# Homework Help: Thermodynamics - heat flow of a rubber band

1. Sep 19, 2010

### mooshasta

1. The problem statement, all variables and given/known data
Consider a rubber band for which the tension, $f$, as a function of temperature $T$ and length $L$ is $f = \kappa T (L+\gamma L^2)$, where $\kappa$ and $\gamma$ are positive constants. Determine the heat flow between it and its surroundings when the rubber band is stretched reversibly and isothermically from length $L_1$ to length $L_2$.

2. Relevant equations
$$dE = TdS + fdL + \mu dN$$

3. The attempt at a solution
I think the question is asking to find $\left( \frac{\partial E}{\partial L} \right)_{T,N}$ and integrate that. I'm not sure how to get this quantity, though, since I don't know what entropy is. Can I use the Helmholtz free energy when calculating heat transfer? I don't think so, since they are not equal...

Thank you for your help!

2. Sep 19, 2010

### Mapes

It's not $\left(\frac{\partial E}{\partial L}\right)_{T,N}$, since some of the energy change dE comes from strain energy, and you wouldn't include that in the heat flow. So you only want to integrate part of dE; which part is it? (Take a look at your equation.)

(Then I expect you'll get into the fun world of Maxwell relations.)

3. Sep 19, 2010

### mooshasta

As I understand it, then, the heat flow should be $\int TdS$, since this is the heat part of the equation. So then should I be integrating:

$$T\int^{L_2}_{L_1} \left( \frac{\partial S}{\partial L}} \right)_{T,N} dL$$

Using a Maxwell relation as you said, I see that $-\left( \frac{\partial S}{\partial L} \right)_{N,T} = \left( \frac{\partial f}{\partial T} \right)_{N,L}$, and then the integral is straightforward.

I am just unsure if that integral above is correct, and if it is, what is the basis on getting that from $\int TdS$ in this context.

Thanks! I appreciate your help.

4. Sep 19, 2010

### Mapes

Looks good to me!

5. Sep 19, 2010

### mooshasta

I still am a little confused how to get to that integral. If S is a function of E, L, and N, doesn't that mean then that:

$$\left( dS \right)_{T,N} = \left( \frac{\partial S}{\partial E} \right)_{L,N} dE + \left( \frac{\partial S}{\partial L} \right)_{E,N} dL$$

But instead I wrote $\left( \frac{\partial S}{\partial L} \right)_{T,N} dL$. How does this match with the above equation?

Thanks again. I'm probably missing something pretty fundamental :)

6. Sep 19, 2010

### Mapes

Let's drop N, since it's constant here. S is a function of two independent variables. We could write

$$dS=\left(\frac{\partial S}{\partial E}\right)_LdE+\left(\frac{\partial S}{\partial L}\right)_EdL$$

or

$$dS=\left(\frac{\partial S}{\partial T}\right)_LdT+\left(\frac{\partial S}{\partial L}\right)_TdL$$

The first isn't of much use; the second simplifies into the integrand above. Does this make sense?

7. Sep 19, 2010

### mooshasta

I think I get it now, thank you. I didn't realize that one of the independent variables could be the conjugate (T).

Thanks!