posted by Sandhya on .
A thin brass ring of inner diameter 10cm at 20degrees C is warmed and slipped over an aluminum rod of diameter 10.01cm at 20degrees C. Assuming the average coefficient of linear expansions are constant, to what temperature must this combination be cooled to separate the parts?
Look up the thermal expansion coefficients of alumimum and brass. They should be in units of degC^-1. It is about 18.7*10^-6 for brass and 22.2 for aluminum. You will find that values depend on temperature ramge and alloy. The aluminum must shrink 0.01 cm more than the brass ring to slip it off. Let dT be the change in temperature from 20 C (positive for an increase).
10.01*22.2*10^-6*dT - 10.00*18.7*10^-6*dT = -0.01
dT(22.2*10^-5 - 18.7*10^-5) = -0.01
dT = -0.01/3.5^10^-5 = -285 C
That would put the final temperature at -265C, whch is near absolute zero.