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When 1 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 273 cal/K 8 x 104 cal/K 80 cal/K |
293 cal/K |
Q = mL = 80×1000 cal = 80,000 cal \(\Delta S = \frac{\Delta Q}{T}\) \(\Delta S = \frac{80,000}{273} = 293 cal/K\) |
