Calculate the entropy change accompanying the following change of state
`H_(2)O (s, 10^(@)C, 1 atm) rarr H_(2)O(l, 10^(@)C, 1atm)`
`C_(P)` for ice `= 9 cla deg^(-1) mol^(-1)`
`C_(P)` for `H_(2)O = 18 cal deg^(-1) mol^(-1)`
Latent heat of fustion of ice `= 1440 cal mol^(-1) at 0^(@)C`.

Explanation

The total process involved three stages and entropy change can be calculated for each stage as follows:
`[DeltaS_(1) rArr` For changing `1` mole of ince from `-10^(@)C, 1atm` to `0^(@)C, 1atm]`
`DeltaS_(1) = int_(-10)^(0) n.(C_(P))/(T).dT`
`=n xx C_(P) xx 2.303 xx "log"(273)/(263)`
`= 1 xx 9 xx 2.303 xx 0.0162`
`= 0.336 cal deg^(-1) mol^(-1)`
`DeltaS_(2) rArr` For melting `1` mol of ice at `0^(@)C` will be as
`DelatS_(2) = (q_(rev))/(T) = (1440)/(273) = 5.25 cal deg^(-1) mol^(-1)`
`DeltaS_(3) rArr` For heating `1` mol of `H_(2)O` from `0^(@)C` to `10^(@)C` at `1 atm`
`DeltaS_(3) = int_(0)^(10) (nC_(P))/(T) dT`
`= 1 xx 18 xx 2.303 "log" (283)/(273)`
`= 0.647 cal deg^(-1) mol^(-1)`
`DeltaS = 0.336 +5.276 + 0.647`
`= 6.258 cal deg^(-1) mol^(-1)`