 m/s2)(50.0 m)
∆U = 0.96 J
1
1
∆U = KEi = 2mvi2 = 2(2.5 kg)(5.7 m/s)2 = 41 J
(41 J)(1.0 kg)
ice melted = 
= 1.2 × 10−4 kg
3.3 × 105 J
1
∆U = KE = 2mv 2
m = 0.25 kg
v=
I Ch. 10–2
1 1
9.3 J)
2∆mU = (2
)0(.2205
kg = 41 m/s
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
2. ms = 0.500 kg
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Section Review, p. 370
Givens
Solutions
4. Ti = 10.0°C
∆U = PEi = mgh = (505 kg)(9.81 m/s2)(50.0 m) = 2.48 × 105 J
m = 505 kg
h = 50.0 m
2
g = 9.81 m/s
4186 J/kg increases water
temperature by 1.0°C
2.48 × 105 J
 (1.0°C)
505 kg
∆T =  = 0.12°C
4186 J/kg
I
Tf = Ti + ∆T = 10.0°C + 0.12°C = 10.1°C
Practice 10C, p. 374
1. mg = 3.0 kg
cp,wmw ∆Tw = cp,gmg ∆Tg
Tg = 99°C
cp,wmw(Tf − Tw) = cp,gmg(Tg − Tf)
cp,g = 129 J/kg • °C
mw = 0.22 kg
Tw cp,wmw + Tg cp,g mg
Tf = 
cp,wmw + cp,gmg
Tw = 25°C
cp,wmw = (4186 J/kg • °C)(0.22 kg) = 920 J/°C
cp,w = 4186 J/kg • °C
Tw cp,w mw = (25°C)(4186 J/kg • °C)(0.22 kg) = 2.3 × 104 J
cp,gmg = (129 J/kg • °C)(3.0 kg) = 390 J/°C
Tg cp,g mg = (99°C)(129 J/kg • °C)(3.0 kg) = 3.8 × 104 J
(2.3 × 104 J)(3.8 × 104 J) + (99°C)(390 J/°C) 6.1 × 104 J
Tf =  = 
1310 J/°C
(920 J/°C + 390 J/°C)
Tf = 47°C
Copyright © by Holt, Rinehart and Winston. All rights reserved.
2. mt = 0.225 kg
cp,wmw ∆Tw = cp,tmt ∆Tt
Tt = 97.5°C
cp,wmw(Tf − Tw) = cp,tmt(Tt − Tf)
mw = 0.115 kg
Tw = 10.0°C
[Twcp,wmw + Ttcp,tmt]
Tf = 
(cp,wmw + cp,tmt)
cp,t = 230 J/kg • °C
cp,wmw = (4186 J/kg • °C)(0.115 kg) = 481 J/°C
cp,w = 4186 J/kg • °C
Tw cp,wmw = (10.0°)(4186 J/kg • °C)(0.115 kg) = 4.81 × 103 J
cp,tmt = (230 J/kg • °C)(0.225 kg) = 52 J/°C
Tt cp,t mt = (97.5°C)(230 J/kg • °C)(0.225 kg) = 5.0 × 103 J
4.81 × 103 J + 5.0 × 103 J 9.8 × 103 J
Tf =  = 
481 J/°C + 52 J/°C
533 J/°C
Tf = 18.0° C
3. mm = 0.032 kg
cp,mmm ∆Tm = cp,c mc ∆Tc
Tm = 11°C
Because cp,m = cp,c , mm ∆Tm = mc ∆Tc .
mc = 0.16 kg
(mm )(Tf − Tm ) = (mc )(Tc − Tf )
Tc = 91°C
mmTf − m mTm = mcTc − mcTf
cp,m = cp,c = cp,w
mmTf + mcTf = mcTc + mmTm
mcTc + mmTm (0.16 kg)(91°C) + (0.032 kg)(11°C)
 = 
Tf = 
mm + mc
0.032 kg + 0.16 kg
15 kg • °C + 0.35 kg • °C 15 kg • °C
Tf =  =  = 79°C
0.19 kg
0.19 kg
Section One—Pupil’s Edition Solutions
I Ch. 10–3
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Givens
Solutions
4. mc = 0.75 kg
I
cp,c mc ∆Tc = cp,w mw ∆Tw
Tc = 36.5°C
∆Tc = Tc − Tf = 36.5°C − 24.4°C = 12.1°C
mw = 1.25 kg
∆Tw = Tf − Tw = 24.4°C − 20.0°C = 4.4°C
Tw = 20.0°C
Tf = 24.4°C
cp,w mw ∆Tw (4186 J/kg • °C)(1.25 kg)(4.4°C)
cp,c =  = 
mc ∆Tc
(0.75 kg)(12.1°C)
cp,w = 4186 J/kg • °C
cp,c = 2500 J/kg • °C
5. mb = 0.59 kg
cp,b mc ∆Tb = cp,w mw ∆Tw
Tb = 98.0°C
∆Tw = Tf − Tw = 6.8°C − 5.0°C = 1.8°C
mw = 2.80 kg
∆Tb = Tb − Tf = 98.0°C − 6.8°C = 91.2°C
Tw = 5.0°C
Tf = 6.8°C
cp,w mw ∆Tw (4186 J/kg • °C)(2.80 kg)(1.8°C)
cp,b =  = 
mb ∆Tb
(0.59 kg)(91.2°C)
cp,w = 4186 J/kg • °C
cp,b = 390 J/kg • °C
6. ∆Ta = 1.0°C
mw = 1.0 kg
∆Tw = 1.0°C
cp,a ma ∆Ta = cp,w mw ∆Tw
cp,w mw ∆Tw
ma = 
cp,a ∆Ta
r = 1.29 kg/m3
cp,w mw ∆Tw 1
1
m
(4186 J/kg • °C)(1.0 kg)(1.0°C)
V = a =   =  3
cp,a ∆Ta
r
1.29 kg/m
r
(1000.0 J/kg • °C)(1.0°C)
cp,w = 4186 J/kg • °C
V = 3.2 m3
cp,a = 1000.0 J/kg • °C
7. ∆Tv = 8.39°C
mw = 101 g
cp,w = 4186 J/kg • °C
∆Tc = 838°C
cp,c = 387 J/kg • °C
cp,c mc ∆Tc = cp,w mw ∆Tw
cp,w mw ∆Tw

mc = 
cp,c ∆Tc
(4186 J/kg • °C)(0.101 kg)(8.39°C)
mc = 
(387 J/kg • °C)(838°C)
mc = 1.09 × 10−2 kg = 10.9 g
Practice 10D, p. 381
1. m = 42 g
∆T1 = melting point − Ti = 0°C − (−11°C) = 11°C
Ti = −11°C (ice)
∆T2 = boiling point − melting point = 100.0°C − 0°C = 100.0°C
Tf = 111°C (steam)
∆T3 = Tf − boiling point = 111°C − 100.0°C = 11°C
Lf = 3.33 × 105 J/kg
Q1 = mcp,ice ∆T1 = (42 × 10−3 kg)(2.09 × 103 J/kg • °C)(11°C) = 970 J
Lv = 2.26 × 106 J/kg
Q2 = mLf = (42 × 10−3 kg)(3.33 × 105 J/kg) = 1.4 × 104 J
cp,ice = 2.09 × 103 J/kg • °C
Q3 = mcp,w ∆T2 = (42 × 10−3 kg)(4186 J/kg • °C)(100.0°C) = 1.8 × 104 J
cp,steam = 2.01 × 103 J/kg • °C
Q4 = mLv = (42 × 10−3 kg)(2.26 × 106 J/kg) = 9.5 × 104 J
Q5 = mcp,steam ∆T3 = (42 × 10−3 kg)(2.01 × 103 J/kg • °C)(11°C) = 930 J
Qtot = Q1 + Q2 + Q3 + Q4 + Q5
Qtot = 970 J + (1.4 × 104 J) + (1.8 × 104 J) + (9.5 × 104 J) + 930 J
Qtot = 1.29 × 105 J
I Ch. 10–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
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Givens
Solutions
2. m = 1.0 kg
Q = mLv = (1.0 kg)(2.01 × 105 J/kg) = 2.0 × 105 J
T = 77 K
Lv = 2.01 × 105 J/kg
∆T = melting point − Ti = 327.3°C − 27.3°C = 300.0°C
3. m = 0.225 kg
Ti = 27.3°C
Q1 = mcp,l ∆T = (0.225 kg)(1.28 × 102 J/kg • °C)(300.0°C) = 8640 J
cp,l = 1.28 × 102 J/kg • °C
Q2 = mLf = (0.225 kg)(2.45 × 104 J/kg) = 5510 J
Lf = 2.45 × 104 J/kg
Qtot = Q1 + Q2 = 8640 J + 5510 J = 1.415 × 104 J
I
melting point = 327.3°C
4. m = (14.0 g/can)(1000
cans) = 14.0 × 103 g =
14.0 kg
∆T = melting point − Ti = 660.4°C − 26.4°C = 634.0°C
Q1 = mcp,a ∆T = (14.0 kg)(8.99 × 102 J/kg • °C)(634.0°C) = 7.98 × 106 J
Q2 = mLf = (14.0 kg)(3.97 × 105 J/kg) = 5.56 × 106 J
Ti = 26.4°C
cp,a = 8.99 × 102 J/kg • °C
Qtot = Q1 + Q2 = (7.98 × 106 J) + (5.56 × 106 J) = 1.354 × 107 J
Lf = 3.97 × 105 J/kg
melting point = 660.4°C
Q1 = cp,s ms ∆Ts
5. mi = 0.011 kg
Ti = 0°C
Q2 = miLf
ms = 0.450 kg
Q3 = cp,w mi ∆Tw
Ts = 80.0°C
By the conservation of energy, Q1 = Q2 + Q3 .
cp,s = cp,w = 4186 J/kg • °C
5
Lf = 3.33 × 10 J/kg
cp,s ms ∆Ts = mi Lf + cp,w mi ∆Tw
cp,s ms (Ts − Tf ) = mi Lf + cp,w mi (Tf − Ti )
cp,s ms Ts − cp,s msTf = mi Lf + cp,w mi Tf − cp,w miTi
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Tf (cp,w mi + cp,s ms ) = cp,s msTs + cp,w m iTi − miLf
cp,s msTs + cp,w m iTi − m i Lf
Tf =  ,
cp,w mi + cp,s ms
where Ti = 0.0°C
(4186 J/kg • °C)(0.450 kg)(80.0°C) + (0 J) − (0.011 kg)(3.33 × 105 J/kg)
Tf = 
(4186 J/kg • °C)(0.011 kg) + (4186 J/kg • °C)(0.450 kg)
(1.51 × 105 J) + (0 J) − (3.7 × 103 J) 1.47 × 105 J
Tf =  =  = 76.2°C
1930 J/°C
46 J/°C + 1880 J/°C
6. mal = 25 kg
Q = mal Lf = mair cp,air ∆T
Tair = 25°C
malLf
(25 kg)(3.97 × 105 J/kg)
∆T =  = 
= 76°C
mair cp,air (130 kg)(1.0 × 103 J/kg • °C)
cp,air = 1.0 × 103 J/kg • °C
Tf = Tair + ∆T = 25°C + 76°C = 101°C
mair = 130 kg
cp,al = 8.99 × 102 J/kg • °C
Lf = 3.97 × 105 J/kg
Section One—Pupil’s Edition Solutions
I Ch. 10–5
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Section Review, p. 382
Givens
Solutions
1. mg = 47 g
I
cp,w mw ∆Tw = cp,g mg ∆Tg
Tg = 99°C
∆Tg = Tg − Tf = 99°C − 38°C = 61°C
Tw = 25°C
∆Tw = Tf − Tw = 38°C − 25°C = 13°C
Tf = 38°C
cp,g = 1.29 × 102 J/kg • °C
cp,g mg ∆Tg (1.29 × 102 J/kg • °C)(47 × 10−3 kg)(61°C)
mw =  = 
cp,w ∆Tw
(4186 J/kg • °C)(13°C)
cp,w = 4186 J/kg • °C
mw = 6.8 × 10−3 kg = 6.8 g
4. m = 15 g = 0.015 kg
Q
7.4 × 103 J
15.8 kJ − 8.37 kJ
a. cp,l =  =  =  = 2 × 103 J/kg • °C
m∆T (0.015 kg)(300°C − 80°C)
(0.015 kg)(200°C)
Q 8.37 kJ − 1.27 kJ 7.10 × 103 J
b. Lf =  =  =  = 4.7 × 105 J/kg
m
0.015 kg
0.015 kg
Q
1.27 × 103 J
1.27 kJ − 0 kJ
c. cp,s =  =  =  = 1 × 103 J/kg • °C
m∆T (0.015 kg)(80°C − 0°C) (0.015 kg)(80°C)
Q
1 × 103 J
796 kJ − 795 kJ
d. cp,v =  =  =  = 7 × 102 J/kg • °C
m∆T (0.015 kg)(400°C − 300°C) (0.015 kg)(100°C)
Q 795 kJ − 15.8 kJ 779 × 103 J
e. Lv =  =  =  = 5.2 × 107 J/kg
0.015 kg
0.015 kg
m
5. Tf = 175°C
∆T = Tf − Tp = 175°C − 21°C = 154°C
Tp = 21°C
Q = mpcp,p ∆T = (0.105 × 10−3 kg/kernel)(125 kernels)(1650 J/kg • °C)(154°C)
cp,p = 1650 J/kg • °C
Q = 3340 J
6. mw = (0.14)(95.0 g)
L v = (0.90)(2.26 × 106 J/kg)
Q = mw Lv = (0.14)(95.0 × 10−3 kg)(0.90)(2.26 × 106 J/kg)
Q = 2.7 × 104 J
Chapter Review and Assess, pp. 387–391
9. TF = 136°F
5
5
5
TC = 9 (TF − 32.0) = 9(136 − 32.0)°C = 9(104)°C = 57.8°C
T = (TC + 273.2)K = (57.8 + 273.2)K
T = 3.31 × 102 K
10. TF = 1947°F
5
5
5
TC = 9(TF − 32.0) = 9(1947 − 32.0) = 9(1915)°C = 1064°C
T = TC + 273.15 = (1064 + 273.15) K = 1337 K
I Ch. 10–6
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
mp = (0.105 g/kernel)
(125 kernels)
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Givens
Solutions
19. F = 315 N
W = (0.14)(Ui )
d = 35.0 m
W = (0.14)(Ui )
20. m = 0.75 kg
vi = 3.0 m/s
29. m = 23 g
W
Fd
(315 N)(35.0 m)
Ui =  =  =  = 7.9 × 104 J
0.14 0.14
0.14
I
a. ∆U = (0.85)(KE) = (0.85)2mv 2
1
1
∆U = 2(0.85)(0.75 kg)(3.0 m/s)2 = 2.9 J
Q
16.6 kJ − 12.0 kJ
4.6 × 103 J
a. cp,l =  = 
= 
−3
m∆T (23 × 10 kg)(340°C − 160°C) (23 × 10−3 kg)(180°C)
cp,l = 1.1 × 103 J/kg • °C
Q 12.0 kJ − 1.85 kJ
10.2 × 103 J

 = 4.4 × 105 J/kg
=
b. Lf =  = 
m
23 × 10−3 kg
23 × 10−3 kg
Q
1.85 kJ − 0 kJ
1.85 × 103 J
= 
c. cp,s =  = 
−3
m∆T (23 × 10 kg)(160°C − 0°C) (23 × 10−3 kg)(160°C)
cp,s = 5.0 × 102 J/kg • °C
Q
857 kJ − 855 kJ
2.0 × 103 J
= 
d. cp,v =  = 
−3
−3
m∆T (23 × 10 kg)(540°C − 340°C) (23 × 10 kg)(2.0 × 102°C)
cp,v = 4.0 × 102 J/kg • °C
Q 855 kJ − 16.6 kJ
838 × 103 J

 = 3.6 × 107 J/kg
=
e. Lv =  = 
m
23 × 10−3 kg
23 × 10−3 kg
30. mr = 25.5 g
Tr = 84.0°C
Copyright © by Holt, Rinehart and Winston. All rights reserved.
mw = 5.00 × 10−2 kg
Qw = Qr − 0.14 kJ
cp,wmw(Tf − Tw) = cp,rmr(Tr − Tf) − 0.14 kJ
Tw = 24.0°C
[cp,r mrTr + cp,wmwTw − 0.14 kJ]
Tf = 
(cp,wmw + cp,r mr)
Qw = Qr − 0.14 kJ
cp,rmr = (234 J/kg • °C)(2.55 × 10−2 kg) = 5.97 J/°C
cp,r = 234 J/kg • °C
cp,r mrTr = (234 J/kg • °C)(2.55 × 10−2 kg)(84.0°C) = 501 J
cp,w = 4186 J/kg • °C
cp,wmw = 4186 J/kg • °C)(5.00 × 10−2 kg) = 209 J/°C
cp,w mwTw = (4186 J/kg • °C)(5.00 × 10−2 kg)(24.0°C) = 5.02 × 103 J
501 J + (5.02 × 103 J) − 140 J 5.38 × 103 J
Tf =  = 
209 J/°C + 5.97 J/°C
215 J/°C
Tf = 25.0° C
31. m1 = 1500 kg
1
∆U = ∆KE = 2 m1(vf − vi )2 = (0.5)(1500 kg)(0 m/s − 32 m/s)2 = 7.7 × 105 J
vi = 32 m/s
Q = ∆U = 7.7 × 105 J
vf = 0 m/s
Q = m2 cp,iron ∆T
cp,iron = 448 J/kg • °C
Q
7.7 × 105 J
∆T =  =  = 120°C
m2 cp,iron (4)(3.5 kg)(448 J/kg • °C)
m2 = (4)(3.5 kg)
32. T = 0.0°C
Q = mLf = (225 × 10−3 kg)(3.33 × 105 J/kg) = 7.49 × 104 J
m = 225 g
Lf = 3.33 × 105 J/kg
Section One—Pupil’s Edition Solutions
I Ch. 10–7
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33. mw = 1.20 × 1016 kg
Q1 = mw cp,w ∆Tw = mw cp,w(Tw − Tf )
Tw = 12.0°C
Q2 = mw Lf
Tf = 0°C
Q1 = (1.20 × 1016 kg)(4186 J/kg • °C)(12.0°C − 0°C)
cp,w = 4186 J/kg • °C
Q1 = 6.03 × 1020 J
Lf = 3.33 × 105 J/kg
Q2 = (1.20 × 1016 kg)(3.33 × 105 J/kg)
Q2 = 4.00 × 1021 J
Qtot = Q1 + Q2 = 6.03 × 1020 J + 4.00 × 1021 J
Qtot = 4.60 × 1021 J
41. TR = 0°R = absolute zero
one Rankine degree =
one Fahrenheit degree
a. TR = TF − (absolute zero in TF)
9
TF = 5TC + 32.0
absolute zero in Tc = −273.15°C
9
absolute zero in TF = 5 (−273.15)°F + 32.0°F
absolute zero in TF = (−491.67 + 32.0)°F = −459.7°F
TR = TF − (−459.7)°F = TF + 459.7°F
TR = TF + 459.7, or TF = TR − 459.7
b. T = TC + 273.15
5
TC = 9(TF − 32.0)
5
T = 9(TF − 32.0) + 273.15
TF = TR − 459.7
5
T = 9(TR − 459.7 − 32.0) + 273.15
5
T = 9(TR − 491.7) + 273.15
5
5
T = 9TR − 9(491.7) + 273.15
5
5
9
T = 9TR , or TR = 5 T
42. mr = 3.0 kg
PEi = ∆U
mw = 1.0 kg
m r gh = cp,w m w ∆T
∆T = 0.10°C
cp,w = 4186 J/kg • °C
cp,w m w ∆T (4186 J/kg • °C)(1.0 kg)(0.10°C)
h =  = 
mr g
(3.0 kg)(9.81 m/s2)
g = 9.81 m/s2
h = 14 m
43. TC = −252.87°C
9
9
a. TF = 5TC + 32.0 = 5(−252.87)°F + 32.0°F
TF = (−455.17 + 32.0)°F = −423.2°F
T = TC + 273.15 = (−252.87 + 273.15) K = 20.28 K
TC = 20.5°C
9
9
b. TF = 5TC + 32.0 = 5(20.5)°F + 32.0°F
TF = (36.9 + 32.0)°F = 68.9°F
T = TC + 273.15 = (20.5 + 273.15) K = 293.6 K
I Ch. 10–8
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
T = 9TR − 273.2 + 273.15
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44. freezing point = 50°TH
(0°C)
a. Set up a graph with Celsius on the x-axis and “Too Hot” on the y-axis. The equation relating the two scales can be found by graphing one scale versus the other
scale and then finding the equation of the resulting line. In this case, the two
known coordinates of the line are (0, 50) and (100, 200).
boiling point = 200°TH
(100°C)
∆y (200 − 50) 150 3
slope = a =  =  =  = 
∆x (100 − 0) 100 2
I
y = ax + b
b = y − ax = 50 − 2 0 = 50
3
3
The values y = TTH , a = 2, x = TC , and b = 50 can be substituted into the equation
for a line to find the conversion equation.
y = ax + b
3
TTH = 2TC + 50
y−b
or x = 
a
TTH − 50
TC = 
3

2
2
TC = 3(TTH − 50)
TC = absolute zero =
−273.15°C
45. TC = −40°C
3
3
b. TTH = 2(TC ) + 50 = 2(−273.15)°TH + 50°TH = (−409.72 + 50)°TH = −360°TH
9
9
TF = 5TC + 32 = 5(−40)°F + 32°F = (−72 + 32)°F
TF = −40°F
Copyright © by Holt, Rinehart and Winston. All rights reserved.
46. A = 6.0 m2
∆T = Tf − Ti = 61°C − 21°C = (4.0 × 101)°C
P/A = 550 W/m2
mw = Vw rw = (1.0 m3)(1.00 × 103 kg/m3) = 1.0 × 103 kg
Vw = 1.0 m3
Q mw c p,w ∆T
(1.0 × 103 kg)(4186 J/kg • °C)(4.0 × 101°C)
∆t =  =  = 
(P/A)(A)
P
(550 W/m2)(6.0 m2)
Ti = 21°C
Tf = 61°C
rw = 1.00 × 103 kg/m3
cp,w = 4186 J/kg • °C
47. mc = 253 g
∆t = 5.1 × 104 s
or (5.1 × 104 s)(1 h/3600 s) = 14 h
cp,a ma ∆Ta = cp,c mc ∆Tc
Tc = 85°C
∆Tc = Tc − Tf = 85°C − 25°C = (6.0 × 101)°C
Ta = 5°C
∆Ta = Tf − Ta = 25°C − 5°C = (2.0 × 101)°C
Tf = 25°C
mc ∆Tc cp,c (0.253 kg)(6.0 × 101°C)(3.87 × 102 J/kg • °C)
ma =  = 
∆Ta cp,a
(2.0 × 101°C)(8.99 × 102 J/kg • °C)
cp,a = 8.99 × 102 J/kg • °C
cp,c = 3.87 × 102 J/kg • °C
ma = 0.33 kg = 330 g
Section One—Pupil’s Edition Solutions
I Ch. 10–9
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48.
TF = 5TC + 32.0
9
TC = T − 273.15
9
TF − 32.0 = 5TC
I
5
T
9 F
− 9 (32.0) = TC
5
5
T
9 F
− 9 (32.0) + 273.15 = T
5
T = TF
5
T
9 F
− 9 (32.0) + 273.15 = TF
5
5
T
9 F
− 17.8 + 273.15 = TF
5
T
9 F
+ 255.4 = TF
4
255.4 = 9TF
9
TF = 4 (255.4)°F = 574.6°F
TF = T
574.6°F = 574.6 K
49. ma = 250 g
mw = 850 g
Q Qa + Qw macp,a∆T + mwcp,w ∆T
∆T
 = 
 =  = (macp,a + mw cp,w ) 
∆t
∆t
∆t
∆t
∆T
 = 1.5°C/min
∆t
Q
 = [(0.250 kg)(899 J/kg • °C) + (0.850 kg)(4186 J/kg • °)](1.5°C/min)
∆t
cp,a = 899 J/kg • °C
Q
 = (225 J/°C + 3.56 × 103 J/°C)(1.5°C/min) = (3.78 × 103 J/°C)(1.5°C/min)
∆t
cp,w = 4186 J/kg • °C
Q
 = 5.7 × 103 J/min
∆t
or (5700 J/min)(1 min/60 s) = 95 J/s
∆Qtea = ∆Qmelted ice
Tice = 0°C
mtea cp,tea ∆Ttea = mice Lf + mice cp,w ∆Tice
Tf = 15°C
mtea = 180 g
mtea cp,tea ∆Ttea
mice = 
Lf + cp,w ∆Tice
mice,tot = 112 g
∆Ttea = Ttea − Tf = 32°C − 15° C = 17°C
cp,tea = cp,w = 4186 J/kg • °C
∆Tice = Tf − Tice = 15°C − 0° C = 15°C
Lf = 3.33 × 105 J/kg
(180 × 10−3 kg)(4186 J/kg • °C)(17°C)
mice = 
3.33 × 105 J/kg + (4186 J/kg • °C)(15°C)
(180 × 10−3 kg)(4186 J/kg • °C)(17°C)
mice = 
3.33 × 105 J/kg + 6.3 × 104 J/kg
(180 × 10−3 kg)(4186 J/kg • °C)(17°C)
mice = 
= 3.2 × 10−2 kg = 32 g
3.96 × 105 kg
mass of unmelted ice = mice,tot − mice = 112 g − 32 g = 8.0 × 101 g
I Ch. 10–10
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
50. Ttea = 32°C