Answer
$F=9,575\ kN$
$Q=-66.8\ kJ$
Work Step by Step
Initial state (saturated steam, 100 kPa) :
$v_1=1.6941\ m³/kg,\ u_1=2505.6\ kJ/kg$
With the volume of 0.05m³: $m=0.02951\ kg$
Final state (saturated mixture, 30°C):
$v_{2,L}=0.001004\ m³/kg,\ u_{2,L}=125.73\ kJ/kg,\ v_{2,G}=32.879\ m³/kg,\ u_{2,L}=2290.2\ kJ/kg$
$P_2=P_{sat}=4.2469\ kPa$
Given that $v_2=v_1$, because the piston is stuck:
$x_2=\frac{v_2-v_{2,L}}{v_{2,G}}=0.05150$
Hence $u_2=u_{2,L}+x.u_{2,G}=243.67\ kJ/kg$
The friction force is given by:
$F=-A\Delta P$
With an area of 0.1 m²:
$F=9.575\ kN$
The heat transfer is given by the energy balance:
$Q=\Delta U$
$Q=-66.8\ kJ$
