Answer
a) $T_2=40.23°F$
b) $70\%$
c) $Q = -4166.33\ Btu$
Work Step by Step
The initial state, saturated R-134a steam at 160 psia, from R-134a tables:
$v_1=0.29339\ ft³/lbm,\ u_1 = 108.51\ Btu/lbm$
Therefore, knowing the volume to be 20 ft³, the mass is:
$m = 68.169\ lbm$
At the final state:
$P_2=50\ psia,\ v_2=v_1$, saturated mixture
$v_{2,L} =0.01252\ ft³/lbm,\ v_{2,G} = 0.94909\ ft³/lbm$
$u_{2,L} =24.824\ Btu/lbm,\ u_{2,G} = 100.05\ Btu/lbm$
a) The final temperature is the saturation temperature: $T_2=40.23°F$
b)From $v_2 = v_{2,L} +x(v_{2,G} -v_{2,L} )$
We have the quality: $x= 0.30$
The ammount of refrigerant that has condensed is: $(1-x)\times 100\% = 70\%$
c) For a closed system stationary at constant volume:
$\Delta U = Q$
$u_2 = u_{2,L} +x(u_{2,G} -u_{2,L} )$
$u_2 = 47.39\ Btu/lbm$
Therefore:
$Q = -4166.33\ Btu$
On a P-v diagram, the process would start at the saturation curve and proceed on a straight vertical line downwards to below the saturation curve.