What happens if I make a hole in a container with superheated water?
If you have water at a temperature of $T=250\ \mathrm{^\circ C}$, you need a pressure of approximately $p=39.2\ \mathrm{atm}=4.0\ \mathrm{MPa}$ to keep it liquid. Liquid water at this temperature and pressure has a specific enthalpy of $h_0=1086\ \mathrm{kJ/kg}$.

If you open the container, the pressure drops to the ambient pressure of $p=1\ \mathrm{atm}=101\,325\ \mathrm{Pa}$. A part of the hot liquid water flashes to steam. The temperature drops to the boiling point ($T=99.974\ \mathrm{^\circ C}$) at the new pressure. A new equilibrium of liquid water and steam is established at the new temperature and pressure. The corresponding specific enthalpy of the steam is $h_\text{steam} = 2676\ \mathrm{kJ/kg}$; the specific enthalpy of the liquid water is $h_\text{liquid} = 419\ \mathrm{kJ/kg}$.

The flashing process is very quick, hence we may assume that there is no significant heat exchange with the environment; i.e. we may assume the following enthalpy balance in order to estimate the vaporized fraction $x$:

$$\begin{align}
h_0 &amp;= x \cdot h_\text{steam} + (1-x) \cdot h_\text{liquid}\\
x &amp;= \frac{h_0 - h_\text{liquid}}{h_\text{steam} - h_\text{liquid}}\\
x &amp;= \frac{1086\ \mathrm{kJ/kg} - 419\ \mathrm{kJ/kg}}{2676\ \mathrm{kJ/kg} - 419\ \mathrm{kJ/kg}}\\
x &amp;= 0.296
\end{align}$$

This means that approximately $29.6\ \%$ of the superheated water flashes to steam when the container is opened.