Let's suppose i have a strong metal container 1 x 1 x 1 cm filled with water at room temperature (25 c). How much pressure would i need to apply from one side of the box container to raise temperature of contained water to > 100 degrees? And how much energy i would get from steam when i release the pressure and let water insta-boil?

I'm trying to figure out what it would take to make water steam-explode using just pressure.


2 Answers 2


The specific heat of water is S = 4.2 MJ/m3/K. The bulk modulus of water is M = 2.2 GPa. Equating the work done on the water with the heat increase in the water: (1/2M)P2 = S.(100-25) giving pressure P = sqrt(2.S.M.(100-25)) = 1.2 GPa !

This is a ridiculous answer suggesting that the water would have to be compressed by a factor of two to reach 100 centigrade. So it's well ouside any linear regime where the result might be remotely credible.

Even for much smaller compressions and smaller temperatures it's unclear how much of that temperature rise would be maintained when the pressure was released. Much of the energy might go into work against whatever mechanism releases the pressure.

I don't have a better answer, but I'm afraid it's all a bit discouraging for your water-steam explosion :-(

  • $\begingroup$ Thanks. Yeah, pressure of several GPa seems a bit high :) Would it be a nice greenenergy source if principle would work? lol $\endgroup$
    – guest86
    Commented Apr 5, 2021 at 9:20

We can estimate the temperature increase upon rapid compression from

$$\left(\frac{\partial T}{\partial P}\right)_S,$$

where $T$ is temperature, $P$ is pressure, and $S$ is entropy. This term represents the temperature rise per unit applied pressure at constant entropy (which idealizes the case of rapid pressurization, with no heat loss to the surroundings).

From the triple product rule,

$$\left(\frac{\partial T}{\partial P}\right)_S=-\left(\frac{\partial T}{\partial S}\right)_P\left(\frac{\partial S}{\partial P}\right)_T.$$

We recognize the first partial derivative on the right-hand side as $T/mc_P$, where $m$ is the mass and $c_P$ is the constant-pressure heat capacity. The second partial derivative on the right-hand side can be shown using a Maxwell relation to be

$$\left(\frac{\partial S}{\partial P}\right)_T=-\left(\frac{\partial V}{\partial T}\right)_P$$

where $V$ is volume. We recognize this to be $V\alpha_V$, where $\alpha_V$ is the volumetric thermal expansion coefficient at constant pressure.


$$\left(\frac{\partial T}{\partial P}\right)_S=\frac{\alpha_V T}{\rho c_P};$$

$$dP=\frac{\rho c_P}{\alpha_V T}dT,$$

where $\rho$ is density. You can numerically integrate this using your favored functions for the temperature dependence of the material properties. Assuming constant material properties as a first pass, I estimate a required pressure of 4 GPa.


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