Entropy Rise: Causes and Effects

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Discussion Overview

The discussion revolves around the concept of entropy, particularly in the context of adiabatic isolated systems and irreversible processes. Participants explore the implications of work done on such systems and the conditions under which entropy changes.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification

Main Points Raised

  • One participant presents the equation dS=\int\frac{dQ}{T} and discusses how entropy increases in an adiabatic isolated system due to irreversible work, suggesting that irreversible losses contribute to an increase in dQ.
  • Another participant agrees with the explanation provided by their thermodynamics teacher but expresses uncertainty about the modern physics perspective on entropy.
  • There is a request for clarification regarding the modern physics interpretation of entropy, indicating a lack of consensus on this aspect.
  • Some participants seek further elaboration on the term "modern physics" in relation to entropy, indicating confusion or a need for more information.

Areas of Agreement / Disagreement

Participants generally agree on the basic principles of entropy as discussed in thermodynamics, but there is uncertainty regarding the modern physics perspective and no consensus on what that perspective entails.

Contextual Notes

The discussion does not resolve the definitions or implications of irreversible losses and their relationship to work and entropy. There are also unresolved questions about the modern physics interpretation of entropy.

HWGXX7
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Hello,

I'd like to get some more accurate idea of entropy in general: dS=\int\frac{dQ}{T}

Given an adiabatic isolated system. Work is irrerversible done onto this system.

Entropy will rise because of the fact that some of the work is transformed into irreversible losses and therefore dQ increases.

A reversible system will not undergo a rise in entropy because of the fact that no losses will occur, so dQ=0 and entropy will remain the same.

Must I also assume that even when the losses are reversible transformable into work again, the transformation is still labelled 'irreversible' . Because losses are inherent factor of a irreversible system (KELVIN) ?

Is this somehow a correct interpretation I have?

ty&grtz
 
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Yeah that is the explanation my thermodynamics teacher gave. I don't know what is modern physics take on entropy. So who is going to exlpain to us?
 
Last edited:
I don't what is modern physics take on entropy.
What did you mean with that?

grtz
 
HWGXX7 said:
What did you mean with that?

I missed a word.
 

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