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 -10 ∘C. Please show work

Answer 1

Heat of fusion of water

Generally we calculate heat of fusion at 0 °C

H2O(l) <====è H2O(s)

But here water is supercooled to -10 °C

H(fusion, -10 °C) = H(fusion, 0 °C) + [Cp(l)-Cp(s)]10 °C

                                    = -6.02 Kj/mol +[75.2-37.7]10

                                    =-6.02 kJ/mol+375 J/mol

                                    =-6.02 kJ/mol+0.375 kJ/mol

                                    =-5.645 kJ/mol

please rate