- #1
Sabian
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Homework Statement
It's basically the classic problem I've seen here a lot.
There are two bodies, both with equal heat capacity ([itex]C_P[/itex]), one at temperature [itex]T_1[/itex] and the other at [itex]T_2[/itex]. They exchange heat using an infinitesimal-reversible Carnot Engine, which will work until thermal equilibrium is reached.
I must find, first, the final temperature of the bodies ([itex]T_F[/itex]) and then the entropy variation of the entire system.
Homework Equations
[itex] dU = \partial Q + \partial W [/itex] (I know Q and W are non-exact differentials but I can't find the [itex]\LaTeX[/itex] code for it.
[itex] dS = \frac {\partial Q}{T} [/itex]
[itex] \eta = 1 - \frac {T_2}{T_1} [/itex] For a Carnot Engine.
The Attempt at a Solution
I've read at least 10 threads in the site about this and still not convinced. Some of the threads affirm the entropy variation of the universe is 0, and that way is pretty easy, but in my exercise I should get the results in the opposed order (moreover, I don't see trivially that it's 0).
Then the other way, which I liked more work it from the first law of thermodynamics and the Carnot efficiency (subtly implying the 2nd Law). But they are made in a total quantities way and, although the solution is fine, I think it's more appropriate to make it in differential form as the quantities of heat vary after each cycle.
Applying the 1st law to the engine should result in
[itex] \partial Q_H + \partial Q_C + W = 0 [/itex]
And the heat quantities are opposite from the one I get by evaluating the bodies (if Q is flowing out of the hotter body, Q is negative, but enters to the engine as a positive amount of heat).
[itex] - C_P dT_c - C_P dT_h = - \partial W [/itex]
But here the work would be the one done ON the engine which sounds weird, so if I consider the work done by the engine
[itex] C_P dT_c + C_P dT_h = \partial W [/itex] But I get some problems while writing [itex] \partial W [/itex] as for a Carnot Engine.
The answer is [itex]T_F = \sqrt {T_1 T_2} [/itex]
Thanks for your time.
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