Answer
$T_{2}=-10.09^{\circ}C$
$\Delta u=-110.57\frac{kJ}{kg}$
Work Step by Step
$P=200kPa$
1) $\upsilon_{1}=\frac{V_{1}}{m}=\frac{12.322m^3}{100kg}=0.12322\frac{m^3}{kg}$
As $\upsilon_{1}\gt \upsilon_{g}$ is a superheated vapor
From table A-13:
$u_{1}=263.09\frac{kJ}{kg}$
2) $\upsilon_{2}=\frac{V_{2}}{m}=\frac{\frac{12.322m^3}{2}}{100kg}=0.06161\frac{m^3}{kg}$
As $0.0007532\frac{m^3}{kg}\lt \upsilon_{2}\lt 0.099951\frac{m^3}{kg}$ is a saturated
$T_{2}=-10.09^{\circ}C$
$x_{2}=\frac{\upsilon-\upsilon_{f,2}}{\upsilon_{g,2}-\upsilon_{f,2}}=\frac{0.06161\frac{m^3}{kg}-0.0007532\frac{m^3}{kg}}{0.099951\frac{m^3}{kg}-0.0007532\frac{m^3}{kg}}=0.6135$
$u_{2}=u_{f}+x_{2}u_{fg}=38.26\frac{kJ}{kg}+0.6135*186.25\frac{kJ}{kg}=152.52\frac{kJ}{kg}$
Finally:
$\Delta u=u_{2}-u_{1}=152.52\frac{kJ}{kg}-263.09\frac{kJ}{kg}=-110.57\frac{kJ}{kg}$
