Answer
$Q_i=486\ Btu$
Work Step by Step
From tables A-11E to A-13E:
Initial ($P_1=160\ psia, x_1=?$): $v_{1,L}=0.01413\ ft³/lbm,\ u_{1,L}=48.11\ Btu/lbm,\ v_{1,G}=v_2,\ u_{1,G}=u_2$
Final ($P_2=160\ psia, x_2=1$): $v_2=0.29339\ ft³/lbm,\ u_2=108.51\ Btu/lbm$
Exitting ($P_2=160\ psia, x_2=1$): $h_e=117.20\ Btu/lbm$
From the material balance:
$m_e=m_1-m_2$
Initial state: $V_{1,L}=0.05V_t$
since $m=V_L/v_L+V_G/v_G,\ V_t=2\ ft³$:
$m_1=13.553\ lbm$
and $mu=u_LV_L/v_L+u_GV_G/v_G,$:
$m_1u_1=1043.2\ Btu$
And at the final state: $m=V/v$, $m_2=6.817\ lbm$
So $m_e=6.736\ lbm$
From the energy balance:
$Q_i-m_eh_e=m_2u_2-m_1u_1$
$Q_i=486\ Btu$
