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 1 atm (100kPa). The density of water and steam are `1000(kg)/(m^3)` and `0.6(kg)/(m^3)` resprectively.
The latent heat of vaporization of water `=2.25xx10^6(J)/(kg)`

Explanation

`V_(omega)=(1)/(1000)m^3`,`V_S=(1)/(0.6)m^3=1.7m^3`
`DeltaV=V_S-V_(omega)=(1.7-10^-3)cong1.7m^3`
`DeltaW=PDeltaV=100xx10^3xx1.7=1.7xx10^6J`
`DeltaQ=mL_S=1xx2.25xx10^6J`
`DeltaU=DeltaQ-DeltaW=(2.25-1.7)xx10^6`
`=2.08xx10^6J`