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When I am heating water on a gas stove, it begins to boil after some time and bubbles of air can be seen escaping out.

However, as soon as I increase the amount of heat in the stove, the rate of escape of air bubbles increases immediately, and as soon as I turn off the stove, the air bubbles stop coming out right then.

In this case, I am boiling, water in a steel utensil. Steel is a good conductor of heat. Then why does this change take place so quickly with the change in heat?

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    $\begingroup$ That's likely not air but steam $\endgroup$ – Thomas Weller Jan 18 '18 at 16:08
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    $\begingroup$ I's surprised by this question. Did you expect the water would somehow keep boiling after you turn off the heat? $\endgroup$ – Dmitry Grigoryev Jan 18 '18 at 20:48
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    $\begingroup$ @DmitryGrigoryev he didn't think it would continue boiling (indefinitely), rather that it wouldn't stop instantly (which it does). $\endgroup$ – WELZ Jan 18 '18 at 22:18
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    $\begingroup$ @WELZ It doesn't stop immediately it stops very fast. It depends largely on how thick the bottom of your pan is. $\endgroup$ – Pieter B Jan 19 '18 at 11:37
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    $\begingroup$ The key to this question may be gas stove (or induction heated stove) as opposed to a closed fire range, electrical hotplate, spiral element, ceramic (radiant) heater types of cooker. The gas flame has essentially no thermal mass when off and so only the heat stored in the pot needs to be used to make steam before boiling ceases and this happens fast as described. $\endgroup$ – KalleMP Jan 21 '18 at 12:18
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In large part because under normal circumstances water doesn't get hotter than boiling - at that point it becomes steam, as you know. You can add heat and boil it away faster, but the water can only get so hot. When you remove the source of heat the water will quickly drop below this threshold. You're right on the knife edge of temperature.

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    $\begingroup$ Key here is that the water doesn't actually need to cool at all. Water at the boiling point won't spontaneously boil. Any heat added to water already at the boiling point, however, will directly be absorbed by converting some of the water at 100C to steam at 100C (a process that takes a considerable amount of energy). When you stop adding heat, the boiling stops. $\endgroup$ – J... Jan 18 '18 at 23:05
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    $\begingroup$ @J... that is called specific heat capacity , if i am not mistaken. $\endgroup$ – roottraveller Jan 19 '18 at 18:15
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    $\begingroup$ @roottraveller That is the amount of heat required per unit (mass/volume/etc) to change the temperature of a material. For boiling the quantity is the latent heat of vaporization - in this case the heat is absorbed but does not change the temperature of the material, it is entirely consumed to change the state of the material. $\endgroup$ – J... Jan 19 '18 at 18:19
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    $\begingroup$ @J... I think it's valid to say there is some cooling, just very little. Given that the heat is coming from the bottom, there is going to be some temperature gradient, even if it's mostly well mixed by convection currents. Also, the water does need to cool the steel pot down, which can rise above 100C on the lower side of the pot. These effects are very minor, but I always get a little itchy when instantaneous transitions start getting mentioned. There's always some dynamic to pay attention to. $\endgroup$ – Cort Ammon Jan 20 '18 at 0:17
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    $\begingroup$ @CortAmmon Yes, but the point is that there needn't be any cooling. If you could control the water to remove all temperature gradients and you could control evaporation, convection, and conduction so that no heat was lost after the burner was turned off you would still notice that boiling would stop immediately. $\endgroup$ – J... Jan 20 '18 at 0:30
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Because the water — and therefore the steel — is not hotter than 100°C

Assuming normal pressure (1 atmosphere), water boils at 100°C. The water cannot become hotter than that because then it turns to steam. Boiling water circulates very effectively. As soon as a steam bubble is formed, it rises, being replaced with liquid water.

This makes it nigh impossible for the steel that is in contact with the water to become hotter than 100 degrees. Even if the steel below it is hotter than that, in the interface between steel and water, the temperature will be very close to 100 degrees.

So because the water is never hotter than 100°C, and because the water very effectively soaks heat from the steel, and because the steel is a good conductor of heat, this combined means the entire steel bottom of the pot never becomes much hotter than 100°C. The water forces the steel to never be much hotter than 100°C.

Side note 1: This is why water cooling works so well, because as long as the water is liquid, you have a max temperature on that is being cooled that is close to 100°C.

Side note 2: this in turn is why you can boil water in a paper or plastic cup over an open flame; the water will force the paper/plastic to not become so hot that it chars/melts.

This means that as soon as you remove the heat source (the burning gas in your case) from the steel, the heat soak that the water does very quickly lowers the temperature of the steel to below 100°C.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Jan 19 '18 at 21:13
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The main reason is that boiling requires rather enormous amounts of energy.1,2 How many minutes does it take to bring, say, a quart or a liter of water to the boil (5 minutes?), compared to actually evaporating it by boiling (30 minutes?). That ratio is a good estimate for the energy it takes for water to evaporate. The energy for evaporation is drawn from the heat in the environment, which is frequently used when cooling is needed in technical and everyday applications.

So the order of events is:

  • The constant influx of heat from the flame keeps the water boiling.

  • The heat influx stops because you turn off the stove.

  • For a small moment there is enough heat in the vessel to boil some more water, which cools down the water proper and the vessel.

  • Because of the energy used for evaporation the water temperature drops below the boiling point, and the water stops boiling. This happens fast because the heat energy stored in the water and the vessel is small compared to the energy needed to evaporate water.

One could perhaps add that water can evaporate very quickly and basically uses the available energy "immediately". The physics are quite interesting because the limiting factor is how fast you can transport the energy into the water. The heat transport from a solid surface (read: metal pot) to liquid water is very good, partly because of the water convection. But when it begins to boil the vapor gets in the way which conducts heat comparatively badly, which leads to the funny effect that there is an optimum temperature for a heating surface above which the heating actually slows down until energy transfer by radiation takes over.

I suspect that one could evaporate a smallish amount of water "instantly" by radiating it with intense microwaves which transport the energy right where it's needed, circumvening, so to speak, all the heat transport issues.


1 Evaporation is a phase transition from the fluid phase ("water") to the gaseous phase ("vapor" or "steam", but be careful with everyday words — water vapor is invisible!). Most phase transitions involve large energy flows. That's the reason ice is an excellent coolant at 0 C (e.g. in your cooler). It's not only the temperature of the coolant, it is the energy needed for the imminent phase transition.

2 This energy is called the "(latent) evaporation heat" or enthalpy. For water it is about 2.2 Kilojoule per gram, if we can trust wikipedia, which would make it 2.2 Megajoule per liter, a.k.a 2.2 Megawattseconds (don't you love metric units?). For Watthours instead of -seconds we divide by the 3600 seconds in an hour to get 611 Watthours, which meets my guess of 30 minutes with a 1 kW cooker pretty nicely :-).

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    $\begingroup$ But a pan full of syrup would take longer to boil, and once off the heat would continue boiling for some little time - as any toffee maker can tell you. $\endgroup$ – CriglCragl Jan 19 '18 at 10:30
  • $\begingroup$ @ColinBennett Why does that happen? Difference in density? $\endgroup$ – Mrigank Pawagi Jan 19 '18 at 10:50
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    $\begingroup$ @MrigankPawagi The syrup transports the heat slower because its higher viscosity slows the convection. The same amount of heat stored in the pan will therefore continue to heat the adjacent syrup (but just the adjacent syrup which cannot transport the heat away as effectively as water can) and keep it boiling for a longer time. Additionally the pan will tend to be hotter because the syrup does not cool as well as water, so there is more energy available after the stove is turned off. $\endgroup$ – Peter A. Schneider Jan 19 '18 at 13:09
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    $\begingroup$ I first thought you made a giant mistake when comparing your original estimate to the calculation in the footnote. But after some research it turned out that a quart is about a litre, Darn, you should have started with SI units right away ... $\endgroup$ – Hagen von Eitzen Jan 20 '18 at 23:04
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Because at room temperature and pressure, the water never exceeds 100C. When you turn the heat higher, more energy is going into vaporising into steam, that is adding the latent heat of vaporisation. The body of the liquid stays the same temperature, only just hot enough to boil.

There is the latent heat in the pan, it's still hot. But exactly because it's a good conductor, that is very rapidly dissipated. There will be a very small lag on a high boil, especially with a heavy bottomed pan. Might be hard to observe though.

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    $\begingroup$ OTOH, if you have an electric stove, the water will continue to boil for quite a few seconds after you turn the heat off, because the red-hot heating element takes time to cool, and so is still transferring heat to the water. $\endgroup$ – jamesqf Jan 18 '18 at 18:41
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Yes the posters who point out the enormous energy of "latent heat of vaporization" ae on the right track.

In detail, what you've got is a pool of water molecules linked to each other by hydrogen bonds. In addition, you have a "skin" to the water where molecules are only "glued' to other molecules of H2O in chains below them. (surface tension)

Finally, in boiling, you have to break all these bonds and launch a single molecule into space as steam.

To further address your question, the rapidity of winding down this process could be pockets of steam down near the heating surface on the stove. Invisible pressure waves of activated water convey the heat to the surface to escape. Don't forget water has a great deal of structure; this accounts for its high boiling point. So when you turn off the heat, the quasi-mechanical "push" from below goes away, the water molecules link up in longer chains, and the entire structure "snaps back" to a regular liquid. Another factor acting here, on a more macroscopic scale, is that bubbles of steam down near the heating surface have to "push" the water out of the way to reach the surface. As soon as you take off the heat, these collapse, further draining enegy from the system. The rapidity of the change may happen because of a rapid inverse chain reaction of excited water recombining into the usual chains (10-20 long)

Its actually a subtle question and an interesting one.

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At 1 atmosphere of pressure the maximum temperature of liquid water is 100 °C. As you add heat the water just evaporates and stays at 100 °C.

J is Joule which is heat.

It takes 4.187 kJ/kg to increase water temperature 1 °C.

It takes 2030 kJ/kg to vaporize. It takes more heat to vaporize water than to heat it from 0 °C to 100 °C.

At 100 °C if you add more heat then it just vaporizes (boils) faster.

When you remove the heat source then water stops boiling immediately.

There will be very small amount of heat that has not yet transferred from the pan to the water but that will be almost immediate (a fraction of a second).

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protected by Qmechanic Jan 18 '18 at 18:28

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