Answer
$V_{tank}=0.005\ m³$
$T_2=45.81°C$
Work Step by Step
At the initial state (60°C, 600 kPa, compressed liquid water):
$v_1\approx v_{sat, 60°C} = 0.001017\ m³/kg$
With a mass of 2.5 kg, the initial volume is:
$V_1=0.0025425\ m³$
Since the final volume, the volume of the tank, is twice the initial:
$V_{tank}=0.005\ m³$
At the final state (10 kPa), the specific volume:
$v_2=0.002034\ m³/kg$
Since this value is intermediary between the saturation specific volumes of liquid and vapor water, the final state is a saturated mixture:
$T_2=45.81°C$
