## Homework Statement

In the entropy equation 1S2 + ∫δQ/T = S2 - S1, what is the difference between 1S2 and ∫δQ/T?

So for example if we have two reservoirs, one hot and one cold and heat is transferred in between we have:

Shot = Q/Thot
Scold = Q/Tcold

Where do the Shot and Scold terms go in that equation?

rude man
Homework Helper
Gold Member
A reservoir has a constant temperature by definition. So what does that enable you to do in evaluating ∫δQ/T ?

And BTW ∫δQ/T = 1S2 only if evaluated over a reversible path. Otherwise the integral is < entropy change.

A reservoir has a constant temperature by definition. So what does that enable you to do in evaluating ∫δQ/T ?

And BTW ∫δQ/T = 1S2 only if evaluated over a reversible path. Otherwise the integral is < entropy change.

Sorry I meant ∫δQ/T = 1S2,generated.

But if ∫δQ/T = 1S2,gen, then that means S2 - S1 = 2∫δQ/T

rude man
Homework Helper
Gold Member
Sorry I meant ∫δQ/T = 1S2,generated.

But if ∫δQ/T = 1S2,gen, then that means S2 - S1 = 2∫δQ/T

No it does not. S2 - S1 = ∫δQ/T over a reversible path. S2 - S1 is the entropy change in ONE of the reservoirs. Since you have two reservoirs at two different temperatures T1 and T2 , each has its own S2 - S1.

Have you figured out how that integral is simplified if you're dealing with entropy changes of reservoirs?

No it does not. S2 - S1 = ∫δQ/T over a reversible path. S2 - S1 is the entropy change in ONE of the reservoirs. Since you have two reservoirs at two different temperatures T1 and T2 , each has its own S2 - S1.

Have you figured out how that integral is simplified if you're dealing with entropy changes of reservoirs?

Yeah I know how to do that. Thanks!