Physics Help Forum second law of thermodynamics

 Aug 23rd 2014, 12:07 PM #1 Junior Member   Join Date: Aug 2014 Posts: 2 second law of thermodynamics Hello everyone My problem is about second law of thermodynamics & reversible work. according to carnot cycle equations we have q/q0=T/T0 which q is absorbed heat from hot reservoir & q0 is exhausted heat to cold reservoir. but according to second law of thermodynamics we have q0 =T0(Si-Se)+q(T0/TH) 1)what is difference between two equations for q0 ? 2) Are two terms T0(Si-Se) & q(T0/TH) Equivalent ?both are q0 .
Sep 5th 2014, 12:51 AM   #2
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 Originally Posted by abcde Hello everyone My problem is about second law of thermodynamics & reversible work. according to carnot cycle equations we have q/q0=T/T0 which q is absorbed heat from hot reservoir & q0 is exhausted heat to cold reservoir. but according to second law of thermodynamics we have q0 =T0(Si-Se)+q(T0/TH) 1)what is difference between two equations for q0 ? 2) Are two terms T0(Si-Se) & q(T0/TH) Equivalent ?both are q0 .
I will check those your formulas later but i think the first is concern with carnot and Second law of thermodynamics. where it is Qh/Th + Qc/Qh = 0 remember it is a matter of impossibility. I will check your second formula and get a conclusion on it cos i dont really understand u from the second one

 Sep 12th 2014, 02:43 PM #3 Member     Join Date: Sep 2014 Location: Brasília, DF - Brazil Posts: 32 Aren't you confused with Exergy analysis?

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