Question

An ice cube of mass 500 g at 0 °C is dropped into an insulated container of 1.0 kg of water that initially is at room temperature (25 °C), and eventually the system reaches equilibrium. The insulator is not perfect, so 20 kJ of heat flows from the room into the water during the process. 3. a. Calculate the entropy increase in the ice that melts into water. b. Calculate the entropy loss of the water that cools down. c. How much entropy is lost by the room in transferring heat to the water? d. What is the net change in entropy of the overall system?

Answer 1

latent heat of ice is 334 J/g, the specific heat of water is 4.186 joule/gram °C and an additional 20KJ from surroundings

the heat required to convert 1kg water at 25 deg to 0 deg is 1000*4.186*25 = 104650 J

==> total heat gained by ice of 104.650+20KJ = 124.65KJ at 273 K ==> entropy change is 124650/273 =456.593 J/K

==> entropy change for surroundings is-20000J/298 K = -67.114 J/K

we know that entropy is dQ/T , dQ = ms$\Delta$T ==> $\Delta$ s = $\frac{ms\Delta T}{T}$ on integrating from 25 degrees to 0 degrees we get change in entropyon calculating we get entropy change for 1 Kg water is -366.784J/K

A entropy increase in ice that melts into water is 456.593 J/K

B entropy decrease in water that melts is 366.784 J/K

C entropy decrease in surroundings is 67.114J/K

D net change in entropy is 456.593-366.784-67.114=22.95 J/K (increase)