Answer
$T_2=96.2°C$
Work Step by Step
From tables A-11 to A-13:
Initial ($P_1=1.4\ MPa, T_1=120°C$): $v_1=0.02039\ m³/kg,\ u_1=323.57\ kJ/kg,\ h_1=352.11\ kJ/kg$
Since $m=V/v$:
Initial ($V_1=0.8\ m³$): $m_1=39.24\ kg$
Final ($V_2=0.5\ m³$): $m_2=0.5/v_2$
From the material balance:
$-m_e=m_2-m_1$
Since the spring is linear, with $P_2=0.7\ MPa$:
$W_{b,i}=-\dfrac{P_1+P_2}2(V_2-V_1)$
$W_{b,i}=315\ kJ$
From the energy balance:
$W_{b,i}-m_eh_e=m_2u_2-m_1u_1$
Assuming $h_e\approx\dfrac{h_1+h_2}2$
$W_{b,i}+(m_2-m_1)\dfrac{h_1+h_2}2=m_2u_2-m_1u_1$
From trial and error:
$T_2=96.2°C,\ u_2=306.43\ kJ/kg,\ h_2=334.51\ kJ/kg,\ v_2=0.04011\ m³/kg$
Hence $m_2=12.47\ kg,\ m_e=26.8\ kg$
