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Heat
Problem C
CALORIMETRY
PROBLEM
In 1906, a 636.73-g diamond was found at the Premier Mine in South
Africa—making it the world’s largest uncut diamond. After being cut, the
diamond pieces were dropped into an insulated water bath containing
SOLUTION
1.00 kg of water. The water temperature increased by 1.30!C, and the dia1. DEFINE
mond’s temperature decreased by 15.54°C. What is the specific heat capacity of diamond? The specific heat capacity of water is 4186 J/kg• !C.
SOLUTION
1. DEFINE
2. PLAN
Given:
md = 636.73 g
∆Tw = 1.30°C
Unknown:
cp,d = ?
mw = 1.00 kg
∆Td = −15.54°C
cp,w = 4186 J/kg • °C
Choose the equation(s) or situation: Use the equation for specific heat capacity to
equate the energy removed from the diamond to the energy absorbed by the water.
Energy absorbed by water = Energy removed from diamond
cp,wmw ∆Tw = −cp,dmd∆Td
Rearrange the equation(s) to isolate the unknown(s):
cp,wmw ∆Tw
cp,d = 
−md∆Td
3. CALCULATE
Substitute values into the equation(s) and solve:
(4186 J/kg• °C)(1.00 kg)(1.30°C)
cp,d =  = 5.50 × 102 J/kg• °C
−(0.63673 kg)(−15.54°C)
The answer can be estimated using rounded values for cp,w (4200 J/kg• °C), md !3 kg", and
1
the ratio of ∆Tw to ∆Td !12". The resulting value for cp,d is then 530 J/kg• °C, which is close
to the calculated value.
2
ADDITIONAL PRACTICE
1. Mixing equal parts of hydrogen peroxide and water to use as a mouthwash disinfects the mouth and whitens teeth. Suppose you mix 15 g of
each into a plastic foam cup. The water’s temperature increases 1.0°C
and the hydrogen peroxide decreases 1.6°C. Disregarding energy transfer as heat to the solution’s surroundings, what is the specific heat capacity of hydrogen peroxide?
Ch. 9–6
Holt Physics Problem Bank
Copyright © by Holt, Rinehart and Winston. All rights reserved.
4. EVALUATE
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NAME ______________________________________ DATE _______________ CLASS ____________________
2. Vinegar, which contains acetic acid, can be used as an effective and
environmentally-friendly household cleanser. Suppose you mix 0.340 kg
of vinegar at 21.0°C with 1.00 kg hot water at 90.0°C in a plastic bucket.
The solution of vinegar and water reaches a final equilibrium temperature of 73.7°C. Disregarding energy transfer as heat to the surrounding
air and bucket, what is the specific heat capacity of vinegar?
3. After eating a hearty stew you cooked over a campfire with your
0.250-kg aluminum-alloy pot, you place the pot in a plastic bucket
containing 1.00 kg of water. The water’s temperature increases 1.00°C
and the temperature of the pot decreases 17.5°C. Disregarding energy
transfer as heat to the surrounding air and bucket, what is the specific
heat capacity of the pot?
4. Suppose you bake cornbread in a 3.0-kg cast iron skillet. After removing the cornbread from the oven, you place the hot skillet in a sink
filled with 5.0 kg of dishwater. The water’s temperature increases
2.25°C, and the temperature of the skillet decreases 29.6°C. Disregarding energy transfer as heat to the surrounding air and sink, what is the
specific heat capacity of the skillet?
5. The water in a swimming pool transfers 1.09 × 1010 J of energy as heat
to the cool night air. If the temperature of the water, which has a specific heat of 4186 J/kg• °C, decreases by 5.0°C, what is the mass of the
water in the pool?
Copyright © by Holt, Rinehart and Winston. All rights reserved.
6. Bismuth’s specific heat is 121 J/kg• °C, the lowest of any nonradioactive metal. What is the mass of a bismuth sample if 25 J raises
its temperature 5.0°C?
7. The temperature of air in a foundry increases when molten metals cool
and solidify. Suppose 4.5 × 107 J of energy is added to the surrounding
air by the solidifying metal. The air’s temperature increases by 55°C,
and the air has a specific heat capacity of 1.0 × 103 J/kg• °C. What is the
mass of the heated air?
8. A 0.190 kg piece of copper is heated and fashioned into a bracelet. The
amount of energy transferred as heat to the copper is 6.62 × 104 J. If
the specific heat of copper is 387 J/kg• °C, what is the change in the
temperature of the copper?
9. A 0.225 kg sample of tin, which has a specific heat of 2.2 × 103 J/kg• °C,
is cooled in water. The amount of energy transferred to the water is 3.9
× 104 J. What is the change in the tin’s temperature?
10. Tantalum is an element used, among other things, in making aircraft
parts. Suppose the properties of a tantalum part are being tested at
high temperatures. Tantalum has a specific heat of about 140 J/kg• °C.
The part, which has a mass of 0.23 kg, is cooled by being placed in
water. If 3.0 × 104 J of energy is transferred to the water, what is the
change in the part’s temperature?
Problem C
Ch. 9–7
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Givens
Solutions
8. ∆KE = KEf − KEi
= −2.15 × 104 J
∆PE + ∆KE + ∆U = 0
There is no change in the height of the sticks, so ∆PE = 0 J.
∆U = −∆KE = −(−2.15 × 104 J) = 2.15 × 104 J
9. vi = 20.5 m/s
∆PE + ∆E + ∆U = 0
vf = 0 m/s
The height of the skater does not change, so ∆PE = 0 J.
m = 61.4 kg
∆KE = KEf − KEi = 0 − 2mvi2
1
1
1
∆U = −∆KF = −(− 2mvi2) = 2(61.4 kg)(20.5 m/s)2 = 1.29 × 104 J
10. ∆KE = KEf − KEi
= −7320 J
∆Uhands = (1 − 0.300)∆U
∆PE + ∆KE + ∆U = 0
The hands don’t change height, so ∆PE = 0 J.
∆U = −∆KE = −(−7320 J) = 7320 J
∆Uhands = (1− 0.300)(7320 J) = (0.700)(7320 J) = 5120 J
Additional Practice C
1. mw = 15 g = 0.015 kg
cp,wmw∆Tw = cp,hpmhp∆Thp
∆Tw = 1.0°C
cp,wmw ∆Tw (4186 J)(kg • °C)(0.015 kg)(1.0°C)
cp,hp =   = 
−mhp ∆Thp
−(0.015 kg)(−1.6°C)
∆Thp = −1.6°C
cp,hp = 2.6 × 103 J/kg • °C
mhp = 15 g = 0.015 kg
cp,w = 4186 J/kg • °C
cp,vmv ∆Tv = −cp,wmw ∆Tw
mw = 1.00 kg
∆Tw = Tf − Tw,i = 73.7°C − 90.0°C = −16.3°C
Tv,i = 21.0°C
∆Tv = Tf − Tv,i = 73.7°C − 21.0°C = 52.7°C
Tw,i = 90.0°C
−cp,wmw ∆Tw
−(4186 J/kg • °C)(1.00 kg)(−16.3°C)
cp,v =  = 
(0.340 kg)(52.7°C)
mv ∆Tv
Tf = 73.7°C
cp,w = 4186 J/kg • °C
3. ma = 0.250 kg
mw = 1.00 kg
∆Tw = 1.00° C
∆Ta = −17.5° C
cp,v = 3.81 × 103 J/kg •°C
cp,wmw ∆Tw = −cp,ama∆Ta
cp,wmw ∆Tw
(4186 J/kg • °C)(1.00 kg)(1.00°C)
cpa = 
= 
−(0.250 kg)(−17.5° C)
−ma∆Ta
cp,a = 957 J/kg • °C
cp,w = 4186 J/kg • °C
cp,wmw ∆Tw = −cp,imi ∆Ti
4. mi = 3.0 kg
mw = 5.0 kg
∆Tw = 2.25° C
∆Ti = −29.6° C
V
(4186 J/kg • °C)(5.0 kg)(2.25°C)
cp,wmw ∆Tw
cp,i = 
= 
−(3.0 kg)(−29.6°C)
−mi∆Ti
cp,i = 530 J/kg • °C
cp,w = 4186 J/kg • °C
V Ch. 9–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
2. mv = 0.340 kg
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Lesson
Givens
Solutions
5. Q = −1.09 × 1010 J
cp,w = 4186 J/kg • °C
∆Tw = −5.0°C
6. cp,b = 121 J/kg • °C
Q = 25 J
∆Tb = 5.0°C
7. Q = 4.5 × 107 J
∆Ta = 55° C
cp,a = 1.0 × 103 J/kg • °C
8. mc = 0.190 kg
4
Q = 6.62 × 10 J
cp,c = 387 J/kg • °C
9. mt = 0.225 kg
cp,t = 2.2 × 103 J/kg • °C
Q = −3.9 × 104 J
10. cp,t = 140 J/kg • °C
mt = 0.23 kg
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Q = −3.0 × 104 J
Q
cp,w = 
mw ∆Tw
Q
−1.09 × 1010 J
mw =   =  = 5.2 × 105 kg
cp,w ∆Tw
(4186 J/kg • °C)(−5.0°C)
Q
cp,b = 
mb ∆Tb
25 J
Q
mb =  =  = 4.1 × 10−2 kg
(121 J/kg • °C)(5.0°C)
cp,b∆Tb
Q
cp,a = 
ma ∆Ta
Q
4.5 × 107 J
ma =  = 
a = 820 kg
cp,a∆Ta
(1.0 × 103 J/kg • °C)(55°C)
Q
cp,c = 
mc∆Tc
6.62 × 104 J
Q
∆Tc =  =  =
mc cp,c (0.190 kg)(387 J/kg • °C)
9.00 × 102°C
Q
cp,t = 
mt∆Tt
−3.9 × 104 J
Q
∆Tt =  = 
= −79°C
mtcp,t (0.225 kg)(2.2 × 103 J/kg • °C)
Q
cp,t = 
mt∆Tt
−3.0 × 104 J
Q
∆Tt =  =  = −930°C
mtcp,t (0.23 kg)(140 J/kg • °C)
Additional Practice D
1. Q = 1.10 × 106 J
m = 5.33 kg
2. Q = 9.6 × 105 J
m = 5.33 kg
3. Q = 3.72 × 105 J
m = 0.65 kg
4. Q = 8.5 × 104 J
Lv = 2.26 × 106 J/kg
Q = mLf
Q 1.10 × 106 J
Lf =  =  = 2.06 × 105 J/kg
m
5.33 kg
Q 9.6 × 105 J
Lf =  =  = 1.8 × 105 J/kg
m
5.33 kg
Q 3.72 × 105 J
Lsubl =  =  = 5.7 × 105 J/kg
m
0.65 kg
Q = mLv
Q
8.5 × 104 J
m =  = 
= 3.8 × 10−2 kg
Lv 2.26 × 106 J/kg
Section Five—Problem Bank
V
V Ch. 9–5