Homework Help: Thermodynamics, transition and cooling off

1. Jul 17, 2013

fluidistic

1. The problem statement, all variables and given/known data
Hello guys, I'm stuck on a problem I've met in an exam and that I will roughly copy out of memory.
There was a substance that had a transition to liquid helium at $T_0=3$K. Also, $M=\frac{nDB}{T}$ (Curie's law was satisfied). Specific heat at constant magnetization was given: $C_M= nAT^3$.
The goal was to cool the substance below 3K if I remember well. In order to do so, they would magnetize the substance isothermally (at $T_0$) up to $M_0$. Then they would demagnetize the substance adiabatically in order for it to cool down.
Assuming that the heat produced by the substance during the isothermal process is Q,
1)Find $M_0 (n,D,Q,T_0)$.
2)Find T(M) in the adiabatic process.
3)Calculate the final temperature of the substance.
2. Relevant equations
(1):$C_M = \left ( \frac{\partial U}{\partial T} \right ) _{M,n}$
(2):$\left ( \frac{\partial U}{\partial M} \right ) _{T,n}=B-T\left ( \frac{\partial B}{\partial T} \right ) _{M,n}$

3. The attempt at a solution
I know that $B=\frac{TM}{nD}$. So using equation (2), I get that $\left ( \frac{\partial U}{\partial M} \right ) _{T,n} =0$. This means that U is a function of T and n, it doesn't depend on M.
From eq. (1) and $C_M= nAT^3$, I get that $U(T,n)=\frac{nAT^4}{4}+f(n)$. In the problem I'm sure we can assume that the substance has a fixed n. So n is a constant.
I apply the 1st law of thermodynamics for the isothermal process, $\Delta U=-Q+W$ where W is the work done BY the system during that process.
But $\Delta U=0$ for the isothermal process, because U only depends on T (and n, but n is fixed).
Therefore the work done ON the system is equal to the heat released BY the system.
Work done by the system is $-\int BdM=-\frac{B_0 ^2 T_0}{2nD}$
So $Q=\frac{B_0^2T_0}{2nD}$. And $M_0=\frac{nDB_0}{T_0}$. So I reached that $M_0 (T_0, n D, Q )=\left ( \frac{nD}{T_0} \right ) ^{3/2} (2Q)^{1/2}$. That result looks so weird to me, I can't believe it is right.
I haven't figured out the other 2 questions yet.
I'd appreciate some feedback on the first question. Thank you.

2. Jul 17, 2013

TSny

Check what you got for this integral. Your result doesn't look right in terms of the placement of nD and T0. Note that B is intensive and M is extensive. So the result of the integration should be extensive (proportional to n).

Wouldn't it save you some work to do the integral by substituting for B in terms of M in the integrand using the Curie law and then integrating with respect to M? That way, the result of the integral will be expressed in terms of M which is what you want.

3. Jul 18, 2013

fluidistic

Indeed. I don't know what I did there now.
However I still do not reach something proportional to n. I reach that the work done by the system is worth $-\frac{TM ^2}{2nD}$. Does this look good? I can't see any mistake. $B=\frac{TM}{nD}$, I don't see how I could get something proportional to n. This would make $M_0 = \sqrt {\frac{2nDQ}{T_0}}$. (I remember that's the answer I gave in my exam but got 0 credit either because it it wrong or because I had not justified that U didn't depend on M. I said "I'll assume that U doesn't depend on M...". I didn't know back then how to show it and I assumed it because otherwise I wouldn't know how to continue further).
I really appreciate your help so far TSny.

4. Jul 18, 2013

TSny

That looks right to me. Think about how Q depends on n.

5. Jul 18, 2013

fluidistic

I see, thank you. Directly proportional to n, so that $M_0$ is extensive.

For part 2), some thoughts: For the adiabatic process, $\Delta U=W$ where W is the work done ON the system.
Work done on the system is $W=\int B dM= \int _{M_0}^0 \frac{TM}{nD}dM$ but T depends on M and I don't have that function. This is what they ask me.
On the other hand I know that $\Delta U = U(T_f )-U(T_0)=\frac{nA}{4} (T _f ^4-T_0 ^4)$. Where that $T_f$ is the one they ask me in the 3) question.
This is where I am now.

6. Jul 18, 2013

TSny

dU = dW = BdM

Write dU in terms of T and dT. Using Curie Law, write BdM in terms of T, M, and dM. Separate variables and integrate.

7. Jul 18, 2013

fluidistic

Oh using dU was bright... Thanks a lot.
I reached that $n^2\frac{DA}{3}(T^3-T_0^3)=\frac{M^2-M_0^2}{2}$ which is what they asked for I believe. For the sake of it, $M(T)=\sqrt {\frac{2}{3}n^2DA (T^3-T_0^3)+M_0^2}$.
I will now think on part 3).

8. Jul 18, 2013

fluidistic

Part 3):
Since they demagnetized, $M(T_{\text{final}})=0$. Using the expression of M(T) that I got for part 2), this lead me to $T_{\text{final}}=\sqrt[3]{T_0^3-\frac{3M_0^2}{2n^2DA}}$. Which seems strange to me because I could in principle set $M_0$ arbitrary big and so $T_{\text{final}}$ could be arbitrary small or not even exist.
Hmm I guess I made some mistake(s)?

9. Jul 18, 2013

TSny

Your answer looks correct to me. Magnetization is limited by its "saturation" value as B becomes strong. Curie's law is not valid for high B or low T. More specifically, you need μB << kT where μ is the magnetic moment per molecule. See: http://en.wikipedia.org/wiki/Curie's_law.

10. Jul 18, 2013

fluidistic

Well, thank you very much. I wish I had knew all this before taking the final exam; what else can I say. You've been extremely helpful to me.

11. Jul 18, 2013

TSny

"extensive" isn't the right word here. I should have just said "proportional to n".