# Need help on Thermal Expansion problem

1. Oct 18, 2004

### andrew410

A brass ring of diameter 10.00 cm at 20 degrees Celsius is heated and slipped over an aluminum rod of diameter 10.01 cm at 20 degrees Celsius. Assuming the average coefficients of linear expansion are constant, a) to what temperature must this combination be cooled to seperate them? b) What if the aluminum rod were 10.02 cm in diameter?

I don't know where to start for this problem. Any help will be much appreciated. Thx in advance!

2. Oct 18, 2004

### Tide

Start by looking up the thermal expansion coefficients for aluminum and brass!

3. Oct 18, 2004

### andrew410

Ok. So far I know that the average coefficient of linear expansion of aluminum is 24*10^-6 and brass is 19*10^-6. I have this equation:
$$\Delta L=\alpha L_{i}\Delta T$$
where $$L_{i}$$ is the initial length. I don't understand how to apply this equation to the two materials.

Last edited: Oct 18, 2004
4. Oct 18, 2004

### Tide

Aside from the problem being somewhat badly stated they want you to determine the temperature required to make the diameter of the ring the same as the diameter of the rod given that the ring starts out at 10.0 cm and the rod at 10.1 cm.

Cooling the two materials will cause them to contract and since the thermal expansion coefficient for aluminum is greater than that of brass then the aluminum rod will shrink more for a given reduction of temperature.

If a material starts off with length (diameter) $L_0$ then its new length, after a temperature change, will be $L = L_0(1+\alpha \Delta T)$.

5. Oct 18, 2004

### andrew410

Thanks a lot for your help.