# 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".