Answer
a) $V_{tank}=0.006476\ m³$
b) $T_2=371.3°C$
c) $\dot{W}_e=1.576\ kW$
Work Step by Step
Initial state for the water (200°C, saturated liquid):
$v_1=0.001157\ m³/kg,\ u_1=850.46\ kJ/kg$
The initial volume of 1.4 kg of water is $V_1=mv_1=0.001619\ m³$
Since the water occupies 25% of the tank initially: $V_{tank}=0.006476\ m³$
b) At the final state $V_2=V_{tank}$, hence:
$v_2=0.004626\ m³/kg$
From steam tables:
$T_2=371.3°C,\ u_2=2201.5\ kJ/kg$
c) The energy balance for this system:
$-W_e=\Delta U\rightarrow \dot{W}_e\Delta t=m(u_2-u_1)$
With $\Delta t=20\ min$:
$\dot{W}_e=1.576\ kW$
