Question #88329
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?
1
Expert's answer
2019-04-22T11:00:49-0400

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

ΔL = αLΔT\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=2mm=2103mL=2\,\text{mm}=2\cdot {{10}^{-3}}\text{m}

ΔL=1.995mm2mm=0.005mm=5106m\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 20oC is (https://en.wikipedia.org/wiki/Thermal_expansion)

α=11.8106K1=11.8106C1\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

5106m = 11.8106C12103mΔT-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

ΔT=511.82103C=212C\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


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