When `1 kg` of ice at `0^(@)C` melts to water at `0^(@)C`, the resulting change in its entropy, taking latent heat of ice to be `80 cal//g` is

Options

A. `8xx10^(4)"cal "K^(-1)`
B. `80" cal "K^(-1)`
C. `293" cal "K^(-1)`
D. `273" cal "K^(-1)`

Answer

C

Explanation

Change in entropy ,
`DeltaS=(ml)/(T)=(1000xx80)/(273)=293"cal "K^(-1)`