## mtimko one year ago 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?

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