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 -13 ∘C.

Answer 1

Water at -13 C is ice ,

so we first heat ice from -13 to 0 , heat required per 1 mole = Molar heat capcity of ice x temp change x moles

                                                                                     = 37.7 x ( 13) x 1 = 490.1 J= 0.49 KJ

Now ice is melted at 0C to water , heat required per mole = enthalphy of fusion x moles = 6.02 x 1 = 6.02 KJ

Now total heta = 6.02+0.49 = 6.51 KJ

Thus heat of fusion at -13 C water is   - 6.51 KJ/mol          ( -ve sign indicates heat required)