Entropy question

1. Nov 21, 2006

leright

ok, so a infinitesimal change in entropy is dS = dQ/T. Differentials are used so that temperature change is neglibible over the change in Q.

Now, I have a homework problem that wants me to express the change in entropy of a rod as a function of length changes and temperature changes. Now, isn't the change in entropy with respect to change in length zero, since dS = dQ/T??? The internal energy changes wrt work done on the rod, but it seems the entropy does not? Can someone explain why it does change with length changes? Also, if entropy does change wrt length changes how would I go about determining the entropy changes along an integration path where T is constant??

the force on the rod is F = -aT^2(L-Lo), where Lo is the initial length.

Thanks a lot.

2. Nov 21, 2006

leright

3. Nov 22, 2006

Cyrus

Maybe you should write the exact problem.

Anywho, I took engineering thermo, so I cant help. We only did systems invloving fluids.

Bed time,

Ciao.

4. Nov 22, 2006

leright

It mentions that the rod is well insulated, which would obviously mean that the system is isenthalpic (constant heat). This means that dU = dW.

If the rod were constant internal energy then any work done would result in a decrease in the heat content, which would correspond to a temperature change, but the conditions of the problem do not imply constna internal energy.

I lose.

5. Nov 22, 2006

leright

Maybe I should just kill myself.

6. Nov 22, 2006

Cyrus

Sounds like a good idea.

7. Nov 22, 2006

leright

haha

no really, I need help.

I figured deltaS = 0.5b((Tf)^2 - (To)^2) = 0, where b is a constant and Tf and Ti are the final and initial temps, respectively. Entropy doesn't seem to have any dependence on length. wtf?

And I thought you were going to bed.

Last edited: Nov 22, 2006
8. Nov 22, 2006

Cyrus

Im working on my heat transfer project, leave me alone! Dead people cant post.

I dont know if this helps, but a change in length of the rod is a result of a change in energy of the rod. It can be thermal energy (expansion) or stress energy (strain).

9. Nov 22, 2006

leright

Yes, I realize this. If the internal energy of the rod was constant then work done on the rod would need to have a corresponding change in heat, and therefore a change in entropy. However, this is NOT a constant internal energy condition...it is a constant enthalpy condition, which simply means that the change in internal energy is equal to the work done on the rod. This, however, does not help me.

hmmm....

10. Nov 22, 2006

leright

ok, let me ask you this. When you stretch a rod, does it change temperature?? If so, why?

This is a constant heat process, btw...but NOT constant U (internal energy)

11. Nov 22, 2006

vanesch

Staff Emeritus

Yes. So if you know the answer, what's the problem ?

If you change the length in a reversible way by applying an external force which will hence only do work, and you don't have any heat transfer, then the process is adiabatic and isentropic.

Of course, if the external force gives rise to an irreversible process (like turning a paddle in a liquid), then there will be an entropy increase (although the process is, strictly speaking, adiabatic but irreversible).

12. Nov 22, 2006

vanesch

Staff Emeritus
Indeed, there will be a change in internal energy equal to the work done by the external force: dU = dQ + dW (if dW is the work done on the system, there can be different sign conventions).
Given that dQ = 0 and dW = F.ds, we have that dU = F.ds

But the internal energy can change while keeping S constant.
As you pointed out correctly, dS = dQ/T (in a reversible process).
Given that dQ = 0, we have that dS = 0, but given that dW is not 0, we have that dU is not 0. There is no contradiction here. S and U are different state functions, and a change in state can result in a change in U and not a change in S.