Answer
$q=2484\ kJ/kg$
Work Step by Step
The state variables are:
State 1 (200°C, saturated vapor): $P_1=1555\ kPa, v_1=0.1272\ m³/kg,\ u_1=2594.2\ kJ/kg$
State 2 (50°C, saturated liquid): $P_2=12.35\ kPa, v_2=0.001012\ m³/kg,\ u_2=209.33\ kJ/kg$
The specific energy balance for this process is:
$q-w_b=\Delta u$
The boundary work for the spring-loaded system:
$w_b=1/2.(P_1+P_2)(v_2-v_1)$
$w_b=-98.9\ kJ/kg$
Plugging in the energy balance: $q=2484\ kJ/kg$
