Answer
$m=2.15\ lbm$
Work Step by Step
From table A-11E to A-13E:
Initial ($P_1=90\ psia, T_1=30°F$): $s_1=0.04750\ Btu/lbm.°R$
Final ($P_2=20\ psia, s_2=s_1$): $x_2=0.1075,\ v_2=0.2555\ ft^3/lbm$
Hence the final mass is:
$m=V_t/v_2,\ V_t=0.55\ ft^3$
$m=2.15\ lbm$
