How Cold to Separate a Heated Brass Ring from an Aluminum Rod?

[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 separate 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!

 

[Tide]

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

 

[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.

 

[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).

 

[andrew410]

Thanks a lot for your help.