S1 S2 S3 S4 S5 1 mol H2O(S) -1 o C ----> H2O(S) 0 o C ----> H2O(l) 0 o C ----->

Question

                  S1                   S2                  S3           S4              S5
1 mol H2O(S) -1oC ----> H2O(S) 0oC ----> H2O(l) 0oC -----> H2O(l) 100 oC -----> H2O(g) 100oC ---> H2O(g) 101oC

S1 = n.Cp.ln(Tf/Ti) = (1mol x 36.9J/mol.K)ln (273.15/272.15) = 0.135 J/K

S2 = n.Hfus/T = (1molx 6.01 x1000 J)/ mol/273.15 K = 21.966 J/K

S3 = n.Cpm.ln(Tf/Ti) = 1mol x 75.5 J/mol*K ln (373.15/273.15) = 23.553 J/K

S4 = n.Hvap/T = (1 mol x 40.7 x1000 J/mol)/373.15 K = 109.07 J/K

S4 = n.Cp.ln(Tf/Ti) = (1mol x 36.5 J/mol.K)ln(374.15/373.15) = 0.977 J/K

Total entropy S = S1 + S2 + S3 + S4 + S4
               = 0.135 J/K+ 21.966 J/K + 23.553 J/K + 109.07 J/K + 0.977 J/K
               = 155.7 J/K
              
Therefore, the total entropy change = 155.7 J/K

Explanation / Answer

The standard enthalpy of fusion of water is 6.01 kJ/ mol, the molar heat capacity at a constant pressure of liquid water can be assumed to be constant at 75.5 J/mol*K, thoughout the liquid phase, and the standard enthalpy of vapoziration is 40.7 kJ/mol at the boiling point. Calculate the total entropy change (in SI units), for heating one mole of water, starting with ice just below the melting point and going to just above the boiling point at standard pressure.

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