Separate Brass & Aluminum Rods: Thermal Expansion

[thenewbosco]

A brass ring of diameter 10.00cm at 20C is heated and slipped over an aluminum rod diameter 10.01cm at 20C. Assuming the average coefficients of linear expansion are constant, to what temperature must the combination be cooled to separate them?

what i was thinking of doing here is setting up two equations for the lengths as functions of temperature and equating them. i don't know what the equations are though that i am supposed to use here. thanks for the help.

 

[code.master]

I think you want to set up an equality with the difference of expansion lengths of radius of different materials.

try putting 0.01cm on one side, and that has to equal the difference (radius of aluminum) - (radius of brass ring). this is found by using the thermal expansion coefficient and the measurments at 20 C. 2.4 x 10^-5 per degree C for alum, 2.0 x 10^-5 per C for brass. this means that (10.00*2.0*10^-5*(20-T)) is the expansion or contraction for the brass ring. you want this value minus the value of the alum to be equal to 0.01.

set this equation up, solve for T, make sure you got the organization of the minus signs and differences correct. if your T is greater than 20, its not making sense. you would have the solution for when the rod would be .01 bigger than the ring. if you go over this Temperature you solve for, the rod always expands more. so make sure you get that right.

Im not sure if that's right, but it might help you.

 

[quicknote]

I would approach this problem by first determining the coefficients of linear expansion for both brass and aluminum. These values can be found in reference materials or calculated using the formula α = (ΔL/LΔT), where α is the coefficient of linear expansion, ΔL is the change in length, L is the original length, and ΔT is the change in temperature.

Once we have these values, we can set up the following equations:
L(brass) = L(aluminum) at 20°C
L(brass) + ΔL(brass) = L(aluminum) + ΔL(aluminum) at final temperature

Substituting the values for the coefficients of linear expansion and the initial lengths, we can solve for the final temperature at which the lengths will be equal:
L(brass) + α(brass)L(brass)(Tf-20°C) = L(aluminum) + α(aluminum)L(aluminum)(Tf-20°C)

Solving for Tf, we get:
Tf = (L(brass) - L(aluminum)) / (α(brass)L(brass) - α(aluminum)L(aluminum)) + 20°C

So in this specific scenario, the combination must be cooled to a temperature of (L(brass) - L(aluminum)) / (α(brass)L(brass) - α(aluminum)L(aluminum)) + 20°C to separate the brass ring from the aluminum rod.