Answer
$39.0^{\mathrm{o}}\mathrm{C}$
Work Step by Step
Apply 17-6, $\Delta L=L_{0}\alpha\Delta T$
with $\alpha_{\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{e}\mathrm{l}} =1.2\times 10^{-5}(\mathrm{C}^{\mathrm{o}})^{-1} \qquad $(from table 17-1),
and given $\Delta L=0.471\mathrm{f}\mathrm{t}$
$\displaystyle \Delta T=\frac{\Delta L}{\alpha L_{0}}$
$=\displaystyle \frac{0.471\mathrm{f}\mathrm{t}}{[1.2\times 10^{-5}(\mathrm{C}^{\mathrm{o}})^{-1}][1671\mathrm{ft}]}$
$=23.5\mathrm{C}^{\mathrm{o}}$
$T_{2}=15.5^{\mathrm{o}}\mathrm{C}+\Delta T=15.5^{\mathrm{o}}\mathrm{C}+23.5\mathrm{C}^{\mathrm{o}}=39.0^{\mathrm{o}}\mathrm{C}$.
