Answer
$T_3=117.3°F$
Work Step by Step
From table A-6:
Steam ($P_1=80\ psia, T_1=400°F$): $h_1=1230.8\ Btu/lbm$
Water ($P_2=80\ psia, T_2=60°F$): $h_2=28.08\ Btu/lbm$
$\dot{m}_1+\dot{m}_2=\dot{m_3}$
$\dot{m}_1=0.05\ lbm/s,\ \dot{m}_2=1\ lbm/s$
$\dot{m}_3=1.05\ lbm/s$
$\dot{m}_1h_1+\dot{m}_2h_2=\dot{m_3}h_3$
$h_3=85.35\ Btu/lbm$
At ($P_3=80\ psia,\ h_3$): compressed liquid, $T_3\approx T_{sat}=117.3°F$
