# Homework Help: Heat Engine Problem

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1. Mar 6, 2015

### SevenTacos

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. The attempt at a solution
I've been stumped at this for hours. I know I simply need to integrate work as dW= dQc + dQh. So i need to define Qc and Qh. Qc = (Cc)(dTc) and Qh= (Ch)(dTh) correct? I am completely at a loss on defining dTc and dTh however. I'm essentially confused because Tc depends on how much energy (q) the engine has deposited into the cold resevoir. When I try to create an equation, I come up with: Tc= Tc0 + (Qc)(C) But to define q, you need Tc. How do you define these variables when they rely on each other!?! Please Help!

2. Mar 6, 2015

### Staff: Mentor

Tc is going to be changing, while Th will be staying the same. So as Tc rises, a smaller fraction of the heat transferred from Th is able to be converted to work. If Tc and Th are the instantaneous values of the temperatures, and dq is transferred to Tc, what is the amount of heat dQ transferred from Th (in terms of Tc, Th, and dq)? How much work dw is done? What is the change in Tc, dTc? What is dw/dTc?

Chet

3. Mar 6, 2015

### SevenTacos

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.

4. Mar 6, 2015

### Staff: Mentor

This is good so far. dw = dQ-dq, so what do you get for that?

Chet

5. Mar 6, 2015

### SevenTacos

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!

6. Mar 6, 2015

### Staff: Mentor

If I take your results, and express dw in terms of dQc, I get:
$$dw=\left(\frac{T_{h0}}{T_c}-1\right)dQ_c$$
Now how is dQc related to the heat capacity C and dTc?
If you substitute this into the equation for dw, what do you get?

Chet

7. Mar 6, 2015

### SevenTacos

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 Tc to Th, because the engine will stop when Tc = Th. The bounds for the second I guess would be Tho to Th? After integration you get (Cc/Ch) ln (Th/ Tc) = ln (Th/Tho), rearranging you get this final factor for Tc:

Tc = Th / (( Th/ Tho) ^ (Ch/Cc))

I feel like this should be right, but it also feels horribly wrong

8. Mar 6, 2015

### Staff: Mentor

This last equation is correct, and it's all you need to finish solving this problem.

Yes. It's horribly wrong. Here's what's wrong:

The problem statement says that you have an infinite heat reservoir at temperature Th. That means that Ch is infinite, and that Th never changes from its initial value Th0. So the rest of the analysis is incorrect.

Getting back to your correct equation for dw, you have:

$$dw=C\left(\frac{T_{h0}}{T_C}-1\right)dT_C$$

This can be integrated immediately from $T_C=T_{C0}$ to $T_C=T_{h0}$ to give you w. Do you know how to integrate this?

Chet

9. Mar 7, 2015

### SevenTacos

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?

10. Mar 7, 2015

### Staff: Mentor

Yes. This is an acceptable final answer to your homework problem. However, consider what happens if we manipulate your final answer mathematically into the following equivalent form:
$$W=CT_h\ln \left(\frac{T_h}{T_{co}}\right)-C(T_h-T_{co})$$
Can you show that this equation is mathematically equivalent to your answer? Can you identify what the second term on the right hand side of this equation represents physically? With that realization, what does the first term on the right hand side represent physically?

Chet

11. Mar 7, 2015

### 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 severely overthought this. Thanks for taking the time to put me through the process I needed to go through

12. Mar 7, 2015

### Staff: Mentor

My pleasure. It's great to work with someone as focused and determined as you are.

Chet

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