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

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Homework Help Overview

The discussion revolves around the entropy equation and the roles of the terms 1S2 and ∫δQ/T, particularly in the context of heat transfer between two reservoirs at different temperatures.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the definitions and implications of the terms in the entropy equation, questioning how the heat transfer terms Shot and Scold relate to the overall entropy change. There is discussion about the conditions under which the integral ∫δQ/T equals the change in entropy, particularly focusing on reversible processes.

Discussion Status

The conversation is active, with participants clarifying definitions and exploring the implications of constant temperature in reservoirs. Some guidance has been offered regarding the evaluation of the integral and its relationship to the entropy change in the reservoirs.

Contextual Notes

There is an emphasis on the distinction between the entropy change of individual reservoirs and the overall system, as well as the conditions required for the integral to represent the entropy change accurately.

theBEAST
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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?
 
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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.
 
rude man said:
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
 
theBEAST said:
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?
 
rude man said:
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!
 

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