Answer
$163.03 \mathrm{~kJ} / \mathrm{kg}$
Work Step by Step
From the Clapeyron equation $$
\begin{aligned}
h_{f g} & =T v_{f g}\left(\frac{d P}{d T}\right)_{\text {sat }} \\
& \cong T\left(v_g-v_f\right)_{@ 400^{\circ} \mathrm{C}}\left(\frac{\Delta P}{\Delta T}\right)_{\text {sat }, 40^{\circ} \mathrm{C}} \\
& =T\left(v_g-v_f\right)_{@ 400^{\circ} \mathrm{C}}\left(\frac{P_{\text {sat } @ 42^{\circ} \mathrm{C}}-P_{\text {sat } @ 38^{\circ} \mathrm{C}}}{42^{\circ} \mathrm{C}-38^{\circ} \mathrm{C}}\right) \\
& =(40+273.15 \mathrm{~K})\left(0.019968-0.0008720 \mathrm{~m}^3 / \mathrm{kg}\right)\left(\frac{(1072.8-963.68) \mathrm{kPa}}{4 \mathrm{~K}}\right) \\
& =163.13 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$ The tabulated value of $h_{f g}$ at $40^{\circ} \mathrm{C}$ is $163.03 \mathrm{~kJ} / \mathrm{kg}$.
