When water is boiled at `2` atm pressure the latent heat of vapourization 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)`]

Options

A. `2.033 J`
B. `2.033xx10^(6) J`
C. `0.167xx10^(6) J`
D. `2.267xx10^(6) J`

Answer

B

Explanation

Total heat given to convert water into steam at `120^(@)C` is, `Q=mL`
`Q=1xx2.2xx10^(6)J=2.2xx10^(6)J`
The work done by the system against the surrounding is `W=pDeltaV`
`W=(2xx1.013xx10^(5))(0.824-0.001)J`
`=0.167xx10^(6)J DeltaU=Q-W=2.033xx10^(6)J`