Thermal Expansion of a steel plate

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Homework Help Overview

The discussion revolves around the thermal expansion of a steel plate with a circular hole and a Pyrex glass marble. The problem involves determining the temperature increase required for the marble to fit through the hole in the plate, given their respective dimensions and the thermal expansion properties of the materials involved.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the thermal expansion formula and the significance of the change in temperature (∆T). There are attempts to calculate the necessary temperature increase based on the difference in expansion between the plate and the marble.

Discussion Status

Some participants have provided calculations and reasoning regarding the temperature change needed for the marble to fit through the hole. There is acknowledgment of confusion regarding the book's answer and the methods used to arrive at it. The discussion is ongoing with various interpretations being explored.

Contextual Notes

Participants note that the initial temperature does not affect the change in temperature (∆T), and there is some uncertainty about the calculations leading to the book's answer of 350. The specific values used for thermal expansion coefficients and dimensions are also under discussion.

Brit412
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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)


Homework Equations


∆L= alpha L(0) ∆T


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
 
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it doesn't 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.
 
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)
 
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
 
Thank you so much for helping me understand that problem
 

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