Recent content by SevenTacos

The two terms represent the heat put into the system Qh, and the heat deposited into the cold reservoir Qc respectively, after all, W = Qh - Qc. And yes, I understand the algebra to clean up that equation. Just have to say Thank you Chet, this problem was much easier than I made it out to be; I...

Yes, the integral comes out to W = C ( Th ln (Tc) - Tc ) which yields a work of C (Th ln (Th) - Th) - C (Th ln(Tco) -Tco ) So W = C (Th ln (Th) - Th) - C (Th ln(Tco) -Tco ) as a final answer?

dQc = (C)(dTc), so dw = (Tho/Tc -1) (C)( dTc) Did some more thought on this: If dQh/dQc = Th / Tc, then (Ch)(dTh)/(Cc)(dTc) = Th / Tc. If this is correct then you can cross multiply and take the integral, looking like Cc/Ch ∫ dTc/ Tc = ∫ dTh / Th. The bounds for the first integral should be...

You would get dw = dQh- dQh(Tc / Th). But I still need to define Tc as a function of Th correct? It's 1 am here, will return in the morning!

I appreciate the reply, but I'm still pretty confused. Using the statement dQh/dQc = Th / Tc, it's clear dQc = (Tc / Th) dQh. I can't really think of anywhere else to go though. I know I can define Qh and Qc in terms of nRTln(v1/v2) but those are the wrong variables here.

Homework Statement You have an infinite heat reservoir with temperature Th. But you’ve only got a finite cool reservoir, with initial temperature Tc0 and heat capacity C. Find an expression for the maximum work you can extract if you operate an engine between these two reservoirs. Homework...