Answer
$q_{out}=14.2\text{ kJ/kg dry air}$
Work Step by Step
The amount of moisture in the air remains constant ($\omega_1 =\omega_2$) as it flows through the cooling section since the
process involves no humidification or dehumidification. The inlet and exit states of the air are completely specified, and the total pressure is $1$ atm. The properties of the air at the inlet state are determined from the psychrometric chart to be $$
\begin{aligned}
h_1 & =76.1 \mathrm{~kJ} / \mathrm{kg} \text { dry air } \\
\omega_1 & =0.0159 \mathrm{~kg} \mathrm{H}_2 \mathrm{O} / \mathrm{kg} \text { dry air }\left(=\omega_2\right) \\
T_{\mathrm{dp}, \mathrm{I}} & =21.3^{\circ} \mathrm{C}
\end{aligned}
$$ The exit state enthalpy is $$
\left.\begin{array}{rl}
P & =1 \mathrm{~atm} \\
T_2 & =T_{\mathrm{dp}, 1}=21.3^{\circ} \mathrm{C} \\
\phi_2 & =1
\end{array}\right\} h_2=61.9 \mathrm{~kJ} / \mathrm{kg} \text { dry air }
$$ From the energy balance on air in the cooling section, $q_{\text {out }}=h_1-h_2=76.1-61.9=14.2 \mathrm{~kJ} / \mathbf{kg~dry~air}$
