When 1 kg of ice at 0 melts to water at 0 , the resulting change in its entropy, taking latent heat of ice to be 80 cal/g, is
- 273 cal/K
- 8 104 cal/K
- 80 cal/K
- 293 cal/K
Answer
(4) 293 cal/K
Solution
Heat absorbed in melting: Q = mL = 1000 g 80 cal/g = 8 104 cal
Melting occurs at constant temperature T = 273 K.
S = QT = 8 104273
293 cal/K
Answer: (4)