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Question

The increase in internal energy of 1 kg of water at 1000C when it is converted into steam at the same temperature and at 1atm (100kPa) will be [The density of water and steam are 1000kg/m3 and 0.6kg/m3 respectively. The latent heat of vapourisation of water is 2.25×106J/kg] :

A
2.08×106J
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B
4×107J
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C
3.27×108J
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D
5×109J
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Solution

The correct option is A 2.08×106J
LV=2.25×106 J/kg
m=1kg
p=1×102pa
volume of water (V)=103m3
volume of steam (V)=10.6m3
Increase in volume=(10.611000)=1.7m3
Work done by system=P(AV)=(1×105)(1.7)=1.7×105j
Q=mlv=2.25×106j
change in internal energy U=QW
U=2.25×1061.7×105
U=2.25×1060.17×106
U=2.08×106j

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