A sample of `0.1 g` of water of `100^(@)C` and normal pressure `(1.013 xx 10^(5) N m^(-2))` requires 54 cal of heat energy to convert to steam at `100^(@)C`. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample is

Options

A. `42.2 J`
B. `84.5 J`
C. `208.7 J`
D. `104. 3 J`

Answer

C

Explanation

`Delta Q = Delta U + Delta W`
`Delta = Delta Q - Delta W = 54 xx 4.2 - P Delta V`
`= 226.8 - 1.013 xx 10^(5) xx (167.1 - 0.1) xx 10^(-6)`
`= 226.8 - 16.9`
`~= 208.7 J`