## Problem 10: A block of ice at 273 K is put in thermal contact with a container of steam at 373 K, converting 25.0 g of ice to water at 273 K while condensing some of the steam to water at 373 K. (a) Find the change in entropy of the ice. (b) Find the change in entropy of the steam. (c) Find the change in entropy of the universe.

Solution

Temperature of ice, Ti = 273 K                    Temperature of steam, Ts = 373 K

Mass of ice, mi = 25.0 g = 0.025 kg             Latent heat of fusion of ice, Hf = 3.333 × 105J

To find                                                  (a) Change in the entropy of ice, ΔSi

(b) Change in the entropy of steam, ΔSs

(c) Change in the entropy of universe, ΔSu

Calculation: We know that ΔS = ΔQ/T. So, we have to find ΔQ first. Also, the heat gained by 25.0 g of ice to convert to water = heat lost by steam to condense into water. The latent heat of fusion of ice is 3.33 × 105 J. Therefore,

Heat gained by 1 kg of ice to convert to water at 273 K = 3.33 × 105 J

∴, heat gained by 0.025 kg of ice to convert to water = 3.33 × 0.025 × 105 J = 0.08325 × 105 J

Similarly, this much heat is lost by the steam to convert to water.

Now, apply the formula, we have

Change in the entropy of steam,

Since heat is emitted in the process, therefore, the entropy is negative.

Change in the entropy of the universe is equal to the sum of the changes in the entropies of the ice and steam. Therefore