Question

A machinist wishes to insert a silver rod with a diameter of 4 mm into a hole with a diameter of 3.993 mm. By how much would the machinist have to lower the temperature (in °C) of the rod to make it fit the hole?

Answer 1

The change in the linear dimension of the rod due to temperature can be estimated to be:

$$\dfrac{\Delta d}{d} = \alpha\Delta T$$

where $d = 4mm$ is a a diameter of the rod, $\Delta d = 3.993 mm-4mm = -0.007mm$ is the diameter change, $\alpha = 18\times 10^{-6} \degree C^{-1}$ is a coefficient of linear thermal expansion and $\Delta T$ is temperature change.

Expressing $\Delta T$, obtain:

$$\Delta T = \dfrac{1}{\alpha}\dfrac{\Delta d}{d} = \dfrac{-0.007}{18\times 10^{-6}\cdot4} \approx -97.2\degree C$$

Answer. The machinist have to lower the temperature by 97.2°C.