Answer
$T_3=228°F$
$x_3=0.415$
Work Step by Step
From tables A-5E and A-6E:
Inlet 1 ($P_1=20\ psia, T_1=65°F$): $h_1=33.08\ Btu/lbm$
Inlet 2 ($P_2=20\ psia$, sat. vapor): $h_2=1156.2\ Btu/lbm$
From the energy balance:
$\dot{m}_1h_1+\dot{m}_2h_2=(\dot{m}_1+\dot{m}_2)h_3$
With $\dot{m}_1=\dot{m}_2$
$h_1+h_2=2h_3$
Hence $h_3=594.6\ Btu/lbm$
At 20 psia saturation : $T_3=228°F,\ h_{3,L}=196.27\ Btu/lbm,\ h_{3,G}=1156.2\ Btu/lbm$
$h_3=h_{3,L}+x_3(h_{3,G}-h_{3,L})$
$x_3=0.415$
