Question

What is the entropy change to the surroundings when a small, decorative ice scuplture at a temperature of 0 degrees C and weighing 456 g melts on a granite tabletop if the temperature of the granite is 12 degrees C and the process occurs reversibly? Assume the final temperature of the water is 0 degrees C. The heat of fusion of ice is 6.01 kJ/mol.

Answer 1

heat of fusion = 6.01 kj/mol

mass of ice =456 gms moles of ice =mass/molecular weight =456/18=25.33 moles

total heat to be added   = moles of water*heat of fusion of ice =25.33*6.01=152.233 Kj

Entropy change of water = heat added/ temperature = 152.233/ (0+273.15)= 0.56 Kj/K

Heat has to be supplied by granite top which is 152.233 Kj

Entropy change of granite= 152.233/ (12+273.15)=-0.53 ( since granite is lossing heat)

entropy change of surroundings = entropy change of water + entropy change of granite= 0.56-0.53=0.03 Kj/K