Question

calculate the volume of the same mass of water at 300 degrees

Answer 1

Assume the water does not evaporate:

$$V=V_0(1+\beta T),\\\space\\ V=\frac m{\rho_0}\Big(1+\Big(210 × 10^{– 6}\Big)×300\Big)=1.063\frac{m}{\rho_0},$$

where $\rho_0$ is the initial density of water.