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Thermal Expansion of two Metals

  1. Oct 11, 2009 #1
    1. The problem statement, all variables and given/known data

    The inner diameter of a steel ring is 2.0000 cm, and the diameter of an aluminum
    disk is 2.0100 cm. Both are at 430 degrees C. At what common temperature will the disk t
    precisely into the hole in the steel ring?

    alpha(Steel) = 1.3e-5
    alpha(Al) = 1.9e-5

    2. Relevant equations
    deltaL/Lo = alpha*deltaT


    3. The attempt at a solution

    What I told myself is this: If these the steel is to fall through the aluminum ring, then the length of its diameter at the temperature of interest must equal the length of the ring's diameter. Therefore, we must solve for the length of each diameter at temperature T and set them equal to each other. Therefore:

    deltaL(steel) = alpha(steel)*Lo(steel)*deltaT
    Lfinal Steel = Lo(steel) + alpha(steel)*Lo(steel)*(T-430) <----Eq.1

    deltaL(Al) = alpha(Al) * Lo(Al) * deltaT
    Lfinal Al = Lo(Al) +alpha(steel)*Lo(Al)*(T-430)<------Eq.2

    Setting Eq.1 and Eq.2 equal to each other, I solved for T.

    The T I got was below absolute Zero.... Now I intuitvtely know that the T must be less than 430 since the coefficient of expansion of Al is higher than Steel, and so it contracts faster. However, I do not see how I could get a value less than 0K...

    Can someone please tell me where I am going wrong? Thank you very much
     
  2. jcsd
  3. Oct 11, 2009 #2

    Delphi51

    User Avatar
    Homework Helper

    I got delta T = 820 degrees, same conclusion.
    Wikipedia has alpha values more divergent - 2.3 for Al and 1.1 to 1.3 for steel (x 10^-5).
     
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