Question

a steel plug has a diameter of 5cm at 30 degrees celsius. At what temperature will it fit exactly into a hole of constant diameter 4.997cm?

Answer 1

By the definition of the linear thermal expansion we have:

$$\Delta L = \alpha L_0 \Delta T,$$

here, $\Delta L = L - L_0$ is the increace in diameter of the steel plug when it is heated, $L_0 = 0.04997 \ m$ is the initial diameter of the steel plug at some temperature $T_0$, $L = 0.05 \ m$ is the diameter of the steel plug at $T = 30 \ ^{\circ}C$, $\alpha = 11 \cdot 10^{-6} \ ^{\circ}C^{-1}$ is the linear expansion coefficient for steel, $\Delta T = T - T_0$ is the change in temperature.

From this formula we can find the change in temperature:

$$\Delta T = \dfrac{L - L_0}{\alpha L_0},$$ $$\Delta T = \dfrac{0.05 \ m - 0.04997 \ m}{11 \cdot 10^{-6} \ ^{\circ}C^{-1} \cdot 0.04997 \ m} = 54.6 \ ^{\circ}C.$$

Finally, we can calculate the temperature at which the steel plug will fit exactly into a hole of constant diameter:

$$T_0 = T - \Delta T = 30 \ ^{\circ}C - 54.6 \ ^{\circ}C = -24.6 \ ^{\circ}C.$$

Answer:

$T_0 = -24.6 \ ^{\circ}C.$