Answer
a) $h_{fg}=89.38\text{ Btu/lbm}$
b) $h_{fg}=96.04\text{ Btu/lbm}$
Work Step by Step
(a) From the Clapeyron equation,
$$ \begin{aligned}
h_{f g} & =T v_{f g}\left(\frac{d P}{d T}\right)_{\text {sat }} \\
& \cong T\left(v_g-v_f\right)_{@ 10^{\circ} \mathrm{F}}\left(\frac{\Delta P}{\Delta T}\right)_{\text {sat, } 10 \% \mathrm{~F}} \\
& =T\left(v_g-v_f\right)_{@ 10^{\circ} \mathrm{F}}\left(\frac{P_{\text {sat } @ 15^{\circ} \mathrm{F}}-P_{\text {sat } @ 5^{\circ} \mathrm{F}}}{15^{\circ} \mathrm{F}-5^{\circ} \mathrm{F}}\right) \\
& =(10+459.67 \mathrm{R})\left(1.7358-0.01200 \mathrm{ft}^3 / \mathrm{lbm}\right)\left(\frac{(29.759-23.793) \mathrm{psia}}{10 \mathrm{R}}\right) \\
& =483.0 \mathrm{psia} \cdot \mathrm{ft}^3 / \mathrm{lbm}=89.38 \mathrm{Btu} / \mathrm{lbm} \quad(0.1 \% \text { error })
\end{aligned}
$$ since $1 \mathrm{Btu}=5.4039 \mathrm{psia} \cdot \mathrm{ft}^3$.
(b) From the Clapeyron-Clausius equation, $$
\begin{aligned}
\ln \left(\frac{P_2}{P_1}\right)_{\text {sat }} & \cong \frac{h_{f g}}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)_{\text {sat }} \\
\ln \left(\frac{23.793 \mathrm{psia}}{29.759 \mathrm{psia}}\right) & \cong \frac{h_{f g}}{0.01946 \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R}}\left(\frac{1}{15+459.67 \mathrm{R}}-\frac{1}{5+459.67 \mathrm{R}}\right) \\
h_{f g} & =\mathbf{9 6 . 0 4} \mathrm{Btu} / \mathrm{lbm}(7.6 \% \text { error })
\end{aligned}
$$ The tabulated value of $h_{f g}$ at $10^{\circ} \mathrm{F}$ is $89.25\ \mathrm{Btu} / \mathrm{lbm}$.
