Answer
a) $V_2=0.007634\ m³$
b) $T_2=120°C$
c) $\dot{W}_e=0.0236\ kW$
Work Step by Step
At the initial state (120°C, saturated liquid):
$v_1=0.001060\ m³/kg,\ h_1=503.81\ kJ/kg,\ P_1=198.67\ kPa$
a) The initial volume of a 1.8 kg mass is $V_1=0.001909\ m³$
Since this volumes quadruples:
$V_2=0.007634\ m³$
b) The final specific volume is:
$v_2=0.004241\ m³/kg$
And since $P_2=P_1$: $T_2=120°C=T_1$,
$x_2=0.00357,\ h_2=511.68\ kJ/kg$
c)
The energy balance for the constant-pressure process:
$-W_e=\Delta H \rightarrow \dot{W}_e\Delta t=m(h_2-h_1)$
For a time period of 10 min:
$\dot{W}_e=0.0236\ kW$
