Answer
$T_{2}=480.3R$
Work Step by Step
From the Clapeyron equation,
$$
\left(\frac{d P}{d T}\right)_{\text {sat }}=\frac{h_{f g}}{T v_{f g}}=\frac{(250 \mathrm{Btu})\left(\frac{5.404 \mathrm{psia} \cdot \mathrm{ft}^3}{1 \mathrm{Btu}}\right) /(0.5 \mathrm{lbm})}{(475 \mathrm{R})\left(1.5 \mathrm{ft}^3\right) /(0.5 \mathrm{lbm})}=1.896 \mathrm{psia} / \mathrm{R}
$$ Using the finite difference approximation,
$$
\left(\frac{d P}{d T}\right)_{\text {sat }} \approx\left(\frac{P_2-P_1}{T_2-T_1}\right)_{\text {sat }}
$$
Solving for $T_2$,
$$
T_2=T_1+\frac{P_2-P_1}{d P / d T}=475 \mathrm{R}+\frac{(60-50) \mathrm{psia}}{1.896 \mathrm{psia} / \mathrm{R}}=480.3 \mathrm{R}
$$
