A sophomore with nothing better to do adds heat to 0.350 kg of ice at 0.0o C until it is all melted. (a) What is the change in entropy of the water? (b) The source of heat is a very massive body at 25.0o C. What is the change in entropy of this body? (c) What is the total change in entropy of the water and the heat source?

Solution 23E Step 1 of 5: A sophomore with nothing better to do adds heat to 0.350 kg of ice at 0.0o C until it is all melted. (a) What is the change in entropy of the water The ice of 0.350 kg gains entropy by absorbing heat from the sophomore and warms 0 from 0 c. Given data, For ice, 0 Temperature, Tice= 0 c T ice 0 + 273 = 273 K Mass of ice, m=0.350 kg 3 Latent heat of fusion, L=334× 10 J/kg Specific heat of water, C= 4190 J/kg.K To find, Heat gained by ice during melting,melt Entropy change in ice during melting, melt Step 2 of 5: To calculate heat gained and entropy change during melting of ice, The heat gain during the melting of 0.350kg ice block is, Q melt mL 3 Substituting m= 0.350 kg and L=334× 10 J/kg 3 Q melt (0.350 kg)(334 × 10 J/kg) Q melt 1.169 × 10 J Change in entropy during melting of ice is given by, S = Qmelt melt Tice 6 Substituting Q melt= 1.169 × 10 J and T ice= 273 K S = 1.169×10 J melt 273 K S melt= 429 J/K Therefore, the change in entropy of water during melting of ice is 429 J/K . Step 3 of 5: (b) The source of heat is a very massive body at 25.0o C. What is the change in entropy of this body 0 0 The source of heat for the melting ice from 0 c is a very massive body at 25 c. In order to calculate change in entropy of this massive body, first we need to calculate the heat lost or gained by massive body. Given, For massive body, Temperature, T = 25 c body T body= 25 + 273 = 298 K To find, Heat lost by massive body, Q body = Entropy change of massive body, S = body