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Total Work out of Three Bodies (Thermodynamics)

by Mathoholic!
Tags: bodies, thermodynamics, work
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Dec17-12, 05:51 AM
P: 49
The problem is:

Having three bodies with thermal capacities (C) as sources of heat to a heat machine, what is the maximum work I can extract from this system, given that the bodies are at temperatures T3, T2 and T1 (T3>T2>T1), leaving them at an equal final temperature?

I tackled this problem by assuming that the amount of heat I can extract from each body is:

Qi=C(Ti-Tf) i=1,2,3

And so:


Now, I'm not sure if I'm breaking the second law of thermodynamics by totally converting heat into work. This is where I'm kind of stuck.

So, if anyone can help me figure out how to solve this and other type-like problems, it'd be great.

Thank you :)
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Dec17-12, 06:08 AM
P: 5,462
Now, I'm not sure if I'm breaking the second law of thermodynamics by totally converting heat into work. This is where I'm kind of stuck.
The second law forbids the total conversion of heat into work in a cyclic process.

It does not forbid it for some part of a process that is not cyclic.

So post details of your proposed system and process for further comment.
Dec17-12, 06:47 AM
P: 49
1Are you saying that I can only fully convert heat into work in a irreversible process?

I'm not sure I can give much more detail about the system. I've already given all there's to know about the problem.

I'm having trouble getting the definitions of reversible process and irreversible process clear in my head. It's all a bit fuzzy and I think that's what is keeping me from solving this problem.

Dec17-12, 07:11 AM
P: 5,462
Total Work out of Three Bodies (Thermodynamics)

Cyclic is not the same as reversible.

You get back to the same conditions (set of thermodynamic values) in each case, but the paths are different.

There are no real reversible processs which must be under equilibrium at all stages.

Cyclic processes may include non equilibrium processes. Irreversible processes are non equilibrium.

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