Question

In: Physics

A brass ring of diameter 10.00 cm at 16.4°C is heated and slipped over an aluminum...

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

Solutions

Expert Solution

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)


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