Mechanical Engineering Principles

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216 MECHANICAL ENGINEERING PRINCIPLES

Table 19.1


Latent heat of Melting
fusion (kJ/kg) point (°C)

Mercury 11.8 − 39
Lead 22 327
Silver 100 957
Ice 335 0
Aluminium 387 660


Latent heat of Boiling
vaporisation (kJ/kg) point (°C)

Oxygen 214 − 183
Mercury 286 357
Ethyl alcohol 857 79
Water 2257 100


Quantity of heat required,


Q=mL=12 kg×335 kJ/kg

=4020 kJor 4 .02 MJ

Problem 11. Calculate the heat required to
convert 5 kg of water at 100°Cto
superheated steam at 100°C. Assume the
latent heat of vaporisation of water is
2260 kJ/kg.

Quantity of heat required,


Q=mL=5kg×2260 kJ/kg

=11300 kJor 11 .3MJ

Problem 12. Determine the heat energy
needed to convert 5 kg of ice initially at
− 20 °C completely to water at 0°C. Assume
the specific heat capacity of ice is
2100 J/(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 tempera-
ture of ice from− 20 °Cto0°C, i.e. sensible heat,


Q 1 =mc(t 2 −t 1 )

=5kg×2100 J/(kg°C)×( 0 −− 20 )°C

=( 5 × 2100 × 20 )J=210 kJ

The quantity of heat needed to melt 5 kg of ice at
0 °C,
i.e. the latent heat,

Q 2 =mL=5kg×335 kJ/kg=1675 kJ

Total heat energy needed,

Q=Q 1 +Q 2 = 210 + 1675 =1885 kJ

Problem 13. Calculate the heat energy
required to convert completely 10 kg of
water at 50°C into steam at 100°C, given
that the specific heat capacity of water is
4200 J/(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 )

=10 kg×4200 J/(kg°C)

×( 100 − 50 )°C=2100 kJ

Latent heat, Q 2 =mL=10 kg×2260 kJ/kg

=22600 kJ
Total heat energy required,

Q=Q 1 +Q 2 =( 2100 + 22600 )kJ

=24700 kJor 24 .70 MJ

Problem 14. Determine the amount of heat
energy needed to change 400 g of ice,
initially at− 20 °C, into steam at 120°C.
Assume the following:
latent heat of fusion of ice=335 kJ/kg,
latent heat of vaporisation of
water=2260 kJ/kg, specific heat capacity of
ice= 2 .14 kJ/(kg°C), specific heat capacity
of water= 4 .2 kJ/(kg°C) and specific heat
capacity of steam= 2 .01 kJ/(kg°C).

The energy needed is determined in five stages:

(i) Heat energy needed to change the temperature
of ice from− 20 °Cto0°C is given by:

Q 1 =mc(t 2 −t 1 )

= 0 .4kg× 2 .14 kJ/(kg°C)×(0–− 20 )°C

= 17 .12 kJ
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