Answer
a) Shown at the image
b) $\Delta V=-0.187m^3$
c) $\Delta U=-1938.8kJ$
Work Step by Step
$P_{1}=1.2MPa$
$T_{1}=70^{\circ}C$
From table A-13:
$\upsilon_{1}=0.019502\frac{m^3}{kg}$
$V_{1}=m\upsilon_{1}=10kg*0.019502\frac{m^3}{kg}=0.19502m^3$
$u_{1}=277.23\frac{kJ}{kg}$
$U_{1}=mu_{1}=10kg*277.23\frac{kJ}{kg}=2772.3kJ$
$P_{2}=1.2MPa$
$T_{2}=20^{\circ}C$
From table A-11:
$\upsilon_{2}=0.0008160\frac{m^3}{kg}$
$V_{2}=m\upsilon_{2}=10kg*0.0008160\frac{m^3}{kg}=0.00816m^3$
$u_{2}=78.85\frac{kJ}{kg}$
$U_{2}=mu_{2}=10kg*78.85\frac{kJ}{kg}=788.5kJ$
Then:
b) $\Delta V=0.00816m^3-0.19502m^3=-0.187m^3$
c) $\Delta U=788.5kJ-2772.3kJ=-1938.8kJ$
