GR8677 #95



Alternate Solutions 
There are no Alternate Solutions for this problem. Be the first to post one! 

Comments 
psychonautQQ 20130915 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 20111108 07:07:24  is it right to say that since the hot reservoir is losing heat its entropy is decreasing?
nngal 20120125 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 20090921 19:37:32  dS is an exact differential so it is zero when integrated around a closed loop   anmuhich 20090326 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 20060725 04:28:34  In the E is indepentend of the working substance as long as it is ideal gas i think
Poop Loops 20081105 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.

 

Post A Comment! 

Bare Basic LaTeX Rosetta Stone

LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces .

type this... 
to get... 
$\int_0^\infty$ 

$\partial$ 

$\Rightarrow$ 

$\ddot{x},\dot{x}$ 

$\sqrt{z}$ 

$\langle my \rangle$ 

$\left( abacadabra \right)_{me}$ 

$\vec{E}$ 

$\frac{a}{b}$ 





The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it... 

