Some of you may have heard of refrigerators powered by natural gas. It sounds odd to use a hot flame to cool something down. This problem doesn't explore the details, just looks at a simple schematic illustration of the principle.

Say that you are in a large room at temperature K. You have access to some ideal heat-engine machines. Someone gives you a pot of hot Entropade at T= K. Entropade conveniently has C_v= *T J/K², a T-dependent heat capacity which makes some math easier. (Note that if T is in K, the units of C_v are the usual J/K.)(Ignore changes in the volume of the Entropade.)

a) If the pot reaches equilibrium with the big room, what will its temperature be?

T = K

b) How much extra U does the Entropade have because it starts off hotter than the room, rather than at room-T?

U_extra = J

c) How much extra S does the Entropade have because it starts off hotter than the room?

S_extra = J/K

d) How much work can you extract from that pot in that room?

W = J

e) Now you decide you would like some cold Entropade. Assuming you used the work to store some mechanical energy, to what T can you now cool the pot of Entropade?

T = K

Comments:

• That pot of stuff has F higher than the equilibrium value if it's hotter than the room OR if it's colder than the room. You were able to use the extra F from the hot pot to cool it below room-T. If it had come in at room-T, you couldn't have cooled it without some work input.