# Thermal Expansion of a copper sphere

"A copper sphere of radius 2.000 cm is placed over a hole of radius 1.990 cm in an aluminum plate at 20 deg. C. At what common temperature will the sphere pass through the hole?"

((∆T(23* 10-6)(0.01990)- .00010 = ∆T(17* 10-6)(0.02)

((∆T(23* 10-6)(0.01990) - .00010 = ∆T(17* 10-6)(0.02)

(∆T x 4.577x10^-7) – 0.00010 = (∆T x 3.4x10^-7)

- 0.00010 = (∆T x 3.4x10^-7) – (∆T x 4.577x10^-7)

-0.00010 = ∆T x -1.177 x 10^-7

∆T = 849.62

Do my equations and calcs look correct? I believed the answer was in the high 700's range....I also dont know how to bring the 20 degrees into it?

HELP!

ehild
Homework Helper
Explain what you did please. Were the thermal expansion coefficients given for 20 °C or for for 0 °C?

ehild

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Sure!

First, I found out that the coefficients of linear expansion were 23* 10^-6 for aluminium and 17* 10^-6 for copper.

Then I formulated a linear equation using these coefficients, to try and find the missing temperature.

The left side of the equation, I also added the difference between the two sides- which is .00010.

I think went about solving the equation, but I am not confident in it being right!

ehild
Homework Helper
Your equations were confusing with that superfluous number of parentheses. The result for ∆T is correct, but add 20°C to get the common temperature, and round it to the significant digits and write out the unit.

ehild

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No I havent used the 20 degrees anywhere..thats the problem I dont know what to do with it.

So how do I work out the coefficients of 20 degrees? I have a table that gives 'typical average values' of each substance, with the coefficient given under the heading a(degrees celcius^-1).

What does that mean?

ehild
Homework Helper
No I havent used the 20 degrees anywhere..thats the problem I dont know what to do with it.

So how do I work out the coefficients of 20 degrees? I have a table that gives 'typical average values' of each substance, with the coefficient given under the heading a(degrees celcius^-1).

What does that mean?

What was the formula you learnt for the linear thermal expansion of solids? Was it the same as this one:

$$L(final)-L(initial)=a(T(final) -T(initial))$$

T(final) -T(initial) is abbreviated as ΔT.

You calculated ΔT. Add the initial temperature to it, and you get the final temperature.

ehild

Thanks very much for your help. I got the idea for the equation from another post as one wasnt given on my assignment.

I am guessing I add 849 degrees to 20 degrees to get my final result?

Thanks again, much appreciated!!

ehild
Homework Helper