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. 273cal/K
B. `8 xx 10^(4)` cal/K
C. 80cal/K
D. 293 cal/K

Answer

D

Explanation

Heat required to melt 1kg ice at `0^(@)C` to water at `0^(@)C` is `Q= m_("ice") L_("ice") = (1kg) (80 "cal/g")`
=(1000g) (80 cal/g), `=8 xx 10^(4)` cal
Change in entropy, `Delta S= (Q)/(T) = (8 xx 10^(4)"cal")/((273K)) = 293` cal/K
Note: In the question paper unit of latent heat of ice is given to be cal/`""^(@)C`. It is wrong. The unit of latent heat of ice is cal/g