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 2.08×106JLV=2.25×10−6 J/kgm=1kgp=1×102pavolume of water (V)=10−3m3volume of steam (V)=10.6m3Increase in volume=(10.6−11000)=1.7m3Work done by system=P(AV)=(1×105)(1.7)=1.7×105j△Q=mlv=2.25×106jchange in internal energy △U=△Q−△W△U=2.25×106−1.7×105△U=2.25×106−0.17×106△U=2.08×106j

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