Question

A silver cube has sides that measure 2.5 meters. The cube is heated from 15°C to 30°C. What is the change in its volume?

Answer 1

The initial volume of the silver cube is:
$V=(2.5 m)^3 = 15.6 m^3$

The volumetric thermal expansion of an object of volume V is given by$\Delta V =3 \alpha _L V \Delta T$
where $\alpha_L$ is the coefficient of thermal expansion and $\Delta T$ the temperature difference.
For silver, $\alpha_L = 18.9 \cdot 10^{-6} m^{-1}K^{-1}$, while the temperature difference in our problem is 
$\Delta T = 30^{\circ}-15^{\circ}=15^{\circ}=15 K$
So we can calculate the change in volume of the silver cube:

$\Delta V = 3 (18.9 \cdot 10^{-6} m^{-1}K^{-1})(15.6 m^3)(15 K)=0.013 m$