Answer
$395^{\mathrm{o}}\mathrm{C}$
Work Step by Step
We use: $\qquad \Delta L=\alpha L_{0}\Delta T ,\qquad$ 16-4,
which expresses the increase in diameter (L=2$\times$radius) of the ball when heated.
Solving for $\Delta T $, reading $\alpha$ from Table 16-1,
$\displaystyle \Delta T=\frac{\Delta L}{\alpha L_{0}}$
$\displaystyle \mathrm{T}-\mathrm{T}_{0}=\frac{\Delta L}{\alpha L_{0}}$
$\displaystyle \mathrm{T}=\mathrm{T}_{0}+\frac{\Delta L}{\alpha\cdot 2r_{0}}$
$\displaystyle \mathrm{T}=22^{\circ}\mathrm{C}+\frac{0.19\times 10^{3}\mathrm{m}}{[17\times 10^{-6}(\mathrm{C}^{\mathrm{o}})^{-1}][2(1.5\times 10^{-2}\mathrm{m})]}=395^{\mathrm{o}}\mathrm{C}$
