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Thermal Expansion of a steel plate

  • Thread starter Brit412
  • Start date
24
0
1. Homework Statement
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. Homework 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
 
11
0
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.
 
24
0
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)
 
11
0
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
(∆T(12E-6)(.01))-.0003=∆T(3.3E-6)(.01003)
and factor out the T, and it will leave you with ∆T=345.2
 
24
0
Thank you so much for helping me understand that problem
 

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