Question

A machinist wishes to insert an iron 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

The dependence of thermal expansion on temperature, substance, and length is determined by the equation

$$\Delta \text{L = }\alpha \text{L}\Delta \text{T}$$

where ΔL is the change in diameter L of rod, ΔT is the change in temperature, and α is the coefficient of linear expansion.

We have

$$L=2\,\text{mm}=2\cdot {{10}^{-3}}\text{m}$$

$$\Delta L=1.995\,\text{mm}-2\,\text{mm}=-0.005\,\text{mm}=-5\cdot {{10}^{-6}}\text{m}$$

The coefficient of linear expansion for iron at 20^oC is (https://en.wikipedia.org/wiki/Thermal_expansion)

$$\alpha =11.8\cdot {{10}^{-6}}\,{{\text{K}}^{-1}}=11.8\cdot {{10}^{-6}}\,{}^\circ {{\text{C}}^{-1}}$$

Substituting the known values into the formula, we get

$$-5\cdot {{10}^{-6}}\text{m = }11.8\cdot {{10}^{-6}}\,{}^\circ {{\text{C}}^{-1}}\cdot \text{2}\cdot \text{1}{{\text{0}}^{-3}}\text{m}\cdot \Delta \text{T}$$

Hence the change in temperature is

$$\Delta \text{T}=-\frac{5}{11.8\cdot 2}\cdot \text{1}{{\text{0}}^{3}}\,{}^\circ \text{C}=-212{}^\circ \text{C}$$

So the machinist would have to cool the iron rod to make it fit the hole by 212 °C