Answer
$\dot Q_{gen, total}=1020\text{ W}$
$\dot Q_{gen, latent}=642\text{ W}$
$\dot m_{moisture }=0.264\text{ g/s}$
Work Step by Step
The total rate of heat generation of the chickens in the breeding room is $$
\dot{Q}_{\text {gen, total }}=\dot{q}_{\text {gen, total }}(\text { No. of chickens })=(10.2 \mathrm{~W} / \text { chicken })(100 \text { chickens })=\mathbf{1 0 2 0}\ \mathbf{W}
$$ The latent heat generated by the chicken and the rate of moisture production are $$
\begin{aligned}
\dot{Q}_{\text {gen, latent }} & =\dot{q}_{\text {gen,latent }}(\text { No. of chickens }) \\
& =(6.42 \mathrm{~W} / \text { chicken })(100 \text { chickens })=642 \mathrm{~W} \\
& =0.642 \mathrm{~kW} \\
\dot{m}_{\text {moisture }} & =\frac{\dot{Q}_{\text {gen, latent }}}{h_{f g}}=\frac{0.642 \mathrm{~kJ} / \mathrm{s}}{2430 \mathrm{~kJ} / \mathrm{kg}}=0.000264 \mathrm{~kg} / \mathrm{s}=0.264 \mathrm{~g} / \mathrm{s}
\end{aligned}
$$
