Temperature and

Question

A machinist wishes to insert a brass 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:

Δd÷d =αΔT

where d = 4mmd=4mm is a a diameter of the rod

Δd=3.993mm−4mm=−0.007mm is the diameter change, 

α=18×10−6°C−1 is a coefficient of linear thermal expansion ΔT is temperature change.

Expressing  ΔT, obtain:

ΔT=1÷α ×dΔ÷d​=-0.007÷18×10−6⋅4​ ≈−97.2°C

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