
#1
Apr1012, 10:18 AM

P: 46

Hello,
I'd like to get some more accurate idea of entropy in general: [tex]dS=\int\frac{dQ}{T}[/tex] 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 [tex]dQ[/tex] increases. A reversible system will not undergo a rise in entropy because of the fact that no losses will occur, so [tex]dQ=0[/tex] 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 



#2
Apr1012, 10:55 AM

P: 383

Yeah that is the explanation my thermodynamics teacher gave. I dont know what is modern physics take on entropy. So who is gonna exlpain to us?




#3
Apr1012, 02:40 PM

P: 46

grtz 



#4
Apr1012, 02:51 PM

P: 383

reason of entropy rise 


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