Answer
$T_2=664\ °C$
$w_s=-887.1\ kJ/kg$ (in)
Work Step by Step
From table A-6:
Initial ($P_1=70\ kPa, T_1=100°C$): $u_1=2509.4\ kJ/kg,\ s_1=7.5344\ kJ/kg.K$
Final ($P_2=4\ MPa, s_2=s_1$): $T_2=664\ °C,\ u_2=3396.5\ kJ/kg$
From the energy balance:
$-W_s=m\Delta U$
$-w_s=\Delta u= u_2-u_1$
$w_s=-887.1\ kJ/kg$ (in)
