Page 233 - Schaum's Outline of Theory and Problems of Applied Physics

218 HEAT [CHAP. 18

SOLVED PROBLEM 18.14
Five kilograms of water at 40 C is poured on a large block of ice at 0 C. How much ice melts?
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Heat lost by water = heat gained by ice
m w c T = m ice L f
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(5kg)[4.185 kJ/(kg· C)](40 C) = (m ice )(335 kJ/kg)
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837
m ice = kg = 2.5kg
335
SOLVED PROBLEM 18.15
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Five hundred kilocalories of heat is added to 2 kg of water at 80 C. How much steam is produced?
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The heat needed to raise the temperature of the water from 80 C to the boiling point of 100 Cis
Q 1 = m 1 c T = (2kg)[1 kcal/(kg· C)](20 C) = 40 kcal
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Hence Q 2 = 500 kcal − 40 kcal = 460 kcal of heat is available to convert water at 100 C to steam at the same
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temperature. Since Q 2 = m steam L ν , the steam produced is
460 kcal
Q 2
m steam = = = 0.85 kg
L ν 540 kcal/kg
SOLVED PROBLEM 18.16
How much heat must be added to 200 lb of lead at 70 F to cause it to melt? The specific heat capacity of
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lead is 0.03 Btu/(lb· F), it melts at 626 F, and its heat of fusion is 10.6 Btu/lb.
Here T = 626 F − 70 F = 556 F. Hence
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Q = mc T + mL f
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= (200 lb)[0.03 Btu/(lb· F)](556 F) + (200 lb)(10.6 Btu/lb)
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= 3336 Btu + 2120 Btu = 5456 Btu
SOLVED PROBLEM 18.17
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Find the minimum amount of ice at −10 C needed to bring the temperature of 500 g of water at 20 C
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down to 0 C.
Here T ice = 10 C and T water = 20 C. Therefore
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Heat gained by ice = heat lost by water
m ice c ice T ice + m ice L f ice = m water c water T water

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m water c water T water (0.5kg)[4.185 kJ/(kg· C)](20 C)
m ice = =
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c ice T ice + L f ice [2.09 kJ/(kg· C)](10 C) + 335 kJ/kg
= 0.12 kg
SOLVED PROBLEM 18.18
A 30.1 g ice cube at 0 C is dropped into 200 g of water at 30 C. What is the final temperature?
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