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 :
Answer & explanation
Correct answer: option 1
Q = mL = 80×1000 cal = 80,000 cal
\(\Delta S = \frac{\Delta Q}{T}\)
\(\Delta S = \frac{80,000}{273} = 293 cal/K\)