# Thermal Expansion of a copper sphere

1. May 24, 2010

### ElBell

"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!

2. May 25, 2010

### ehild

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

ehild

Last edited: May 25, 2010
3. May 25, 2010

### ElBell

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!

4. May 25, 2010

### ehild

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

Last edited: May 25, 2010
5. May 25, 2010

### ElBell

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?

6. May 25, 2010

### ehild

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

7. May 26, 2010

### ElBell

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!!

8. May 26, 2010