What happens if I make a hole in a container with superheated water? Superheated water is achieved by heating water above 100 degree Celsius under high pressure. 

Now, suppose the temperature at which the water is getting heated is around 250 Celsius and sufficient pressure is provided. If, I now make a small-hole in the container, the pressure will decrease obviously, but what will happen to the superheated water?

Will it come out in the form of vapor with lot of energy or some explosion might take place or just about something else?
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.