Question

*A 20°C a steel ball has a diameter of 0.7cm while the diameter of a hole in a aluminum plate is 0.699cm at what temperature the same for both will the ball just pass through the hole?*

Linear expansivity of steel 11e-6

Linear expansivity of aluminum 23e-6

Answer 1

Let's first write the lenghts of the steel ball and the hole in aluminum plate:

$$L_{Fe}=L_{Fe,0}[1+\alpha_{Fe}(T_f-T_i)],$$ $$L_{Al}=L_{Al,0}[1+\alpha_{Al}(T_f-T_i)].$$

When the ball just pass through the hole, $L_{Fe}=L_{Al}$ and we can write:

$$L_{Fe,0}[1+\alpha_{Fe}(T_f-T_i)]=L_{Al,0}[1+\alpha_{Al}(T_f-T_i)].$$

From this equation we can find the temperature, $T_f$, at what the ball just pass through the hole:

$$T_f=\dfrac{(\alpha_{Fe}L_{Fe,0}-\alpha_{Al}L_{Al,0})T_i+L_{Al,0}-L_{Fe,0}}{\alpha_{Fe}L_{Fe,0}-\alpha_{Al}L_{Al,0}},$$

$T_f=\dfrac{(11\cdot10^{-6}\ ^{\circ}C^{-1}\cdot0.007\ m-23\cdot10^{-6}\ ^{\circ}C^{-1}\cdot0.00699\ m)\cdot20^{\circ}C+0.00699\ m-0.007\ m}{11\cdot10^{-6}\ ^{\circ}C^{-1}\cdot0.007\ m-23\cdot10^{-6}\ ^{\circ}C^{-1}\cdot0.00699\ m},$

$$T_f=139.4^{\circ}C.$$