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One gram of water (1 cm^3) becomes 1671 ...

One gram of water `(1 cm^3)` becomes `1671 cm^3` of steam when boiled at a constant pressure of 1 atm `(1.013xx10^5Pa)`. The heat of vaporization at this pressure is `L_v=2.256xx10^6J//kg`. Compute (a) the work done by the water when it vaporizes and (b) its increase in internal energy.

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`DeltaQ=Deltau+Deltaw`
m=1gm
`L_(v)=540cal//gm`
`DeltaQ=1xx540=540`
`540=Deltau+10^(5)xx(1671-1)xx10^(-6)`
`540=Deltau+167`
`Deltau=540-167`
`=373kcal`
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