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)`]

Options

A. `2.033 J`
B. `2.033 xx 10^(6)J`
C. `0.167 xx 10^(6)J`
D. `2.267 xx 10^(6)J`

Answer

B

Explanation

Total heat given to convert water into steam at `120^(@)C` is `Q` = `"m"L = 1xx 2.2 xx 10^(6) = 2.2 xx 10^(6)J`
The work done by the system against the surrounding is
`PdeltaV = 2 xx 1.013 xx 10^(5) (0.824 -0.001) = 0.167 xx 10^(6)J therefore DeltaU= Q-W = 2.033 xx 10^(6)J`