Answer
$\dot{m}=0.0280\ kg/s$
$\dot{V}_2=0.0284\ m³/s$
Work Step by Step
From the energy balance:
$\dot{m}(h_1+\mathcal{V}_1^2/2)+\dot{W}_e=\dot{m}(h_2+\mathcal{V}_2^2/2)$
$\dot{W}_e=\dot{m}(c_p(T_2-T_1)+(\mathcal{V}_2^2-\mathcal{V}_1^2)/2)$
Given $\dot{W}_e=1.5\ kW,\ c_p=1.005\ kJ/kg.K,\ T_2=353\ K,\ T_1=300\ K,\ \mathcal{V}_2=21\ m/s,\ \mathcal{V}_1\approx0$:
$\dot{m}=0.0280\ kg/s$
For ideal gases:
$\dot{m}=\frac{P_2\dot{V}_2}{RT_2}$
In this case $P_2=100\ kPa,\ R=0.287\ kJ/kg.K$:
$\dot{V}_2=0.0284\ m³/s$
