Calculate the increase in internal energy of 1 kg of water at 100 degrees C when it is converted into steam at the same temperature and at 1atm.Te density of water and steam are 1000kg/m^2 and 0.6 kg m^ -3 respectively. The Latent heat of vapourisation of water is 2.25 X 10^6 J kg^ -1.
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Answer:
Explanation:L
V
=2.25×10
−6
J/kg
m=1kg
p=1×10
2
pa
volume of water (V)=10
−3
m
3
volume of steam (V)=
0.6
1
m
3
Increase in volume=(
0.6
1
−
1000
1
)=1.7m
3
Work done by system=P(AV)=(1×10
5
)(1.7)
=1.7×10
5
j
△Q=ml
v
=2.25×10
6
j
change in internal energy △U=△Q−△W
△U=2.25×10
6
−1.7×10
5
△U=2.25×10
6
−0.17×10
6
△U=2.08×10
6
j
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