Question

A machinist wishes to insert a brass rod with a diameter of 2 mm into a hole with a diameter of 1.995 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

$\Delta T=\frac{\Delta l}{\alpha l}=\frac{5\cdot 10^{-6}}{19\cdot 10^{-6}\cdot 2\cdot 10^{-3}}=132°C.$