Answer
a) $T_3=296.6\ K=23.6°C$
b) $\dot{Q}_r=0.691\ kW$
Work Step by Step
From table A-17:
Inlet 1 ($T_1=7°C$): $h_1=280.13\ kJ/kg$
Inlet 2 ($T_2=34°C$): $h_2=307.23\ kJ/kg$
Room ($T_r=24°C$): $h_r=297.18\ kJ/kg$
Energy balance for the mixing:
$\dot{m}_1h_1+\dot{m}_2h_2=(\dot{m}_1+\dot{m}_2)h_3$
Since $\dot{m}_2/\dot{m}_1=1.6$
$h_1+1.6h_2=2.6h_3$
So $h_3=296.81\ kJ/kg$
and from the table $T_3=296.6\ K=23.6°C$
For the first inlet ($P_1=105\ kPa,\ \dot{V}_1=0.55\ m³/s,\ R=0.287\ kJ/kg.K,\ T_1=280\ K$}:
$\dot{m}_1=\frac{P_1\dot{V}_1}{RT_1}$
$\dot{m}_1=0.7186\ kg/s$
$\dot{m}_3=1.868\ kg/s$
For the room:
$\dot{Q}_r=\dot{m}_3(h_r-h_3)$
$\dot{Q}_r=0.691\ kW$
