Answer
a) Show at the image
b) $T=151.83^{\circ}$
c) $\Delta V=-0.1422m^3$
Work Step by Step
From table A-6
$\upsilon_{1}=0.42503\frac{m^3}{kg}$
b) From table A-5:
$T=151.83^{\circ}$
c) $\upsilon=\upsilon_{f}+x(\upsilon_{g}-\upsilon_{f})=0.001093\frac{m^3}{kg}+0.5*(0.37483\frac{m^3}{kg}-0.001093\frac{m^3}{kg})=0.1879615\frac{m^3}{kg}$
Then:
$\Delta \upsilon= 0.1879615\frac{m^3}{kg}-0.42503\frac{m^3}{kg}=-0.2371\frac{m^3}{kg}$
$\Delta V= m\Delta \upsilon= 0.6kg*-0.2371\frac{m^3}{kg}=-0.1422m^3$
