Answer
See the image
Work Step by Step
a) From table A-12E
$T=65.89^{\circ}F$
As $33.391\frac{Btu}{lbm}\lt h\lt 112.22\frac{Btu}{lbm}$ is a saturated mixture.
$x=\frac{h-h_{f}}{h_{fg}}=\frac{78\frac{Btu}{lbm}-33.391\frac{Btu}{lbm}}{78.83\frac{Btu}{lbm}}=0.57$
b) From table A-11E
$P=29.759psia$
$h=h_{f}+xh_{fg}=16.979\frac{Btu}{lbm}+0.6*88.403\frac{Btu}{lbm}=70.0208\frac{Btu}{lbm}$
As $0\lt x\lt 1$ is a saturated mixture.
c) From table A-11E
$h=15.308\frac{Btu}{lbm}$
As $T\lt T_{sat}$ is a compressed liquid
d) From table A-13E
$T=160^{\circ}F$
As $h\gt h_{g}$ is a superheated vapor
e) From table A-11E
$P=161.16psia$
$h=117.25\frac{Btu}{lbm}$
As $x=1$ is a saturated vapor
