Entropy Rise: Causes and Effects

AI Thread Summary
In an adiabatic isolated system where irreversible work is done, entropy increases due to energy losses, represented by dQ increasing. Conversely, in a reversible system, no energy losses occur, leading to dQ equaling zero and no change in entropy. The discussion raises the question of whether losses that can be transformed back into work are still considered irreversible. It highlights the inherent nature of losses in irreversible systems, referencing the Kelvin statement of thermodynamics. The conversation seeks clarification on modern interpretations of entropy in physics.
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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