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A Problem on Thermal Equilibrium

  1. Dec 10, 2005 #1
    The problem is

    An aluminum rod with cross-sectional area 0.0400 cm^2 and length 80.00 cm at a temperature of 140.0 Celcius is laid alongside a copper rod of cross-sectional area 0.0200 cm^2 and length 79.92 cm at temperature T. The two rods are laid alongside each other so that they are in thermal contact. No heat is lost to the surroundings, and after they have come to thermal equilibrium, they are observed to be the same length. Calculate the original temperature T of the copper rod and the final temperature of the rods after they come to equilibrium.

    I tried using the MCAT equation (Q = mc X Delta T) but I can't seem to derive a formula that works for the lengths of TWO objects in thermal equilbrium. In my textbook there is a formula that says that

    The Change in the Length = Alpha X Initial Length X The Change in Temperature

    I'm not sure if this helps, but I have worked hard on this problem, and asked several people for advice, ultimately gaining no progress. My book has the answers, but I need to know how to solve this one, not just the answer. But if it helps, the answer is:

    89.7 Degrees Celcius for the Original Temperature of Copper.
    119.2 Degrees Celcius for the final temperature of the two rods after they come to equilbrium.

    Thanks to all that help me.
  2. jcsd
  3. Dec 10, 2005 #2


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    Eq. 1) Q = m cp [itex]\Delta\,T[/itex]


    Eq. 2) [itex]\Delta l[/itex] = [itex]\alpha\,\Delta\,T[/itex], where [itex]\alpha[/itex] is the thermal expansion coefficient of the material.

    Let Th = the initial 140°C.

    One has to find Teq (i.e. equilibrium), and Tc (i.e. cold). And one has two equations.

    The heat lost from Al has to equal heat absorbed by copper.

    [itex]\Delta T[/itex] (Al) = Th - Teq
    [itex]\Delta T[/itex] (Cu) = Tc - Teq

    If one needs to find mass, it is just density * volume.

    One also knows that the length at Teq is the same for both rods, so that gives a second equation with Th, Teq, and Tc by virtue of Eq. 2.
  4. Dec 10, 2005 #3
    But how does one find [tex]T_c[/tex]?
    Also, I regret mentioning that I was given the following values:
    [tex] P_a [/tex] = 2.7 X 10^3 Kg / m^3
    [tex] C_a [/tex] = 910 J/Kg-k
    [tex] P_c [/tex] = 8.4 X 10 ^3 Kg/ m^3
    [tex] C_c [/tex] = 390 J/ Kg-k
    Coefficient of Linear Expansion for Aluminum:
    2.4 X 10^(-5)
    Coefficient of Linear Expansion for Copper:
    1.7 X 10^(-5)
    Last edited: Dec 10, 2005
  5. Dec 10, 2005 #4


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    If one writes the thermal equilibrium equation in terms of Tc and Teq, and one writes the equation for the equivalent length in terms of the same temperatures, then one has two equations and two unknowns, which one wishes to find.

    One can use direct substitution, to find either temperature.
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