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Thermal Expansion of a steel plate

  1. Jan 27, 2009 #1
    1. The problem statement, all variables and given/known data
    A steel plate has a circular hole with a diameter of 1.000 cm. In order to drop a Pyrex glass marble 1.003 cm in diameter through the hole in the plate, how much must the temperature of the system be raised? (Assume the plate and the marble always have the same temperature)

    2. Relevant equations
    ∆L= alpha L(0) ∆T

    3. The attempt at a solution
    Use the above equation for both the marble and plate, but I don't know what temperature to use for ∆T
  2. jcsd
  3. Jan 27, 2009 #2
    it doesnt matter what temperature it starts at, DELTA T is just the change in temperature, so if it went from 0 to 10 or 900 to 910 DELTA T would still be 10.
  4. Jan 27, 2009 #3
    I'm still confused. The book says the answer is 350 but I don't see how they arrived at that answer. Here's what I did so far:

    ∆L (plate) = (12E-6) times (0.01m)
    ∆L (marble) = (3.3E-6) times (0.01003m)
  5. Jan 27, 2009 #4
    so, the difference in length between the P and the M is .00003m.
    so what youre looking for will be the temperature at which the ∆Lp is different by ∆m by .00003m. (since when the plate is .00003 meters bigger, the marble will fit through.)

    so ∆Lp-.00003m = ∆m
    (it is minus the difference because you really want the plate to be .00003m bigger, so if you substract it, it would make the terms equal
    then you can plug in for the rest
    and factor out the T, and it will leave you with ∆T=345.2
  6. Jan 27, 2009 #5
    Thank you so much for helping me understand that problem
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