Question

Given that the heat of fusion of water is -6.02 kJ/mol, that the heat capacity of H2O(l) is 75.2 J/mol·K and that the heat capacity of H2O(s) is 37.7 J/mol·K, calculate the heat of fusion of water at -74°C.

Answer 1

Solution :-

heat of fusion of water is -6.02 kJ/mol = 6020 J/mol

heat capacity of H2O(s) is 37.7 J/mol·K

first solid water will change from -74 C to 0 C then it will melt at 0 C

So Q total = q ice + q fus

               = (n*c*delta T)+ delta H fus

               = (1 mol * 37.7 J per mol K *74 C)+ 6020 J per mol

               = 8810 J per mol

converting J to kJ

8810 J per mol * 1 kJ / 1000 J = 8.81 kJ/ mol

So the answer is 8.81 kJ/ mol