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. `84.5J`
B. `104.3J`
C. `42.2J`
D. `208.7J`

Answer

D

Explanation

`Delta Q = 54cal = 54xx4.18joule = 225.72` joule
`DeltaW = P[V_(steam)-V_(water)]`
[for water 0.1 gram = 0.1 c c]
`=1.013xx10^(5)[167xx10^(-6)-0.1xx10^(-6)]"joule"`
`=1.013xx167xx10^(-1)=16.917 "joule"`
By first law of thermodyamics
`implies DeltaU=DeltaQ-DeltaW=225.72-16.917`
`DeltaU = 208.8 "Joule"`.