Answer
$P_{a}= 13.14\text{ psia}$
$w=0.0169\text{ lbm H}_2\text{O/lbm dry air}$
$h=39.01\text{ Btu/lbm dry air}$
Work Step by Step
(a) The partial pressure of dry air can be determined from $$
\begin{aligned}
& P_v=\phi P_g=\phi P_{\text {sat } @ 85^{\circ} \mathrm{F}}=(0.60)(0.5966 \mathrm{psia})=0.358\ \mathrm{psia} \\
& P_a=P-P_v=13.5-0.358=13.14\ \mathrm{psia}
\end{aligned}
$$ (b) The specific humidity of air is determined from $$
\omega=\frac{0.622 P_v}{P-P_v}=\frac{(0.622)(0.358 \mathrm{psia})}{(13.5-0.358) \mathrm{psia}}=0.0169\ \mathrm{lbm} \mathrm{H_{2 }} \mathrm{O} / \mathrm{lbm} \text { dry air }
$$ (c) The enthalpy of air per unit mass of dry air is determined from $$
\begin{aligned}
h & =h_a+\omega h_v \equiv c_p T+\omega h_g \\
& =\left(0.24 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\left(85^{\circ} \mathrm{F}\right)+(0.0169)(1098.3 \mathrm{Btu} / \mathrm{lbm}) \\
& =\mathbf{3 9 . 0 1} \text{ Btu/lbm dry air}
\end{aligned}
$$
