Answer
$160 m^3$
Work Step by Step
We know that $\Delta V = V_0 \beta \Delta T$.
For a hemisphere of radius $R$, the volume is $V = \frac{2}{3} \pi R^3 = \frac{2}{3} \pi (27.5)^3 = 4.356 \times 10^4\ m^3 $.
For aluminium, $\beta = 7.2 \times 10^{-5}\ {^{\circ}} C^{-1}$. Then $\Delta V = V_0 \beta \Delta T = (4.356 \times 10^4\ m^3)(7.2 \times 10^{-5})[35-(-15)] = 160\ m^3$.
