At `100^(@)C` the volume of `1 kg` of water is `10^(-3) m^(3)` and volume of `1 kg` of steam at normal pressure is `1.671 m^(3)`. The latent heat of steam is `2.3 xx 10^(6) J//kg` and the normal pressure is `10^(5) N//m^(2)`. If `5 kg` of water at `100^(@)C` is converted into steam, the increase in the internal energy of water in this process will be

Options

A. `8.35 xx 10^(5) J`
B. `10.66 xx 10^(6) J`
C. `11.5 xx 10^(5) J`
D. zero

Answer

B

Explanation

b. Heat required to convert 5 kg of water into steam
`Delta Q = mL = 5 xx 2.3 xx 10^(6) = 11.10^(6) J`
Work done in expanding volume,
`Delta W = P Delta V`
`= 5 xx 10^(5) (1.671 - 10^(-3)) = 0.835 xx 10^(6) J`
Now by the first law of thermodynamics `Delta U = Delta Q - Delta W`
`implies Delta U = 11.5 xx 10^(6) - 0.835 xx 10^(6) J`