Answer
$h=179.9\frac{kJ}{kg}$
Work Step by Step
$\upsilon=\frac{V}{m}=\frac{9m^3}{300kg}=0.03\frac{m^3}{kg}$
From table A-11:
As $0.0007929\frac{m^3}{kg}\lt \upsilon\lt 0.049466\frac{m^3}{kg}$ is a saturated mixture
$x=\frac{\upsilon-\upsilon_{f}}{\upsilon_{fg}}=\frac{0.03\frac{m^3}{kg}-0.0007929\frac{m^3}{kg}}{0.049466\frac{m^3}{kg}-0.0007929\frac{m^3}{kg}}=0.6$
$h=h_{f}+xh_{fg}=65.42\frac{kJ}{kg}+0.6*190.8\frac{kJ}{kg}=179.9\frac{kJ}{kg}$
