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GR8677 #95
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Thermodynamics$\Rightarrow$}Carnot Cycle

The Carnot cycle is the cycle of the most efficient engine, which does NOT have $e=1$ (unless $T_1=0$), but rather, $dS=0$---this means that entropy stays constant. Choice (C) is thus false.

The efficiency of the Carnot cycle is dependent on the temperature of the hot and cold reservoir. The hot reservoir has decreasing entropy because it gets cooler as the cycle proceeds. From writing down the thermodynamic relations for isothermal and adiabatic paths and matching P-V boundary conditions, one can determine that $Q_1/Q_2 = T_1/T_2$. The efficiency is thus $e=\frac{Q_2-Q_1}{Q_2}=1-Q_1/1_2 = 1-T_1/T_2$.

Alternate Solutions
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psychonautQQ
2013-09-15 16:15:45
entropy is a state variable, therefore it is path independent. Since you end up at the same spot as you started, it must be the same. KhanAcademy thermal videos FTW :P!
livieratos
2011-11-08 07:07:24
is it right to say that since the hot reservoir is losing heat its entropy is decreasing?
 nngal2012-01-25 11:29:56 Yes, I think it's right. In KL the hot reservoir's entropy is decreasing while the system's entropy is increasing. In MN the system's entropy increases in the same amount so in a cycle there's no net change in the entropy of the system.
daverigie
2009-09-21 19:37:32
dS is an exact differential so it is zero when integrated around a closed loop
anmuhich
2009-03-26 09:52:40
The whole point of a Carnot cycle is to show the theoretical maximum efficiency. By definition this means that the entropy of the system stays the same because entropy is what makes you lose efficiency. Also, this is a theoretical system, so no change in entropy is possible.
tera
2006-07-25 04:28:34
In the E is indepentend of the working substance as long as it is ideal gas i think
 Poop Loops2008-11-05 23:41:43 No, it is always independent of the substance. The problem is you can't always get a substance to work in a Carnot cycle. But if you can, then it doesn't matter what you're using.

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