Respuesta :
Answer:
[tex]T_f=-263.2135^{\circ}C[/tex]
Explanation:
Given:
- initial temperature of brass ring, [tex]T_{Bi}=20.7^{\circ}C[/tex]
- initial temperature of aluminium rod, [tex]T_{Ai}=20.7^{\circ}C[/tex]
- initial diameter of brass ring, [tex]d_{Bi}=10.00\ cm[/tex]
- initial diameter of aluminium rod, [tex]d_{Ai}=10.01\ cm[/tex]
We have:
- coefficients of linear expansion for Al, [tex]\alpha_A=22.2\times 10^{-6}\ ^{\circ}C[/tex]
- coefficients of linear expansion for Al, [tex]\alpha_B=18.7\times 10^{-6}\ ^{\circ}C[/tex]
(a)
Initial circumference of the brass ring:
[tex]l_{Bi}=\pi.d_{Bi}[/tex]
[tex]l_{Bi}=\pi\times 10\ cm[/tex]
Initial circumference of the aluminium rod:
[tex]l_{Ai}=\pi.d_{Ai}[/tex]
[tex]l_{Ai}=\pi\times 10.01\ cm[/tex]
According to condition:
The final circumference of the two must be at least equal to separate them.
[tex]l_A_f=l_B_f[/tex]
[tex]l_{Ai}+\Delta l_{Ai}=l_{Bi}+\Delta l_{Bi}[/tex]
[tex]l_{Ai}+l_{Ai}.\alpha_A.\Delta T=l_{Bi}+l_{Bi}.\alpha_B.\Delta T[/tex]
[tex]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[/tex]
[tex]\Delta T=-283.9135[/tex]
Now, the final temperature:
[tex]T_f=20.7-283.9135[/tex]
[tex]T_f=-263.2135^{\circ}C[/tex]
is the temperature to which the combination must be cooled to separate the two metals.
