EXAMPLE 20-6 Entropy change in melting. A 1.00-kg piece of ice at o'C melts very slowly to water at 0C. Assume the ice is in contact with a heat reser. voir whose temperature is only infinitesimally greater than 0 C. Determine the entropy change of (a)the ice cube and(b) the heat reservoir SoLUTION(a) The process is carried out at a constant temperature T=273K and is done reversibly, so we can use Eq 20-12 do 1 △S;。= TT do Since the heat needed to melt the ice is 0= mL, where the heat of fusion L=797 kcal/kg=3.33 X 10 J/kg, we have mL(1.00 kg)(797 kcal/kg) △Sice=T 273K 0.292 kcal/K or 1220J/K (b)Heat @- mL is removed from the heat reservoir, so(since T=273 K and is constant △Sres =-0.292kcal/K Note that the total entropy change, ASice ASres, is zero