When 2 kg of ice at 0°C melts to water at 0°C, the resulting change in its entropy, taking latent heat of ice to be 80 cal/g is :

293 cal/K 8 x 104 cal/K 273 cal/K 586 cal/K

586 cal/K

Q = mL = 2000 x 80 cal = 160,000 cal \(\Delta S = \frac{\Delta Q}{T}\) \(\Delta S = \frac{160,000}{273} = 586 cal/K\)