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Basic Thermal Conductivity Question

  1. Sep 18, 2007 #1


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    edit: not sure why this was moved to homework - it isn't.

    Hi :) I'm a bit embarrassed to ask this, but I hope someone can set me straight. As a second year university chem student I feel I should know this.

    In this question about thermal conductivity, I realize that it is a subject to itself, however I just want to stick to something very basic I'm trying to remember from my chemistry class. I forgot something but am not exactly sure what it is.

    We mostly worked with q = s(t1-t2)

    I believe we somehow used that to solve problems that I no longer remember how to do. They were when two masses of different mass, temperature, and elements were placed in thermal contact, what will the final temperature of each be? Of course they would both be the same temperature, but I don't remember how to get that with the specific heat formula.

    I tried to figure it out on my own, and made up a sample problem, I think I failed though.

    50g of Al at 80C is placed in thermal contact with 20gFe at 60C. Al has a higher specific heat (.901 J/g vs .449 for Fe), and more mass, so it has a lot more energy per degree C. It will be linearly related, so once I figure out how much more energy per degree I can set up an equation.

    For abbreviation, all Al components are referred to as 1, and Fe as 2. s = specific heat. k = unknown constant

    (m1 *s1)*delta t = k(m2*s2)*delta t -->
    (m1*s1)/(m2*s2) = k = 5.01679 when the numbers are plugged. k is the factor of the energy difference per degree. Al is about 5 times higher, so for every degree that Al loses to Fe, it will go ~5x farther to heat it.

    Then I made the equation for final temperature:
    80C - 1C x = 60C + 5.01679x* 1C -->
    6.01679x = 20 C x = 3.32403
    Plugging the value of x backs in gets 76.67586 C as the final temperature for both.

    The number looks OK to me, however, when I plug that in to find the total energy (q = s(delta t) for both, then adding), I have about 600 more J of energy than I started with. I am pretty sure my math is fine, so could someone tell me my mistake?
    Last edited: Sep 18, 2007
  2. jcsd
  3. Sep 18, 2007 #2
    Quick question, as a second year chem student have you done either differential equations or linear algebra?
  4. Sep 18, 2007 #3


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    Differential equations yes. Linear algebra... I think there is a class dedicated to that? If that's what you meant, no, but I have done the basics they put into college algebra and trig level courses.
  5. Sep 18, 2007 #4
    Alright, well I never deal with chemistry, so I'm going to tackle it from the perspective of differential equations. It actually starts off with a PDE, but yours is the really easy kind of one spatial dimension, good thing you didn't want it for two pyramids.

    [tex]\frac{\partial T}{\partial t} = k \frac{\partial^2 T}{\partial x^2}[/tex]

    Because you only want what happens way in the future, when temperature stops changing with respect to time, also known as the equilibrium solution, you can just take out the time dependency. So...

    [tex]0 = k \frac{\partial^2 T}{\partial x^2}[/tex]

    Ultimately you get the diff eq

    [tex]\frac{d^2 F}{dx^2} = 0[/tex]

    F has the solution of
    [tex]F = C_1 x + C_2[/tex]

    Lets make up some boundary values. We know that the rod comes in contact with the 80 degree aluminum, which I think we can assume might as well be a bath because Al has much better thermal properties (this is where your chemistry should pick up).
    F (x=L,t) = 80
    dT/dx (x,t) = 0

    So, with boundary conditions
    C1*L + C2 = 80
    C1 = 0

    The temperature will be

    [tex]T(x,t)_{lim_{t -> \infty}} = 80[/tex]

    This should probably be a lot more rigorous, but should give a general idea.
    Last edited: Sep 18, 2007
  6. Sep 18, 2007 #5


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    A different method than I had looked for, but a much better one. Thanks for making it easy :)
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