## Physics: Principles with Applications (7th Edition)

The sphere is $.975$% larger.
First, let's calculate the original volume of the aluminum sphere. $$V=\frac{4}{3}(3.14)r^{3}$$ $$V=\frac{4}{3}\times3.14\times(8.75cm/2)^{3}$$ $$V=350.6cm^{3}$$ Then, let's calculate the change in volume due to thermal expansion. $$\Delta V = bV_{0}\Delta T$$ $$V_{f}-350.6cm^{3}=(75\times10^{-6})(350.6cm^{3})(160^{\circ}C-30^{\circ}C)$$ $$V_{f}=354.0cm^{3}$$ Therefore, the percent change is $$\frac{354.0mL-350.6mL}{350.6mL}=.975 \%$$