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Calculating temperature of a system

  1. Nov 30, 2009 #1
    1. The problem statement, all variables and given/known data
    The first part of the problem says this figure shows a one-dimensional row of 5 microscopic objects each of mass 6e-26 kg, connected by forces that can be modeled by springs of stiffness 30 N/m (so the effective stiffness of the inner springs is 4ks = 120 N/m; approximate the end springs this way, too). These 5 objects can move only along the x axis.

    http://www.webassign.net/csmech/10-52.jpg
    (this is the picture of the system)

    The first part of the question asks to calculate the entropy of the system at different quanta and I calculated the entropy of the system with 3 quanta to be 4.9089e-23 J/K and with 4 quanta to be 5.8659e-23 J/K, and got that part correct.

    The part I need help with reads Calculate to the nearest degree the average absolute temperature of the system when the total energy is in the range from 3 to 4 quanta. (You can think of this as the temperature when there would be 3.5 quanta of energy in the system, if that were possible.)



    2. Relevant equations

    1/T = dS/dEint (definition of temperature) where S is the entropy and Eint is the internal energy.
    Eint = q*hbar*sqrt(ks/m) where q is the number of quanta, hbar is the reduced planks constant, ks is the spring constant, and m is the mass.

    3. The attempt at a solution

    I figured in order to solve for T in the equation defining temperature, I had to take the difference in Entropy between 4 and 3 quanta which is simply 5.8659e-23 J/K - 4.9089e-23 J/K = 9.57e-24 J/K. Then find dEint and divide. My problem arrises when finding Eint for the system when it has 4 and 3 quanta respectively. In the equation Eint = q*hbar*sqrt(ks/m), q is obvious and hbar is a constant, but what value should I use for ks and m? I think I should simply add all of the masses up to get 5*6e-26=3e-25kg, but I am genuinely confused as to what to use for the spring constant of this system. Should I simply take ks and multiply it by the number of springs? The sentence at the end of the first part that reads "the effective stiffness of the inner springs is 4ks = 120 N/m; approximate the end springs this way, too" makes absolutely no sense to me, and I feel like it is an indication as to how to find the spring constant of the system.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 30, 2009 #2
    Might I suggest an approach using the equipartition theorem?
     
  4. Nov 30, 2009 #3
    I'm not quite sure what that is, our physics book doesn't talk about it haha. But I'm not doubting that is what I am supposed to use. Could you explain how I would go about using that in this situation?
     
  5. Nov 30, 2009 #4
    I'd be happy to. The Equipartition Theorem says that, for certain systems*, the average total internal energy of the system is equal to f*Kb*T/2. Kb is the Boltzmann constant, T is temperature. The tricky part of this is f-it refers to the NUMBER OF DEGREES OF FREEDOM ON WHICH THE ENERGY DEPENDS QUADRATICALLY. So in other words, if part of the energy of a system depends on a parameter squared, then that parameter is one of the degrees of freedom counted in f.

    It's tricky, so I'll give you an example: Imagine a ball hanging from a massless spring. Assume it's one dimensional. This system's energy is as follows: U= 1/2*k*x^2 + 1/2*m*v^2. Note that the energy depends on v quadratically and on x quadratically. That's 2 parameters on which the energy depends quadratically, so the f of this system is 2.

    You're going to have to figure out how to apply that to your system. Hint, each ball will have its own set of degrees of freedom-you'll need to add them ALL together. Also, watch out: there are two springs for each ball, does that mean that each spring contributes 2 degrees of freedom to f?


    ________
    * The systems for which the Equipartition Theorem applies are those in which ALL the terms in the energy depend on a parameter quadratically, or else are constant. This system applies. Also, please note that the equipartition theorem gives the time averaged internal energy, so at any given moment, you wouldn't expect U=f/2*Kb*T.
     
  6. Nov 30, 2009 #5
    Okay, I read the question again, and I realized that my response may not have been too helpful. I mistakenly thought the problem gave you the numerical value of one quanta of energy. You can still use it, but you'd have to figure out q.

    As it so happens, the value of q depends on the fundamental frequency of the oscillator. A quantum mechanical SHO has energy levels E(n) = hbar*w*(1/2 + n), which should give you an idea of what q of what q is in an SHO. Can you see how that relates to your equation for Eint? Why are Ks and m in your Eint equation?
     
  7. Dec 1, 2009 #6
    in E(n) =hbar*w*(1/2+n) isn't w equal to sqrt(Ks/m)?
     
  8. Dec 1, 2009 #7
    and for f in the equipartition theorem, would i be correct in assuming if each spring contributes 2 degrees of freedom to f than the total f for the system would be 4+4+4+4+4=20?
     
  9. Dec 1, 2009 #8
    Yes on the degrees of freedom. Almost in the case of the frequency of a harmonic oscillator. The issue here is that you have two springs working on each ball, so the effective ks isn't 30 N/m. What is it?
     
  10. Dec 1, 2009 #9
    is the effective ks 4*30N/m so 120N/m?
     
  11. Dec 1, 2009 #10
    I figured it out, thank you very much for the help!
     
  12. Dec 2, 2009 #11
    My pleasure!
     
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