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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 temperture and at 1 atm (100 kPa) will be:
(The density of water and steam are 1000kg/m3 and 0.6kg/m3 respectively. The latent heat of vapourisation of water is 2.25 106 J/kg.)

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

The correct option is A 2.08×106 J
Latent heat of vaporisation of water , H=2.25×106 J/kg
H=2.25×106 J/kg
and ,
Work done =Pext(V2V1)


(a) Now, volume of water V=(massdensity) =(md)=11000m3, since 1 m3=1000 L
=1 L
(b) Volume of steam =(massdensity) 10.6 m3=10.6×1000 L=1666.67 L
So, using the relationship,
H=U+PV
2.25×106=U+1[1666.671]1.01325×105
U=22.5×1051.68×105=20.8×105=2.08×106 J

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