Question

Calculate the entropy change, in melting of ice at 0˚C.(heat of fusion of ice is 6.02 kJ/mol)

Answer 1

This heat is absorbed at the constant temperature T=0+273.15=273.15 K.

The entropy increase in the process:

$ΔS = \frac{ΔH}{T} \\ = \frac{6.02 \times 10^3}{273.15} \\ = 22.04 \;J/mol\;K$