Answer
$w = -8.40\ Btu/lbm$
$q = -71.09\ Btu/lbm$
Work Step by Step
Initial state (100°F, saturated vapor):
$v_1=0.34074\ ft³/lbm,\ h_1=116.22\ Btu/lbm$
Final state (100°F, saturated liquid):
$v_2=0.01386\ ft³/lbm,\ h_2=45.130\ Btu/lbm$
Saturation pressure: $P=138.93\ psia$
Work done by the system
$w=P(v_2-v_1) = -45.41\ psia.ft³/lbm\times \frac {1\ Btu}{5.40395\ psia.ft³}$
$w = -8.40\ Btu/lbm$
Energy balance for constant-pressure, stationary closed-system:
$q=\Delta h$
$q = -71.09\ Btu/lbm$
