# Is there a chemical that when added to water makes the water flash evaporate?

Im wondering if there is a chemical that you can add to water that will make it evaporate extremely fast, either by raising the temperature of the mix or some other method.

• – Loong Sep 14 '16 at 19:35

Yes, the idea is to add a solvent that forms a lower boiling point azeotrope with water.

See "section 8.2 Azeotropic Drying" in The Synthetic Organic Chemist's Companion for specific suggestions such as benzene.

In principle, the evaporation of water is a purely physical process. In order to evaporate water, the considered water sample has to be heated to its boiling temperature at the given pressure, and then further heat must be supplied, which corresponds to the enthalpy of vaporization.

For example, the specific enthalpy of liquid water at an initial temperature of $T_0=25\ \mathrm{^\circ C}$ and a pressure of $p_0=1\ \mathrm{bar}$ is $h_0=104.9\ \mathrm{kJ\ kg^{-1}}$. The boiling temperature of water at this pressure is $T_\mathrm b=99.6\ \mathrm{^\circ C}$. The specific enthalpy of liquid water at this boiling point is $h_\text{liquid}=417.5\ \mathrm{kJ\ kg^{-1}}$; the specific enthalpy of steam is $h_\text{steam}=2674.9\ \mathrm{kJ\ kg^{-1}}$, which corresponds to a specific enthalpy of vaporization of $\Delta_\text{vap}h=2257.4\ \mathrm{kJ\ kg^{-1}}$. Therefore, in order to evaporate the water, the specific enthalpy of the cold water has to be increased by \begin{align} \Delta h&=h_\text{steam}-h_0\\[6pt] &=2674.9\ \mathrm{kJ\ kg^{-1}}-104.9\ \mathrm{kJ\ kg^{-1}}\\[6pt] &=2570.0\ \mathrm{kJ\ kg^{-1}} \end{align} The enthalpy of the water can by increased by any kind of heat transfer. The heat may be supplied by a purely physical heat source (e.g. a sufficient amount of a red-hot substance that is dropped into the water); the heat may also be supplied by an exothermic chemical reaction or any other exothermic process (e.g. nuclear fission), provided that the transferred heat is sufficient for the evaporation of the considered amount of water.

Anyway, even if the heating process may be very fast, this is not flash evaporation. Flash evaporation occurs when a superheated liquid undergoes a quick reduction in pressure (e.g. when the liquid is passing through a small opening, or when the pressure vessel suddenly fails).

For example, water at an initial pressure of $p_0=150\ \mathrm{bar}$ and temperature of $T_0=300\ \mathrm{^\circ C}$ is liquid. Liquid water at this temperature and pressure has a specific enthalpy of $h_0=1338.3\ \mathrm{kJ/kg}$. When the pressure vessel suddenly breaks, the pressure drops to the ambient pressure of $p=1\ \mathrm{bar}$. A part of the hot liquid water flashes to steam. The temperature drops to the boiling point ($T=99.6\ \mathrm{^\circ C}$) 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 $h_\text{steam}=2674.9\ \mathrm{kJ\ kg^{-1}}$; the specific enthalpy of the liquid water is $h_\text{liquid}=417.5\ \mathrm{kJ/kg}$. Since the flashing process is very quick, there is no significant heat exchange with the environment. Therefore, the vaporized fraction $x$ may be estimated according to the following enthalpy balance:

\begin{align} h_0&=x\cdot h_\text{steam}+(1-x)\cdot h_\text{liquid}\\[6pt] x&=\frac{h_0-h_\text{liquid}}{h_\text{steam}-h_\text{liquid}}\\[6pt] &=\frac{1338.3\ \mathrm{kJ/kg}-417.5\ \mathrm{kJ/kg}}{2674.9\ \mathrm{kJ/kg}-417.5\ \mathrm{kJ/kg}}\\[6pt] &=0.408 \end{align}

This means that approximately $40.8\ \%$ of the superheated water flashes to steam when the pressure vessel suddenly breaks.