Heat Q flows spontaneously from a reservoir at 394 K

[copitlory8]

Heat Q flows spontaneously from a reservoir at 394 K into a reservoir that has a lower temerature T. Because of the spontaneous flow, thirty percent of Q is rendered unavailable for work when a Carnot engine operates between the reservoir at temperature T and reservoir at 248 K. Find the temperature T.

First of all... I don't even understand the problem.

I know the change in entropy equals the heat (Q) divided by the temperature (T). I also know that the work unavailable (W) equals the temp of the coldest reservoir multiplied by the change of entropy of the universe.

I would appreciate any help! Thanks

 

[jbsteele]

.The temperature T is the temperature of the reservoir into which Q is flowing. The equation you need to solve is: 0.3Q = (T - 248K) * ΔS where ΔS is the change in entropy of the universe due to the flow of heat Q. Once you solve for T, you will have the temperature of the reservoir.