When water is boiled at `2` atm pressure the latent heat of vaporization is `2.2xx10^(6) J//kg` and the boiling point is `120^(@)C` At `2` atm pressure `1` kg of water has a volume of `10^(-3)m^(-3)` and `1` kg of steam has volume of `0.824m^(3)` . The increase in internal energy of `1` kg of water when it is converted into steam at `2` atm pressure and `120^(@)C` is [`1` atm pressure `=1.013xx10^(5)N//m^(2)`]

Answer

`10.5^(@)C`