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

Answer

B

Explanation

Change in heat (`DeltaQ`)-54 cal = `54xx4.18`=225.72.J
Change in work done `(Delta w)=P[v-v_w]`
`implies Delta w=1.013xx10^5 [167.1xx10^(-6) -0.1xx10^(-6)]` [as 1g = 1 cc]
`implies Delta w = 16.917 J`
Since, `DeltaQ=DeltaU + Deltaw`
`implies Delta U=225.72-16.917`
`implies DeltaU=208.8 J`