# Thermodynamics Question

1. Dec 6, 2006

### tigerguy

1. The problem statement, all variables and given/known data
Hi, I'm in desperate need for help. Any guidance in the right direction will be appreciated:

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

2. Relevant equations

W=TS

3. The attempt at a solution
0.30Q = TS, where S = Q*/T

I know the T is in between and the Q value from the drop of 348 to T, multiplied by 0.30 will give you the work needed. I'm just so confused; any help will be appreciated! Thanks!

2. Dec 6, 2006

### OlderDan

I have no idea what the reference to "spontaneous flow" is supposed to mean. I do know that a carnot engine can only convert a fraction of the heat input to work and must output some of the heat it takes in. The ratio of heat input to worked depends on the temperatures of the resevoirs. What is that relationship?

3. Dec 6, 2006

### Andrew Mason

Like Dan I am confused by the "because of the spontaneous flow" reference. The efficiency of the Carnot cycle depends only upon the temperatures of the reservoirs between which it operates. I would ignore that part of the question.
I don't think this is relevant and, besides, it is not generally true. From the first law: $dQ = TdS = dU + dW$, so $W = T\Delta S$ only where $\Delta U = 0$
As Dan says, find T from the given efficiency of the Carnot cycle. Q is irrelevant to the calculation.

AM