Answer
a) $P_3=1555\ kPa,\ x_3=0.00062$
b) $W_b=-70.45\ kJ$
c) $Q_{1-2}=-622.9\ kJ$
d) $Q_{1-3}=-690.8\ kJ$
Work Step by Step
From steam tables:
Initial state ($P_1=3.5\ MPa,\ T_1=T_{sat}+7.4°C$):
$T_{sat}=242.6°C \rightarrow T_1=250°C,\ v_1=0.05875\ m³/kg,\ u_1=2623.9\ kJ/kg$
Second state($P_2=P_1$, saturated liquid):
$v_2=0.001235\ m³/kg,\ u_2=1045.4\ kJ/kg$
Third state ($v_3=v_2, T_3=200°C$):
$P_3=1555\ kPa,\ x_3=0.00062,\ u_3=851.55\ kJ/kg$
Only in the process 1-2 boundary work is done (2-3 is constant-volume):
$W_b=P_1.m(v_2-v_1)$
Given the mass of 0.35 kg:
$W_b=-70.45\ kJ$
The energy balance for the 1-2 process:
$Q_{1-2}-W_b=\Delta U_{1-2}$
Hence $Q_{1-2}=-622.9\ kJ$
For the 1-3 process:
$Q_{1-3}-W_b=\Delta U_{1-3}$
Hence $Q_{1-3}=-690.8\ kJ$
