Answer
$T_4=206.11\ °C$
$\dot{Q}_w=97.26\ kW$
Work Step by Step
From tables A-4 and A-5:
Water inlet ($ T_1=15°C,\ x_1=0$): $h_1=62.98\ kJ/kg$
Steam outlet ($P_2=2\ MPa, x_2=1$): $h_2=2798.3\ kJ/kg$
The combined energy balance:
$\dot{m}_1h_1+\dot{m}_3h_3=\dot{m}_1h_2+\dot{m}_3h_4+\dot{Q}$
Since $h_4=h_3+c_p(T_4-T_3),\ \dot{m}_3=15\dot{m}_1,\ \dot{Q}=0.1\dot{m}_3(h_3-h_4)$
$h_2-h_1=13.5c_p(T_3-T_4)$
Given $c_p=1.045\ kJ/kg.K,\ T_3=400°C$:
$T_4=206.11\ °C$
Since $\dot{m}_3=15\dot{m}_1,\ \dot{m}_3=32\ kg/min$
$\dot{m}_1=0.0356\ kg/s$
$\dot{Q}_w=\dot{m}_1(h_2-h_1)=97.26\ kW$
