Answer
a)$T=227.92^{\circ}F$
b)$H=1216.4Btu$
c)$m_{g}=0.245lbm$
$m_{f}=4.755lbm$
Work Step by Step
$\upsilon=\frac{V}{m}=\frac{5ft^3}{5lbm}=1\frac{ft^3}{lbm}$
As $0.01683\frac{ft^3}{lbm}\lt \upsilon\lt 20.093\frac{ft^3}{lbm}$ is a saturated mixture.
From table A-5E:
a)$T=227.92^{\circ}F$
b)$x=\frac{\upsilon-upsilon_{f}}{\upsilon{g}-\upsilon_{f}}=\frac{1\frac{ft^3}{lbm}-0.01683\frac{ft^3}{lbm}}{20.093\frac{ft^3}{lbm}-0.01683\frac{ft^3}{lbm}}=0.048971$
$h=h_{f}+xh_{fg}=196.27\frac{Btu}{lbm}+0.048971*959.93\frac{Btu}{lbm}=243.28\frac{Btu}{lbm}$
$H=5lbm*243.28\frac{Btu}{lbm}=1216.4Btu$
c)$m_{g}=xm_{t}=0.048971*5lbm=0.245lbm$
$m_{f}=m_{t}-m_{g}=5lbm-0.245lbm=4.755lbm$
