Question

calculate the change of entropy when 27.3 grams of ice at 0 degree C are converted into water at 0 degree C at normal pressure .That latent heat of fusion of ice is 80 calories per gram. ​

Answer 1

The change of entropy when 27.3 grams of ice at 0 degree C are converted into water at 0 degree C at normal pressure is 8 cal/K

1. change in heat is given by
2. ΔQ = mL
3. here m is mass of the ice and L is the latent heat of fusion of ice 80 cal/g
4. ΔQ= 27.3 g x 80 cal/g= 2184 cal
5. The change of entropy is calculated using formula given below
6. ΔS = ΔQ/T=mL/T
7. here ΔS is the change in entropy and T is temperature in kelvin
8. ΔS = 2184/273= 8 cal