CHAPTER 20 HEAT ENERGY AND TRANSFER
EXERCISE 106, Page 226
1. Convert the following temperatures into the Kelvin scale: (a) 51C (b) - 78C (c) 183C
Kelvin temperature, K = C + 273
(a) When Celsius temperature = 51C, K = 51 + 273 = 324 K
(b) When Celsius temperature = - 78C, K = - 78 + 273 = 195 K
(c) When Celsius temperature = 183C, K = 183 + 273 = 456 K
2. Convert the following temperatures into the Celsius scale: (a) 307 K (b) 237 K (c) 415 K
If K = C + 273 then C = K - 273
(a) When Kelvin temperature = 307 K, C = 307 – 273 = 34C
(a) When Kelvin temperature = 237 K, C = 237 – 273 = - 36C
(a) When Kelvin temperature = 415 K, C = 415 – 273 = 142C
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EXERCISE 107, Page 228
1. Determine the quantity of heat energy (in megajoules) required to raise the temperature of 10 kg
of water from 0C to 50C. Assume the specific heat capacity of water is 4200 J/(kg C).
Quantity of heat energy, Q = mc(t 2 - t 1 )
= 10 kg  4200 J/(kg C)  (50 - 0)C
= 2100000 J or 2100 kJ or 2.1 MJ
2. Some copper, having a mass of 20 kg, cools from a temperature of 120C to 70C. If the
specific heat capacity of copper is 390 J/(kg C), how much heat energy is lost by the copper ?
Quantity of heat energy, Q = mc(t 2 - t 1 )
= 20 kg  390 J/(kg C)  (70 - 120)C
= 20  390  - 50 = - 390000 J or 390 kJ
Hence, the heat energy lost by the copper = 390 kJ
3. A block of aluminium having a specific heat capacity of 950 J/(kg C) is heated from 60C to its
melting point at 660C. If the quantity of heat required is 2.85 MJ, determine the mass of the
aluminium block.
Quantity of heat, Q = mc(t 2 - t 1 ), hence,
2.85  10 6 J = m  950 J/(kg C)  (660 - 60)C
i.e.
2850000 = m  950  600
from which,
mass, m =
2850000
kg = 5 kg
950  600
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4. 20.8 kJ of heat energy is required to raise the temperature of 2 kg of lead from 16C to 96C.
Determine the specific heat capacity of lead.
Quantity of heat, Q = mc(t 2 - t 1 ), hence:
20.8  10 3 J = 2 kg  c  (96 – 16)C where c is the specific heat capacity,
20800 = 2  c  80
i.e.
from which, specific heat capacity of lead, c =
20800
= 130 J/(kg C)
2  80
5. 250 kJ of heat energy is supplied to 10 kg of iron which is initially at a temperature of 15C. If
the specific heat capacity of iron is 500 J/(kg C) determine its final temperature.
Quantity of heat, Q = mc(t 2 - t 1 ), hence,
250  10 3 J = 10 kg  500 J/(kg C)  (t 2 - 15)C
from which, (t 2 - 15) =
250 103
= 50
10  500
Hence, the final temperature, t 2 = 50 + 15 = 65C
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EXERCISE 108, Page 229
1. Some ice, initially at - 40C, has heat supplied to it at a constant rate until it becomes
superheated steam at 150C. Sketch a typical temperature/time graph expected and use it to
explain the difference between sensible and latent heat.
See Section 20.4 and Figure 20.1 on page 228 of textbook. Just replace the - 30C at A with - 40C
and replace 120C at F with 150C.
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EXERCISE 109, Page 230
1. How much heat is needed to melt completely 25 kg of ice at 0C. Assume the specific latent heat
of fusion of ice is 335 kJ/kg.
Quantity of heat required, Q= mL = 25 kg  335 kJ/kg
= 8375 kJ or 8.375 MJ
2. Determine the heat energy required to change 8 kg of water at 100C to superheated steam at
100C. Assume the specific latent heat of vaporisation of water is 2260 kJ/kg.
Quantity of heat required, Q = mL = 8 kg  2260 kJ/kg
= 18080 kJ or 18.08 MJ
3. Calculate the heat energy required to convert 10 kg of ice initially at - 30C completely into
water at 0C. Assume the specific heat capacity of ice is 2.1 kJ/(kg C) and the specific latent
heat of fusion of ice is 335 kJ/kg.
Quantity of heat energy needed, Q = sensible heat + latent heat.
The quantity of heat needed to raise the temperature of ice from - 30C to 0C
i.e. sensible heat, Q 1 = mc(t 2 - t 1 ) = 10 kg  2100 J/(kgC)  (0 - - 30)C
= (10  2100  30) J = 630 kJ
The quantity of heat needed to melt 10 kg of ice at 0C,
i.e. the latent heat, Q 2 = mL = 10 kg  335 kJ/kg = 3350 kJ
Total heat energy needed, Q = Q 1 + Q 2 = 630 + 3350 = 3980 kJ = 3.98 MJ
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4. Determine the heat energy needed to convert completely 5 kg of water at 60C to steam at
100C, given that the specific heat capacity of water is 4.2 kJ/(kg C) and the specific latent heat
of vaporisation of water is 2260 kJ/kg.
Quantity of heat required = sensible heat + latent heat.
Sensible heat, Q 1 = mc(t 2 - t 1 ) = 5 kg  4.2 kJ/(kg C)  (100 - 60)C
= 840 kJ
Latent heat, Q 2 = mL = 5 kg  2260 kJ/kg = 11300 kJ
Total heat energy required, Q = Q 1 + Q 2 = (840 + 11300) kJ
= 12140 kJ or 12.14 MJ
EXERCISE 110, Page 232
Answers found from within the text of the chapter, pages 225 to 232.
EXERCISE 111, Page 233
1. (d) 2. (b) 3. (a) 4. (c) 5. (b) 6. (b) 7. (b) 8. (a) 9. (c) 10. (b) 11. (d) 12. (c) 13. (d)
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