Sometimes my wife has a pan of water 'boiling furiously'. Is the extra heat (wasted in my opinion) actually making any difference, apart from reducing the amount of water in the pan - which could be done by pouring some away?

  • $\begingroup$ I've deleted some comments which were answering the question and their replies. Remember that comments are meant for suggesting improvements to the post, requesting, clarification, or occasionally linking to related resources. $\endgroup$
    – David Z
    Commented Oct 19, 2017 at 21:21

6 Answers 6


The bottom line is that for all practical purposes turning up the heat won't make any difference. However there are a few subtleties to consider.

Water boils when the chemical potential of the water is equal to the chemical potential of the water vapour at the same temperature. One of the factors that determine the chemical potential of the water vapour is the pressure, or more precisely the partial pressure of the water vapour in the air.

If you take a bowl of water at room temperature and put it in a sealed chamber then the water evaporates until the partial pressure of the water vapour is about 0.03 atmospheres. If you increase the temperature to say 50ºC then the chemical potential of the water increases so the partial pressure of the water vapour rises to match - to about 0.12 atmospheres. As long as the partial pressure of the water vapour can keep rising it will stop the water boiling.

But when heating water in the open the partial pressure of the water vapour can't rise above 1 atmosphere so when the temperature reaches 100ºC the partial pressure of the water vapour can no longer increase to stop the water boiling, and heating the water more vigorously just causes increased evaporation and doesn't raise the temperature. The temperature of the water remains locked at 100ºC.

However this assumes the water and the vapour are in equilibrium, and in real life the water/steam system may be near equilibrium but won't actually be at equilibrium. We inevitably get some superheating and the water temperature can rise slightly above 100ºC. The more vigorously you heat the water the greater will be the degree of superheating and the higher will be the temperature of the water.

So turning up the gas will increase the water temperature slightly above 100ºC. However we are talking about a very small increase - a few degrees at most - and this is unlikely to make any significant difference to the cooking time. Hence my opening comment that for all practical purposes turning up the heat will make no difference.

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    $\begingroup$ Depends on what you mean by "make a difference". The water's temperature won't go up, but it IS taking on more heat, evaporating faster, etc., and these things may have some effect on food. $\endgroup$ Commented Oct 17, 2017 at 22:46
  • $\begingroup$ By raising the partial pressure of water vapor above the hot water you raise it's boiling point. (Of course, the effect is fractions of a degree, but it happens.) $\endgroup$
    – Hot Licks
    Commented Oct 18, 2017 at 12:55
  • $\begingroup$ Unfortunately my wife would not agree with you, and will continue wasting our gas, but that's another story..!! $\endgroup$
    – Tim
    Commented Dec 31, 2023 at 17:20

Water is not going to have a temperature significantly higher than 100° C, even if you top up the gas (or whatever).

But if there is not just water, the final result can be different. For example, in a Pilaf rice (like a Paella) it is very important to time the thing so that the broth is "finished" just when the rice is cooked --- and you do not want to remove it physically because it has to release the flavors to the rice. So the quantity of heat will have an effect here.

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    $\begingroup$ +1 for pointing out the culinary scenario in which controlling the rate of evaporation is important, and not equivalent to just using less water or pouring some out. $\endgroup$ Commented Oct 17, 2017 at 16:30
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    $\begingroup$ So it's more about heat transfer rate than about the total amount of heat delivered to the water, right? $\endgroup$ Commented Oct 18, 2017 at 12:33

Providing heat to boiling water does not increase its temperature. The boiling temperature of water is approximately $100^\circ$C (at sea level and lower at higher altitudes) and the heat added to boiling water (the technical term for this is Latent Heat) will only cause water molecules in liquid phase to turn into gas phase. However, if one wants to keep the pan and the water at $100^\circ$C then one needs to provide heat constantly otherwise the pan and the water would cool down to room temperature. So, if the only reason is to get rid of the water then boiling is waste of energy.

  • $\begingroup$ Assuming it is only water you are throwing away. E.g. pasta water contains starch.. $\endgroup$
    – Communisty
    Commented Oct 17, 2017 at 8:01
  • $\begingroup$ So, could I assume that simmering, where the water is kept at 100C is not going to differ from 'boiling furiously' as far as whatever is in the pan, apart from, obviously, the water, is concerned. Basically, would just keeping the water boiling be as effective? $\endgroup$
    – Tim
    Commented Oct 17, 2017 at 8:34
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    $\begingroup$ as soon as the water is boiling it will is at 100C, irrespective of boiling being furious or calm. Being furious or calm will only effect the time of total water fully evaporate which in turn cooking time. I know some people cook rice with a lot of water and boil it furiously or some people cook with less water but boil it calmly. $\endgroup$ Commented Oct 17, 2017 at 8:53
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    $\begingroup$ @physicopath: fuŝaĵo: ‘effect’ estu ‘affect’ $\endgroup$
    – Mike Jones
    Commented Oct 20, 2017 at 14:29

I would say you could in fact decrease the cooking time on an item with non-negligible mass (get to that later) somewhat significantly by turning up the stove. When water starts boiling, it starts moving relative to the cooked item, meaning the heat transfer from the water to the item can increase dramatically (forced convection versus natural convection/conduction) with the same temperature difference between the water and the item. When you turn up the heat, more water evaporates and therefore the speed of water movement increases, leading to more convection and more heat transfer to the item, despite the fact that the water is very nearly the same temperature.

This assumes the item does not quickly reach the temperature of the water (at which point heat transfer effectively ceases). Spaghetti might do that in seconds, but if you were going to boil, say, a chicken, you could substantially reduce the time for the center to reach the desired temperature.

This is analagous to increasing the speed of a fan that's using room temperature air to cool a hot chunk of metal: the ambient temperature stays the same but the heat transfer increases causing the temperature of the metal to drop faster.

  • $\begingroup$ "especially since water is nearly incompressible and very little natural convection goes on" I'm not sure how much that has to do with it. Water still changes density significantly in response to temperature; so convection is still driven by the process. AFAIK the natural convection in a pot of water is still pretty decent before the boiling sets in. $\endgroup$
    – JMac
    Commented Oct 17, 2017 at 19:31

Since most of the water is being cooled by the environment, 100ºC water will only be in the bottom of the pan. Increasing the heat will actually make a difference, since bottom water will boil faster and it will transfer more heat to top cool water, before being cooled by ambient temperature.

  • $\begingroup$ when the water is boiling the bubbles are formed at the bottom of the pan travelling up to the air, that results a constant stirring and mixing. Do you think in a small amount of water such as in a pan where there is a constant mixing you would have such a high temperature gradient? $\endgroup$ Commented Oct 17, 2017 at 9:34
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    $\begingroup$ I actually don't know if the temperature gradient will be high or low, but it seems to me that increasing the amount of bubbles in the bottom of the pan will definitely increase the temperature of the top water (eventhough this increment can be minimal). $\endgroup$
    – Guille
    Commented Oct 17, 2017 at 9:58
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    $\begingroup$ Other way around mostly. At the water surface, the water is boiling at 1 atmosphere. That's where the temperature (by definition) is 100C. Below the surface, the pressure is slightly higher due to the weight of the water above. But since the depth will be limited to a few centimeters, you won't get a pressure much above 1 Atm andtherefore not much of a rise in temperature. And that's ignoring the pressure variations due to turbulence. $\endgroup$
    – MSalters
    Commented Oct 17, 2017 at 10:19

Is the extra heat (wasted in my opinion) actually making any difference,

It is if you put a lid on the pot to retain the heat.

apart from reducing the amount of water in the pan

"all" the vigorous boiling"

  • agitates the water (which can be important for cooking)
  • indicate a hot burner, which will quickly bring the water back up to temperature if you add something cooler.

Bottom line: your Significant Other might be wasting energy, or she might be doing the right thing for that particular recipe.

  • $\begingroup$ Actually it indicates a hot pot bottom, which is often caused by a hot burner (but not always; see induction stoves) $\endgroup$
    – Yakk
    Commented Oct 17, 2017 at 19:32
  • $\begingroup$ So what's 'SO'? $\endgroup$
    – Tim
    Commented Oct 18, 2017 at 6:29
  • $\begingroup$ @Tim "significant other". $\endgroup$
    – RonJohn
    Commented Oct 18, 2017 at 7:24
  • $\begingroup$ So, putting a lid on will retain the heat. Will that increase the temperature? I know it'll stop a lot evaporating. $\endgroup$
    – Tim
    Commented Mar 12, 2019 at 11:31
  • $\begingroup$ @Tim it'll -- up to a point -- increase the temperature of the system that is "covered pan full of water". After that point, though, the stove will heat it as much as it can, while steam escapes from lip of the rim and heat from the sides of the pan. $\endgroup$
    – RonJohn
    Commented Mar 12, 2019 at 13:28

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