Calculate the increase in internal energy of `1 kg` of water at `100^(@)C` when it is converted into steam at the same temperature and at `1atm` (`100 kPa`). The density of water and steam are `1000kg m^(-3)` and `0.6 kg m^(-3)` respectively. The latent heat of vaporization of water `=2.25xx10^(6) J kg^(1)`.

Answer

`2.08xx10^(6)J`

Explanation

Volume of `1 kg` of water `=1/(1000)m^(3)= 0.001m^(3)`
Volume of `1 kg` of steam `=1/(0.6)m^(3)= 1.7m^(3)`
`:.` Increase in volume,
`DeltaV= 1.7-0.001= 1.7,^(3)`
Work done by the system `=pDeltaV`
`DeltaW=(100xx10^(3))xx1.7= 1.7xx10^(5)J`
`:. DeltaU= DeltaQ-DeltaW= mL- DeltaW`
`= 1xx2.25xx10^(6)-1.7xx10^(5)`
`=2.08xx10^(6)J`