1.

In the process : `H_(2)O(s, -10^(@)C, 1atm)rarrH_(2)O(1, 10^(@)C, 1 atm)` `C_(p)` for ice = 9 cal `deg^(-1) mol^(-1), C_(p)` for `H_(2)O=18` cal `deg^(-1)mol^(-1)`. Latent heat of fusion of ice = 1440 cal `mol^(-1)` at `0^(@)C`. The entropy change for the above process is 6.258 cal. `deg^(-1)mol^(-1)` Give the total number of steps in which the third law of thermodynamics is used

Answer» Correct Answer - B
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)=underset(-10)overset(0)intn(C_(p))/(T)dT=1xx9xx2.3xxlog.(273)/(263)=0.336" cal deg"^(-1) mol^(-1)`
Step 2 (using the second law of thermodynamics) :
`H_(2)O(s)(0^(@)C, 1 atm)rarrH_(2)O(l)(0^(@)C, 1 atm)`
`DeltaS_(2)=(q_(rev))/(T)=(1440)/(273)=5.258" cal deg"^(-1)mol^(-1)`
Step 3 (using the third law of thermodynamics) :
`H_(2)O(l)(0^(@)C, 1 atm)rarrH_(2)O(l)(10^(@)C, 1 atm)`
`DeltaS_(3)=underset(0)overset(10)intn(C_(p))/(T)dT=1xx18xx2.3xxlog.(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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