Question

A brass ring of diameter 10.00 cm at 20.7°C is heated and slipped over an aluminum rod of diameter 10.01 cm at 20.7°C. Assume the average coefficients of linear expansion are constant. (a) To what temperature must the combination be cooled to separate the two metals

Answer 1

Answer:

$T_f=-263.2135^{\circ}C$

Explanation:

Given:

• initial temperature of brass ring, $T_{Bi}=20.7^{\circ}C$

• initial temperature of aluminium rod, $T_{Ai}=20.7^{\circ}C$

• initial diameter of brass ring, $d_{Bi}=10.00\ cm$

• initial diameter of aluminium rod, $d_{Ai}=10.01\ cm$

We have:

• coefficients of linear expansion for Al, $\alpha_A=22.2\times 10^{-6}\ ^{\circ}C$

• coefficients of linear expansion for Al, $\alpha_B=18.7\times 10^{-6}\ ^{\circ}C$

(a)

Initial circumference of the brass ring:

$l_{Bi}=\pi.d_{Bi}$

$l_{Bi}=\pi\times 10\ cm$

Initial circumference of the aluminium rod:

$l_{Ai}=\pi.d_{Ai}$

$l_{Ai}=\pi\times 10.01\ cm$

According to condition:

The final circumference of the two must be at least equal to separate them.

$l_A_f=l_B_f$

$l_{Ai}+\Delta l_{Ai}=l_{Bi}+\Delta l_{Bi}$

$l_{Ai}+l_{Ai}.\alpha_A.\Delta T=l_{Bi}+l_{Bi}.\alpha_B.\Delta T$

$10.01+10.01\times 22.2\times 10^{-6}\times \Delta T=10.00+10.00\times 18.7\times 10^{-6}\times \Delta T$

$\Delta T=-283.9135$

Now, the final temperature:

$T_f=20.7-283.9135$

$T_f=-263.2135^{\circ}C$

is the temperature to which the combination must be cooled to separate the two metals.