I've got a question about the boiling of water. I'm a first year physics student and from the Netherlands.

I've searched alot about the boiling of water and this confused me. Everyone said something else about the cause of the boiling. Let me explain it further.

Let's say you want to cook some eggs. You put on the gas. 1. As the temperature of the water increases, the evaporation increases. 2. When the evaporation increases, the vapor pressure will increase too. When the vapor pressure is equal to the external pressure, there will form a bubble.

So, my question is: is this a chain/link of causes? So the first link causes the next one? So the temperature increase causes the evaporation to increase which causes the vapor pressure to increase which causes the forming of a bubble (the actual boiling)?

I doubt if it is a link of causes (the one thing causes the other) because they happen at the same time. And in my opinion a cause happens BEFORE the consequence.

https://www.youtube.com/watch?v=CfagHzOtIDM In this video they say boiling has more causes. What are these causes?

When the boiling point is reached, there happen two things: - 100 degree celcius - vapor pressure that equals the external pressure

Are they both causes of the boiling (coming up of the bubbles)? Or is the vapor pressure that equals the external pressure the cause?

  • $\begingroup$ You're boiling the water using an electric plate? So the source of heat energy is restricted to the bottom of your pot? That induces some temperature gradient that may well explain what you are seeing. Put the pot into an oven where it is heated from all sides almost equally to see the difference. $\endgroup$ Jan 14, 2016 at 19:01
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    $\begingroup$ No, I was looking for the actual cause of the boiling of water. Is it the temperature or the vapor pressure that equals the external pressure? Or are they both causes? $\endgroup$ Jan 14, 2016 at 19:09
  • $\begingroup$ What exactly do you mean by "boiling"? Forming bubbles, for one, or, more generally, changing state from liquid to gas? Note that the liquid/gas transition is not only governed by temperature alone, but also by pressure. At see level, water boils at 100°C, but on mount Everest 60°C is enough. On Mars, 40°C are enough (don't start a law suit if the actual figures are somewhat different). Keyword is en.wikipedia.org/wiki/Phase_diagram $\endgroup$ Jan 14, 2016 at 19:16
  • $\begingroup$ Yes, I mean the formation or nucleation of bubbles. Do you say that vapor pressure is the true cause of the formation of bubbles? Or the combination of temperature (depending on the location) AND vapor pressure? Can i say that the temperature (100 or i.e. 60 at mount everest) causes the vapor pressure to equal the external pressure? Or isnt it a cause because they happen at the same time? $\endgroup$ Jan 14, 2016 at 19:24
  • $\begingroup$ I have to admit that I don't have a complete, satisfying understanding of that. But I am good enough to throw some buzzwords at you ;-) en.wikipedia.org/wiki/Superheating is one of it. $\endgroup$ Jan 14, 2016 at 19:29

5 Answers 5


When the vapor pressure is equal to the external pressure, there will form a bubble.

Not true. Instead, when the vapor pressure is equal to the external pressure, then any existing bubbles will begin growing continuously.

And, if no bubbles are already present, then the water will superheat far above the boiling temperature, yet no bubbles will appear. For vapor pressure to exist within water, first gas pockets must exist within water. We need gas-fluid interfaces. Without these, all boiling takes place silently, at the surface where water touches air.

Beware of common misconceptions. Boiling-bubbles are typically not seeded by dirt or contamination. ("Mythbusters" show got it wrong![1]) Instead, boiling is seeded by existing micro-bubbles trapped in small crevices. So, yes, your coffee heated in a microwave oven can superheat and explode, even though coffee is very impure water. If the surface of your ceramic mug lacks air-filled micro-scratches, the coffee will not boil until its temperature is raised far above 100C. But liquids may refuse to boil even when up against a very rough surface, if that surface has been previously wetted with water over 100C. The hot water fills the microscopic roughness with steam, which then condenses, removing any air pockets that let the rough surface act like a "seed" for roiling boil. A fully-wetted rough surface won't prevent superheating.

On a typical stove with a metal pot, the metal bottom will be heated far above 100C, even though the water has not yet approached 100C. The metal surface will be covered with spontaneous steam pockets, but these bubbles cannot grow, since they're right against cooler, under-100C water. When the main volume of water reaches 100C, seed-bubbles are already present on the hot metal, so the pot will immediately begin a visible boil. But without this large temperature excursion at the metal surface, visible boiling may not commence.

Often with new glass cookware, (with no scratches,) and with water heated on a gas stove (with no tiny hot-spots,) the water won't boil. Instead it superheats far above 100C, then unexpectedly produces a few spontaneous micro-bubbles, and exhibits the boiling-explosions called "bumping." The explosions may splash boiling water out of the container. Sometimes they're violent enough to shatter glass. To prevent this, use 'boiling stones' sold in laboratory supply catalogs. Or with microwaved coffee, provide a dry, air-filled wooden stir-stick. These cause a roiling boil at 100C, which cools the bulk liquid and halts superheating.

The 'boiling stones' commonly used to prevent bumping-explosions in laboratory glassware won't work if cooled and then quickly re-used. This happens because all their internal crevices become filled with water. No small bubbles, no "boiling seeds." Restore your 'boiling stones' before re-use, by allowing them to dry thoroughly so the air again fills all of their internal pockets. Or, use a 150C drying oven to rapidly boil off their trapped water. Similarly, a sodden, well-boiled wooden stick in your microwave coffee may stop working. So turn it over and use the dry end!

[1]Big caveat: if micro-bubbles aren't present anywhere, then superheated water will continue to rise in temperature, finally boiling spontaneously with explosive violence. The "seed" or "trigger" for this boiling can be: intrusion of a dry object, vibration and sound waves, thermal fluctuations, ionization from background radiation, and yes, suspended particles, surface roughness, and contamination. Mythbusters(tm) did have it right, but only regarding extremely superheated water. When we first eliminate all surface micro-bubbles and then raise the temperature far past boiling, eventually something will trigger a steam-explosion. The center of this explosive vapor-production might be a tiny grain of dirt, or a pencil tapping against the container, or even a cosmic-ray strike.

  • $\begingroup$ wbeaty is right, and if you are designing thermal inkjet printheads and get the nanoscale surface texture of the microscopic heaters wrong, the bubble nucleation process (which propels the ink droplets out the nozzles in such a device) will occur at the wrong superheat temperature and the ejected droplets of ink will have the wrong size and velocity. $\endgroup$ Nov 11, 2017 at 7:01
  • $\begingroup$ @wbeaty can u please explain to a high school student{which is me} how on adding "ideal solute" would decrease this inside bubble vapor pressure and in turn would increase the boiling point? I used to think it somehow attracts water molecules more strongly than the intermolecular force between water molecules themselves but then I thought if its ideal solute and it won't change the bonding so it shouldn't change the vapor pressure so what's going on? I know it would change the concentration but how "inside bubble pressure depends on the concentration"? $\endgroup$ Jul 19, 2022 at 6:41
  • $\begingroup$ also the inside bubble pressure is due to both vapors and air right? so I think the pressure inside the bubbles would be equal to 1atm even when the temperature is not 100 degrees,,,,because air was already there?? also why at boiling the the temperature of water becomes constant? how this relation that "vapor pressure=external pressure" indicates that the temperature would become constant at boiling point{i.e when the bubble begins to grow}? $\endgroup$ Jul 19, 2022 at 6:46
  • $\begingroup$ I think it should increase continously and produce mor and more inside bubble pressure?? some says at this temperature water molecules changes move apart and changes state ,,,so if that's the case why not all water molecules becomes vapor simultaneously at 100 degrees? and also why they even become vapors at other temperatures when there temeprature is not even 100 degrees? $\endgroup$ Jul 19, 2022 at 6:54
  • $\begingroup$ 3) why they don't change into vapors and form inside bubbles when there are no nucleation sites,,,I mean at 100 degrres water molecules should still separate and form bubbles right? $\endgroup$ Jul 19, 2022 at 7:00

While the answer of wbeaty is very interesting in showing points relevant in practice, I think all the answers are still missing an important and simple theoretical point, which you should consider to understand the process.

vapour pressure does mean two different things as used above. First, the pressure, the existing water vapour would have (if it were alone) - let's call it the partial pressure of the water. And secondly, the maximal possible vapour pressure (a function of temperatue) - let's call it the saturation pressure.

The latter is easy to understand qualitatively:
If you have a surface between water and vapour, you have two random processes: particles hop out of the surface into the gas, and particles from the gas get caught in the water. After some time, there will be equillibrium of these processes. The first process depends mainly on temperature (how fast are the water particles), the second mainly on the pressure (how often do vapour particles hit the surface).
There has thus to be some function $p(T)$ or $T(p)$ which describes these two at equillibrium. This is the abovementioned saturation pressure.

Over every water surface there is some vapour, and after a while it has the saturation pressure - if nothing moves to blow it away. If you blow on your soup to cool it, that's the point: to remove vapour, reduce its partial pressure, and thus enable the water to evaporate further.
Note: the net pressure of the air is always atmospheric, you just change the part of it produced by water and the other part produced by the rest gases.

Now, what happens if the saturation pressure gets greater than the atmospheric?
There is no need to blow anymore! The vapour can just push the atmosphere away, it doesn't have to wait, it just takes the place it need to evaporate into. That's boiling.

PS: concerning the concept of partial pressure.
The necessary asumption for this concept is, that we deal with ideal gases. The particles do not interact, they just bump against the walls and thus produce the pressure. The net pressure is just the sum of the pressures each part of the gas produces. I.e., each part of the gas would have if it were alone at that temperature in that volume.

For an ideal gas you know: $pV=nRT$
So since $p$ is proportional to $n$ it means, that $n_1$ of oxygen and $n_2$ of nitrogen will produce the sum of the pressures each part does by itself.
Mark: this is not self-evident, it's a concequence of the proportionality between $p$ and $n$.


It's the vapor pressure equaling the external pressure that is the cause, and not the temperature (except to the extent that it determines the vapor pressure). When the vapor pressure is equal to the external pressure, bubbles can begin forming under the liquid, and the atmosphere can be pushed back at the upper surface of the liquid to accommodate the additional volume of the bubbles within the liquid. The heating and temperature changes that take place before the temperature reaches the required value (for the vapor pressure to equal the external pressure) does not matter.

  • $\begingroup$ Thanks for the reply. You say the external pressure is the cause. Does this mean that first the temperature reaches the required value (boiling point) and later the vapor pressure equals the external pressure. Is this right? So the temperature is the cause of the vapor pressure to equal the external pressure, the external pressure is the cause of the forming of bubbles. Is the reaching of the temperature level the exact moment that vapor pressure is at the required value? Of is it that first the temperature reaches the value, and later the vapor pressure reaches the value? $\endgroup$ Jan 15, 2016 at 6:25
  • $\begingroup$ The temperature and vapor pressure reach the required value at the same time. Before this, the temperature is lower, and the vapor pressure is not high enough to match the external pressure. The vapor pressure is a unique function of temperature, and so the temperature and vapor pressure vary in tandem. $\endgroup$ Jan 15, 2016 at 12:13
  • $\begingroup$ Okay, that's clear. But why isn't the temperature a cause too then? You say they vary in tandem. Aren't they both causes then? Or is the temperature the cause of the vapor pressure, and the vapor pressure the cause of the bubbles to form? $\endgroup$ Jan 15, 2016 at 17:13
  • $\begingroup$ The latter is the correct mechanism. $\endgroup$ Jan 15, 2016 at 17:39
  • $\begingroup$ Okay, thanks. Do the water vapor's leave the water to join the bubble before the reaching of 100C? Because the vapor pressure is reached at 100C at the exact same time, right? So the joining of the bubble (increasing of the vapor pressure) has to occur before the 100C. Because if they would leave the bulk AT the time of reaching 100C, the vapor pressure equalling the external pressure would occur a bit later than reaching 100C. When the temperature is below 100C, the bubbles CAN grow but they will collapse because of the external pressure, right? $\endgroup$ Jan 15, 2016 at 21:48

It seems like you are complicating the process a bit too much. As noted in the earlier response: boiling occurs when the vapor pressure of the liquid reaches the surrounding pressure (e.g. of atmosphere or atmosphere + pressure of overlying liquid (heating usually done from bottom)). Evaporation, in the context of initiation of boiling, serves only to carry heat away and reduce the temperature rate of rise ("a watched pot never boils"). You can attain boiling in two ways: heat the liquid or reduce the overlying pressure. Put water into a sealed container and draw a vacuum on it, and it will boil at room temperature... Water at room temperature has a vapor pressure (a few mm Hg pressure) - by heating it you are only raising it's vapor pressure until it reaches the atmospheric pressure (at your current elevation - at higher elevation, water boils at a lower temperature). Once boiling is achieved, the temperature will not increase further - because you are removing heat through evaporation as fast as you are putting heat into the liquid from the external heat source.

There is one other added complexity as well: in order for a vapor bubble to form, the vapor pressure must overcome the surface tension of the water in order to make a "hole" in the water. When the water is very pure, and the container surface is very smooth, it can be difficult for vapor bubbles to form and the water can become superheated - this is why, in chemistry lab, boiling chips are placed in beakers - the ceramic chips provide nucleation points for vapor bubbles to form and avoid superheating the water.


Boiling, liquid starts to vapourize which means sufficient heat is provided break the inter-molecular bonds. the temperature at which the inter molecular bond breaks will be called as boiling point of that liquid.Breaking of the bonds results in the formation of bubbles i.e liquid is broken into gases.


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