Answer
$T_2=10.7°C$
Work Step by Step
The energy balance for the radiator:
$Q=\Delta U$
From steam tables:
Initial state ($P_1=200\ kPa,\ T_1=200°C$): $v_1=1.08049\ m³,\ u_1=2654.6\ kJ/kg$
Final state ($P_2=100\ kPa,\ v_2=v_1$):
$v_{2,L}=0.001043\ m³/kg,\ v_{2,G}=1.6941\ m³/kg,\ u_{2,L}=417.40\ kJ/kg,\ u_{2,G}=2088.2\ kJ/kg$
The steam quality is: $x=\frac{v_2-v_{2,L}}{v_{2,G}} \rightarrow x=0.6376$
Hence $u_2=1748.7\ kJ/kg$
And since the volume is 0.15m³ $m=V_1/v_1=0.01388\ kg$
Therefore, from the energy balance: $Q=-12.58\ kJ$
For the air:
Volume of air is the volume of the room: 3x4x6=72m³
$PV=mRT$
Given $P_1=100\ kPa,\ R=0.287\ kJ/kg.K,\ T_1=280K$
we get $m=89.60\ kg$
The shaft work of the 120W fan after 45 min is $-W_s=\dot{W}_s\Delta t = 324\ kJ$
The energy balance for this system is:
$Q-W_s=\Delta U$
Plugging in the variables ($Q=12.58\ kJ,\ c_v=1.005\ kJ/kg.K$):
and solving for the final temperature: $T_2=10.7°C$
