Help on Entropy Problem?

Question

A heat engine operates between a high-temperature reservoir at 610 K and a low temperature reservoir at 320 K. In one cycle, the engine absorbs 6400 J of heat from the high temperature reservoir and does 2200J of work. What is the net change in entropy as a result of this cycle.

I did what you guys said and still got the wrong answer.

Answer 1

Entropy S = Q / T. The entropy of the "hot" reservoir goes down by ( 6400 J ) / 610 K. The entropy of the "cool" reservoir increases by ( 4200 J ) / 320 K.

4200 / 320 - 6400 / 610
13.125 - 10.492
2.633 J / K

As expected, entropy increased in this process; the entropy lost by the hot side was less than the entropy gained by the cool side.

Answer 2

Ok, since the engine absorbs 6400J from the high temp side, then does 2200J of work, it must expell 6400-2200J of heat to the low temp side, since in one cycle dU =0

entropy will then be given by S1 +S2
= 2200/610 + 4200/320