Answer
See explanation
Work Step by Step
From the psychrometric chart (Fig. A-31) we read
(a) $\omega=0.0165\ \mathrm{lbm}\ \mathrm{H}_2 \mathrm{O} / \mathrm{lbm}$ dry air
(b) $h=37.8\ \mathrm{Btu} / \mathrm{lbm}$ dry air
(c) $T_{\mathrm{wb}}=74.3^{\circ} \mathrm{F}$
(d) $T_{\mathrm{dp}}=71.3^{\circ} \mathrm{F}$
(e) $\boldsymbol{v}=14.0 \mathrm{ft}^3 / \mathrm{lbm}$ dry air
