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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?

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    $\begingroup$ Be very certain, that if you have superheated water with a lot of pressure and you make a hole in the container, this is very dangerous. Please be careful. $\endgroup$
    – Martin - マーチン ♦
    Commented May 25, 2015 at 8:29
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    $\begingroup$ I know that, that's why I'm asking that what will exactly happen? Some websites (i guess it was wikipedia) say that superheated water is very stable and won't show any explosive sort of behaviors unless you shake it or add something like sugar. But I doubt that previous statement is correct. Anyways, would be great if someone could tell me what exactly would happen. Thanks anyway. :-) Edit: Thanks for your concern and I can assure you that I won't do that unless I get some expert advice. :) $\endgroup$
    – Anoneemus
    Commented May 25, 2015 at 8:34
  • $\begingroup$ "Superheated" doesn't mean the liquid is under sufficient pressure to keep it from boiling. It means that the liquid should be boiling according to the charts, but it is not. It's a metastable state, that can occur if the container is very clean, and the liquid is free of any solid particles. youtube.com/watch?v=SC_NtH8vWSc $\endgroup$
    – Solomon Slow
    Commented Nov 29, 2015 at 21:48

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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 $x$:

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

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

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    $\begingroup$ I would not be inclined to stand anywhere near that. $\endgroup$
    – user15489
    Commented May 25, 2015 at 11:30
  • $\begingroup$ One thing that i guess you understand wrong i.e. I'm not going to open the whole container. Instead I'll be making a tiny hole(say 0.5 cm in diameter) or opening this tiny hole. So, will the water vapor still come out at same pressure? If yes then how much. NOTE: the water is still getting heated, even after opening this tiny hole. You can ask me for more clarification. And thanks :) $\endgroup$
    – Anoneemus
    Commented May 25, 2015 at 14:01
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    $\begingroup$ @Dahano If the opening is small, the flow rate of the escaping fluid (liquid water and steam) is limited. For the given example, a simple opening ($\zeta=1$) with a diameter of 0.5 cm causes an initial mass flow rate of about 0.3 kg/s. Nevertheless, the initial and the final state (temperature and pressure) are not affected by the size of the opening. However, the enthalpy balance is a simplified approximation since it ignores the work that is required to expand the steam against the ambient pressure. $\endgroup$
    – user7951
    Commented May 25, 2015 at 15:43
  • $\begingroup$ @Loong From my understanding, will something like this would happen --> The water say, 15litre, is superheated @ 250C and now I open the tiny hole; the water will come out in form of vapor with lot of energy, which would be enough to push something (and eventually 3rd law of newton)~ Will something like this happen? If yes, then how much push that energy could create(in kg) and how long would this last? PS: I know I've been asking a lot of questions but it would be seriously great if you can answer them all. Thanks a ton! :-) $\endgroup$
    – Anoneemus
    Commented May 25, 2015 at 17:09
  • $\begingroup$ @Dahano In principle, your idea is correct. A volume of $V_0=15\ \mathrm{l}$ of liquid water at a temperature of $T_0=250\ \mathrm{^\circ C}$ and a pressure of $p_0=39.2\ \mathrm{atm}=4.0\ \mathrm{MPa}$ corresponds to a mass of $m_0=11.98\ \mathrm{kg}$. After relaxation to atmospheric pressure, we have a mixture of $V_\text{liquid}=8.8\ \mathrm{l}$ liquid water and $V_\text{steam}=5925\ \mathrm{l}$ steam. Therefore, a large fraction of this mixture is pushed through the opening out of the small container. The container and its remaining content are pushed into the opposite direction. $\endgroup$
    – user7951
    Commented May 25, 2015 at 20:26

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