Answer
$\Delta V=0.0168m^3$
Work Step by Step
1) As $T\gt T_{sat}$ is a is a superheated vapor
From table A-13:
$\upsilon_{1}=0.33608\frac{m^3}{kg}$
2) As $T\gt T_{sat}$ is a is a superheated vapor
From table A-13:
$\upsilon_{2}=0.50410\frac{m^3}{kg}$
Then the change of volume is:
$\Delta V=m\Delta \upsilon=0.1kg*(0.50410\frac{m^3}{kg}-0.33608\frac{m^3}{kg})$
$\Delta V=0.0168m^3$
