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Question

Calculate the increase in internal energy of 1 kg of water at 100 C when it is converted into steam at the same temperature and at 1 atm (100 kPa). The density of water and steam are 1000 kgm3 and 0.6 kgm3 respectively. The latent heat of vapourisation of water = 2.25×106 J kg1

A
4.16 × 106 J
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B
2.08 × 106 J
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C
1.04 × 106 J
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D
3.04 × 106 J
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Solution

The correct option is B 2.08 × 106 J
The volume of 1kg of water
= 11000 m3
and 1 kg of steam = 10.6 m3
The increase in volume
= 10.6 m3 11000 m3= (1.7 0.001) m3 = 1.7 m3
The work done by the system is pΔV
= (100 kPa)(1.7 m3)= 1.7 × 105 J
The heat given to convert 1 kg of water into steam
= 2.25 × 106 J
The change in internal energy is
ΔU = ΔQ ΔW = 2.25 × 106 J 1.7 × 106 J = 2.08 × 106 J

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