Answer
a) $\dot{W}_e=3.151\ kW$
b) $\Delta T=5.0°C$
Work Step by Step
For the house:
$m=\frac{P_1V}{RT_1}$
Given $P_1=98\ kPa,\ V=120\ m³,\ R=0.287\ kJ/kg.K,\ T_1=15°C$:
$m=142.3\ kg$
From the energy balance:
$Q-W=\Delta U$
$\Delta t(\dot{W}_e+\dot{W}_s-\dot{Q})=mc_v(T_2-T_1)$
Since $\dot{W}_s=200W,\ \dot{Q}=150\ kJ/min,\ c_v=0.718\ kJ/kg.°C,\ T_2=25°C,\ \Delta t=20 min$:
$\dot{W}_e=3.151\ kW$
For the ducts ($\dot{m}=40\ kg/min$):
$\dot{W}_e+\dot{W}_s=\dot{m}c_p\Delta T$
$\Delta T=5.0°C$
