In: Physics
A brass ring of diameter 10.00 cm at 16.4°C is heated and slipped over an aluminum rod of diameter 10.01 cm at 16.4°C. Assume the average coefficients of linear expansion are constant.
(a) To what temperature must the combination be cooled to separate the two metals?
? °C
(b) What if the aluminum rod were 11.76 cm in diameter?
? °C
For aluminum, coefficient of thermal linear expansion = 23.1 x 10^-6 / C
and for brass, coefficient of thermal linear expansion = 19.0 x 10^-6 / C
Initial dia of Al, D(Al) = 10.01 cm
Initial dia of brass, D(brass) = 10.0 cm
Suppose is the change in temperature.
So, to separate the two metals, final length of the two metals will be the same.
Means,
10.01 + 23.1 x 10^-6 x 10.01 x = 10.0 + 19.0 x 10^-6 x 10.0 x
=> (23.1 x 10^-6 x 10.0 - 19.0 x 10^-6 x 10.0) x = - 0.01
=> = -0.01 / (231 - 190.0) x 10^-6 = -10^4 / 41.0 = -245.04 deg C
Hence final temperature of the combination = 16.4 - 245.04 = -228.64 deg. C (Answer)
(b) In this case, dia of aluminium rod = 11.76 cm
So, solve like above -
11.76 + 23.1 x 10^-6 x 11.76 x = 10.0 + 19.0 x 10^-6 x 10.0 x
=> (23.1 x 10^-6 x 11.76 - 19.0 x 10^-6 x 10.0) x = - 1.76
=> = -1.76 / (271.7 - 190.0) x 10^-6 = -(1.76 x 10^4) / 81.7 = -215.4 deg C
Hence final temperature of the combination = 16.4 - 215.4 = -199.0 deg. C (Answer)