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Thermodynamics, transition and cooling off

  1. Jul 17, 2013 #1

    fluidistic

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    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. jcsd
  3. Jul 17, 2013 #2

    TSny

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    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.
     
  4. Jul 18, 2013 #3

    fluidistic

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    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.
     
  5. Jul 18, 2013 #4

    TSny

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    That looks right to me. Think about how Q depends on n.
     
  6. Jul 18, 2013 #5

    fluidistic

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    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.
     
  7. Jul 18, 2013 #6

    TSny

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    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.
     
  8. Jul 18, 2013 #7

    fluidistic

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    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).
     
  9. Jul 18, 2013 #8

    fluidistic

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    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)?
     
  10. Jul 18, 2013 #9

    TSny

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    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.
     
  11. Jul 18, 2013 #10

    fluidistic

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    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.
     
  12. Jul 18, 2013 #11

    TSny

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