What is the role of Shot and Scold in the entropy equation?

[theBEAST]

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:

S_hot = Q/T_hot
S_cold = Q/T_cold

Where do the S_hot and S_cold terms go in that equation?

 

[rude man]

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.

 

[theBEAST]

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 = ₁S_2,generated.

But if ∫δQ/T = ₁S_2,gen, then that means S₂ - S₁ = 2∫δQ/T

 

[rude man]

Sorry I meant ∫δQ/T = ₁S_2,generated.

But if ∫δQ/T = ₁S_2,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?

 

[theBEAST]

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!