1.

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`.

Answer» Step 1. (using the third law of thermodynamics):
(For changing
`H_(2)O(s) (-10^(@)C, 1 atm) rarrH_(2)O(s, 0^(@)C1 atm)`
`DeltaS_(1) =int_(-10)^(0) n(C_(P))/(T)dT = 1 xx 9 xx 2.3 xx log(273)/(263)`
`= 0.336 cal deg^(-1) mol^(-1)`
Step 2 (using the second law of thermodynamics):
`H_(2)O(s) (0^(@)C, 1atm) rarr H_(2)O(l) (10^(@)C, 1atm)`
`DeltaS_(2) = (q_(rev))/(T) = (1440)/(273) = 5.258 cal deg^(-1) mol^(-1)`
Step 3 (using the third law of themrodynamics):
`H_(2)O(l) (0^(@)C, 1atm) rarr H_(2)O(l)(10^(@)C, 1 atm)`
`DeltaS_(3) =int_(0)^(10) n(C_(P))/(T) dT = 1 xx 18 xx 2.3 xx log (283)/(273)`
`= 0.647 cal deg^(-1) mol^(-1)`
`DeltaS = DeltaS_(1) +DeltaS_(2) +DeltaS_(3) = 0.336 +5.258 +0.647`
`=6.258 cal deg^(-1) mol^(-1)`


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