Answer
$T_{1}=0.605^{\circ}C$
$H_{1}=545kJ$
$T_{2}=21.55^{\circ}C$
$H_{2}=846.4kJ$
Work Step by Step
$\upsilon=\frac{V}{m}=\frac{14L*\frac{1m^3}{1000L}}{10kg}=0.0014\frac{m^3}{kg}$
1) As $0.0007734\frac{m^3}{kg}\lt \upsilon\lt 0.0680575\frac{m^3}{kg}$ is a saturated mixture.
From table A-12:
$T_{1}=0.605^{\circ}C$
$x_{1}=\frac{\upsilon-\upsilon_{f,1}}{\upsilon_{g,1}-\upsilon_{f,1}}=\frac{0.0014\frac{m^3}{kg}-0.0007734\frac{m^3}{kg}}{0.0680575\frac{m^3}{kg}-0.0007734\frac{m^3}{kg}}=0.009313$
$h_{1}=h_{f,1}+x_{1}h_{fg,1}=52.65\frac{kJ}{kg}+0.009313*198.195\frac{kJ}{kg}=54.50\frac{kJ}{kg}$
$H_{1}=mh_{1}=10kg*54.50\frac{kJ}{kg}=545kJ$
2) As $0.0008198\frac{m^3}{kg}\lt \upsilon\lt 0.034335\frac{m^3}{kg}$ is a saturated mixture.
From table A-12:
$T_{2}=21.55^{\circ}C$
$x_{2}=\frac{\upsilon-\upsilon_{f,2}}{\upsilon_{g,2}-\upsilon_{f,2}}=\frac{0.0014\frac{m^3}{kg}-0.0008198\frac{m^3}{kg}}{0.034335\frac{m^3}{kg}-0.0008198\frac{m^3}{kg}}=0.017312$
$h_{2}=h_{f,2}+x_{2}h_{fg,2}=81.50\frac{kJ}{kg}+0.017312*180.95\frac{kJ}{kg}=84.63\frac{kJ}{kg}$
$H_{2}=mh_{2}=10kg*84.63\frac{kJ}{kg}=846.4kJ$
