Question

An aluminum plug (α = 2.3 x 10-5/C°) has a diameter of 10.003 cm at 40.0 °C. At what temperature (in Celcius) will it fit precisely into a hole of constant diameter 10.000 cm?

Answer 1

Yo can use thermal expansion for linear dimensions
\[\frac{ \Delta L }{ L_0 }=\alpha \Delta T\]
\[\frac{ L-L_0 }{ L_0 }=\alpha(T-T_0)\]
Let \(L_0=10.003~cm\) at \(T_0=40°C\) and \(L=10.000~cm\) at unknown temperature \(T\).

\[\frac{ 10.000~cm-10.003~cm }{ 10.003~cm }=(2.3 \times 10^{-5}/°C)(T - 40°C)\]

fyi: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/thexp.html#c1