Question

A carpenter spent his morning working using a steel hammer under the heat of the sun. The volume of the hammer is 10 cubic inches at 21 degrees Celcius. After his work, what was the volume of the hammer if the temperature at noon is at 33 degrees Celcius? 

Answer 1

Given:

$V_0=10\:\rm in^3$

$t_0=21^{\circ}\rm C$

$t_1=33^{\circ}\rm C$

$\beta=35*10^{-6} \frac{1}{^{\circ}\rm C}$

The volume thermal expansion equation says

$$V=V_0(1+\beta\Delta t)$$

Hence,

$$V=10*(1+35*10^{-6}*(33-21))=10.0042\:\rm in^3$$