Question

An ice cube of mass $5kg$ is put on a tray at just above $0^{\circ }C$. The ice melts by absorbing energy from the surroundings as heat. What is the change in entropy (in kJ/K) of the sample of ice? (Latent heat of ice = 78 cal/g and 1 cal = 4.2 J)

Answer 1

Heat absorbed by 5 kg ice $=\frac{78\times 5000\times 4.2}{1000}kJ=1638kJ$
So, entropy change of ice $=\frac{1638}{273}kJ/K=6kJ/K$