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I'm debating about this problem with my friends (they and I majored physics. But I think it's not a trivial question.)

The problem is :
The water in the pot is boiling by the gas stove, and there is a steel ladle in it. When the temperature of the water reaches 100°C, can the temperature of the ladle be higher than 100°C?

At this time, it is assumed that the ladle does not come into contact with the pot but only comes into contact with boiling water. Of course, put aside thermal energy of the pot, because this is not problem about the pot.

The clear thing is that, the entire system is opened system, so the situation is not thermal equilibrium, between the boiling water and the ladle. But after that, I can't explain and analyze this problem.

Can someone explain it to me? and give me an answer?

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    $\begingroup$ @StevanV.Saban The temperature of steam and boiling water are the same at 1 atm, thus on average it will have the same temperature as boiling water. $\endgroup$ Commented Mar 21 at 15:59
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    $\begingroup$ The only perhaps-not-so-trivial question is whether the gas above the boiling water is cooler or hotter than the water. (It may be hotter because it consists of vapor which may be hotter than boiling; it may contain air heated by the flame below the pot). But usually I'd assume the vapor is cooler, partly because of the evaporation energy needed and because it mixes with colder air). In any case, the different temperature of the handle sticking out will be conducted downward along the handle and will cause it to have a slightly different temperature than the water. $\endgroup$ Commented Mar 22 at 9:06
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    $\begingroup$ A question which puzzles me a bit: Since you majored in physics, which mechanism would you suggest that can make the ladle hotter than its surroundings? $\endgroup$ Commented Mar 22 at 9:11
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    $\begingroup$ @Peter-ReinstateMonica, My thought exactly! I'm not understanding how this can't be understood with a just a basic understanding of how heat is transferred. I have never taken any physics, yet I know this... Now, gas vs electric may may some difference because a small pot on a large burner will allow some flame to creep around the side, but this specific thing wasn't mentioned as a variable at the source of confusion. $\endgroup$ Commented Mar 22 at 16:02
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    $\begingroup$ In practice, ladle handles sticking out of the pot on a gas stove can get extremely hot because of the gas flame coming around the pot. This is the source of many a surprise burn.... but this isn't what you mean, I think $\endgroup$
    – Joel Keene
    Commented Mar 22 at 20:16

6 Answers 6


The second law of thermodynamics says (among other formulations) that heat flows from high temperatures to low temperatures, but never the other way around.

In your illustration, you have the ladle in contact with the boiling water and with the air. The steady-state temperature of the ladle will therefore be between the temperature of the water and the temperature of the air. If you were heating the water from the top with a flamethrower, or with the broiler element of an oven, then the steel might conduct some heat down into the water. (Note that steel actually has poor thermal conductivity, as metals go.) But you are clear that the heat source is below the pot, and therefore that the air above the pot is your system's cold reservoir.

You might imagine engaging in some phase-change shenanigans. For example, if your pot were a closed pressure cooker with a ladle inside, the water would boil at some higher temperature, and the ladle would approach that temperature as well. If the seal in the pressure cooker fails, the water will produce new vapor (doing work on its surroundings) until the temperature has fallen to the boiling point at ambient pressure. So not only would you have the lid of your pressure cooker wedged into your kitchen ceiling or attic — you would also have a ladle in your water at the previous boiling temperature, vigorously boiling any unspilled hot water as it cools.

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    $\begingroup$ Yes, perhaps the key to understanding this problem is to realize that heat flows from hotter bodies to colder ones, so that at the very instant the ladle becomes hotter than the surrounding water, the excess heat flows back into the water, i.e. the ladle is at a steady state. $\endgroup$ Commented Mar 21 at 15:31
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    $\begingroup$ Or, for the student who is ready to think a little more about statistical mechanics rather than about thermodynamics, one could point out that there are constant, random, microscopic energy flows in both directions between two objects, regardless of the temperature difference between them. However, if one of the objects is hotter than the other, then the fluctuations where energy flows from the hotter object to the cooler object overwhelmingly outnumber the fluctuations where the energy flows in the other direction. At equilibrium, fluctuations in each direction are equally likely. $\endgroup$
    – rob
    Commented Mar 21 at 15:50
  • $\begingroup$ Yeah, I almost forgot the meaning in thermodynamics of the second law, also I didn't think about entropy connecting thermodynamic situation. Thank you. $\endgroup$ Commented Mar 22 at 2:37

The pot as it is in contact with the fire can have hot spots, however, the water is at a temperature of $100^\circ$C. If the ladle is completely surrounded by boiling water, then the ladle will quickly rise to the temperature of boiling water, i.e. it will reach equilibrium with its surroundings, the water. It cannot get hotter than this as it is not in contact with anything capable of transmitting more energy than that possessed by the boiling water. If part of the ladle is in contact with air, then the ladle can actually cool by giving off some of its heat to the air so that the exposed portions of the ladle can actually be cooler than the boiling water. So in the case of complete immersion but no contact with the pot, any point in the ladle satisfies $T=100^\circ$C, and in the case that there is some exposure to air: $T\leq 100^\circ$ C.

It is perhaps worth noting that $100^\circ$C is the temperature of phase change of water into steam at 1atm of pressure, thus the temperature of steam will be the same as that of the boiling water. Of course there are phenomena like superheating, however, on average, the boiling point will be the observed temperature. In fact the technique of water or oil bathing is commonly used in the laboratories of chemists to control reaction temperatures.

  • $\begingroup$ Then, Could I think that There's no thermal energy transferred from the water to the ladle? Does it mean that Input energy from gas stove to water is used to only phase change of water when the ladle is isolated from air? $\endgroup$ Commented Mar 21 at 14:43
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    $\begingroup$ The ladle comes to equilibrium with the water. As the water continuously has energy transferred to it because the system is not closed, then the ladle has energy continuously transferred to it. If there was no continuous transfer of energy, then the water and ladle would cool to room temperature. $\endgroup$ Commented Mar 21 at 14:46
  • $\begingroup$ But I thought that This is not thermal equilibrium. Because the water is exchanging heat with gas stove. So It's not situation fitted with zeroth law. $\endgroup$ Commented Mar 21 at 14:53
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    $\begingroup$ The system is not closed so it is not in equilibrium, this does not mean that the system does not approach a steady state. Kind of like wind resistance, the system is not closed yet a falling object reaches a terminal velocity. Here the system is not closed yet the water reaches the boiling point and so does the ladle. $\endgroup$ Commented Mar 21 at 14:56
  • $\begingroup$ Umm.. It's so tricky. I understood what you mean, but I can't understand its mechanism and why the T of the ladle can't exceed 100 degrees. $\endgroup$ Commented Mar 21 at 15:07

There will of course be a temperature gradient along the ladle, and near the surface the temperature of the ladle will be cooler than the water temperature. Steel is a good conductor of heat and some of that heat is conducted up the handle, where the handle is cooled by the surrounding air.

However, the bottom of the ladle will be at a higher temperature than 100C. This is because there is a temperature gradient in the water. Water in contact with the bottom of the pan, which will be above 100C is superheated. The bubbles of water vapor that are formed there are also above 100C. This is especially true if the water is at a "rolling boil". The water has to transfer heat from the pan to the air; this mainly by conduction and convection and of course the bubbles of steam. Convection currents can only exist when there is a temperature difference between the bottom and top of a fluid. Rayleigh_Benard Obviously the top of the water is at least at 100C (It is boiling) so this implies that the temperature at the bottom is higher.

There is also a pressure difference due to the depth of water. If the water in the pan is 6" deep, the pressure at the bottom will be about 0.5 psi higher. Assuming standard atmospheric at the surface, with this additional pressure the boiling point of water is 100.5 degC.

Of course the metal ladle will never be at a higher temperature than the water it is in. It will in fact be slightly lower on average, as there needs to be a temperature gradient present to drive heat into the ladle.

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    $\begingroup$ Amazing that only one answer mentioned the temperature gradient in the water. Every cook (and, I imagine, submarine crewman) knows this! $\endgroup$
    – Fattie
    Commented Mar 22 at 3:17
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    $\begingroup$ Can you quantify the size of this temperature gradient? My understanding is that, if there are liquid-vapor interfaces available (e g. at bubble surfaces), then phase changes at the interface bring the temperature to the boiling point very rapidly. My intuition is sub-kelvin fluctuations, but I don't know whether decikelvin or millikelvin are more reasonable. I have always been successful with the approximation that a material undergoing a phase change is at uniform temperature throughout, but I haven't needed high-precision temperatures at short length and time scales. $\endgroup$
    – rob
    Commented Mar 22 at 3:27
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    $\begingroup$ This answer is qualitatively good, but quantitatively rather off. Steel is only a good conductor of heat if compared to e.g. wood and plastics, but not good enough to conduct a significant amount of energy away from a boiling-water pot (evaporation does way more). And the gradient in the water is, as rob comments, very small. Since water has such a vast latent heat of evaporation, a part that is hotter than 100°C (even by a lot) will only result in steam very slightly hotter than 100° as long as there's any liquid water present. This changes only with high pressure. $\endgroup$ Commented Mar 22 at 15:32
  • $\begingroup$ @leftaroundabout Compared to water. Steel is about 50w/(mK) water is about 0.7w/(mK) at 100C. The question is about the temperature of the ladle. My point was that near the surface heat will be conducted up the handle of the ladle to the cooler part of the handle in the ambient air. If the air above the pot is any cooler! $\endgroup$
    – Rich
    Commented Mar 22 at 22:34
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    $\begingroup$ Thanks for editing the answer to quantify the gradient. I think you are saying that the boiling temperature increases by about 3K for every meter of depth? That is a bigger effect than I expected. But it didn't seem big enough to be useful in the kitchen, as @Fattie suggests. $\endgroup$
    – rob
    Commented Mar 22 at 23:39

In order to get to the actual question, let's assume you didn't mean 100°C but the actual boiling temperature, which will differ somewhat with time (because the atmospheric pressure varies with the weather) and quite a bit with location (because the atmospheric pressure depends strongly on the altitude).

Furthermore, there is a pressure gradient in the water. Due to the higher pressure at the bottom of the pot (at 1 m it will be 110 kPa instead of 100 kPa) the water there will have a higher boiling point, leading to a certain vertical temperature gradient; but because hotter water is lighter, there is considerable convection which prevents proper stratification, in addition to the turbulence caused by the rising vapor bubbles. Let's for simplicity assume that the water temperature, with fluctuations, is similar across the pot, or at least that your question is whether at any point the ladle can be hotter than the water at that point.

There are three mechanisms which can make the ladle hotter.

  1. The gas above the boiling water is hotter than the water. (One possible explanation could be that the flame has heated the air in the space in which the pot is situated.) That heats the exposed part of the ladle, and the steel conducts that heat downward, heating adjacent immersed parts of the ladle so that those are hotter than the water. Because of the cooling effect of the water and because steel doesn't conduct heat that well, this effect would not extend far.

  2. The ladle absorbs radiation from the flame (if the pot is made from a transparent material like borosilicate glass) or from the bottom of the pot, which may well be hotter than the boiling point. While water absorbs most infrared, it doesn't absorb all of it; some will reach the ladle, and a fraction of that will be absorbed by the ladle and heat it. Radiation from the bottom of the pot may be quite pronounced if the flame is strong; while the water will cool the bottom of the pot, it is entirely possible that it becomes much hotter than the water — for example, it may start to glow so that a water vapor layer emerges. (This is actually an undesired effect when one tries to heat water quickly because it leads to a reduction in overall heat transfer into the water even though the heating surface gets hotter.)

  3. The bubbles of water vapor rising from the bottom of the pot may contain gas which is hotter than the surrounding water. I am wording carefully here — in principle, such hot vapor would evaporate surrounding water, cooling down in the process until it reaches the condensation point. But the process may still be ongoing when the bubble reaches the ladle, and the vapor in the bubble may not be perfectly well mixed and thus contain hotter vapor in its middle. When the vapor bubble hits the ladle, it transfers heat to it which will make the ladle marginally hotter than the surrounding liquid water. This effect should be more pronounced in violently boiling water with a high density of vapor bubbles.

  • $\begingroup$ My thoughts exactly. Mainly radiation, not from the bottom of the pot that should be hotter to boil $\endgroup$ Commented Mar 24 at 9:17

The system is not at equilibrium, so there is energy flowing between the various parts of it (and also energy flowing into and out of the system as a whole).

But thermal energy still flows only from warmer bodies to colder bodies. Even in a non-equilibrium state, one object cannot be hotter than the things it is exchanging thermal energy with. And the rate at which thermal energy flows between objects depends on the difference in temperature between them (with greater difference leading to faster transfer, from the hotter object to the colder). That is enough to deduce an answer to your question.

There is a lot of thermal energy coming from the flame, which is much hotter than 100° C. The pot is much colder than the flame, so thermal energy flows into the pot.

Thermal energy also flows out of the pot, into the surrounding air and into the water. We'll imagine that we have an essentially infinite reservoir of cool air that can easily carry away the heat, but won't worry about the exact temperature; if it's liveable for the human operating the stove it's definitely much less than 100°! So the surrounding air remains at a lower temperature than the pot and energy will continue to flow out of the pot into the air.

Energy is also flowing from the pot into the water. We know it must because energy is also leaving the water into the air (both by direct contact and in the form of energy carried away by the water that evaporates into steam). If thermal energy were not flowing into the water from the pot, then the water would be cooling down and would not remain boiling at 100° C.

So the pot must be at least a bit hotter than 100°. If it were cooler then it would actually be taking thermal energy from the 100° water instead of supplying energy to it, and if it were exactly 100° then the energy flow between the pot and the water would be zero. In either case the water would cool down until its temperature were low enough below the pot that the rate of energy transfer from the pot to the water is enough to replace the energy the water is losing to the air, rather than remaining boiling. So if we've reached a steady boil, the pot is hotter than 100°. (You can see this by boiling water in a kettle, then turning on the heat under a cold pot and pouring the boiling water in; the water will stop boiling as it first loses energy heating up the cold pot, and only returns to the boil once the gas flame has heated the pot sufficiently)

Finally consider the ladle. As you've drawn it, it is in contact with the water and with the surrounding air. So it cannot reach a temperature hotter than both of them; if it were it would be losing heat to its surroundings. At a steady state (where things will remain roughly the same until we alter the gas setting or too much water evaporates), the average temperature of the ladle must be near the temperature of the water, but a little below. The air is definitely colder, so the exposed part of the ladle will be transferring heat energy to the air. For it to remain a roughly steady average temperature, the rate of energy flow from the water into the ladle must be equal to the rate of energy flow from the ladle to the cooler air. So it must be below 100° or there would be no energy flowing from the water to the ladle to replace the energy lost to the air. The exact average temperature this will be depends on how fast heat can conduct from the underwater parts of the ladle to the exposed handle (it's not actually all one temperature), what the ladle is made of, the exact temperature and composition of the air, etc etc. But we can see that these two flows have to be equal and opposite, or we wouldn't have a steady state yet.

If the ladle were cooler than this steady state temperature (as when you first put it in the water), then the difference in temperature from the 100° water would be larger and the underwater part would heat up faster than the exposed part can lose energy to the air (indeed if the ladle is colder than the air, the air-exposed handle will be heating it up too!). So it will heat up, reducing the temperature difference with the water and the energy flow, until those two flows are in balance and we do reach that steady state.

If the ladle were hotter than the steady state temperature then it's losing energy to the air faster than it's gaining energy from the water, so it would be cooling down. So there's no mechanism that could even maintain a temperature hotter than the steady state temperature, let alone a mechanism that could allow it to reach a hotter temperature in the first place.

The feature of this system that allows it to reach a known pseudo-equilibrium (where things reach and hold a particular temperature for a while, even though lots of energy is flowing through the system and we don't even know exactly how much energy is coming from the flame) is the boiling point of water. If the water didn't evaporate then it would just keep heating up above 100°. So the pot wouldn't stop heating up when the flame's heating effect is equal to the cooling effect of 100° water; the water's temperature would rise and the cooling effect would be less so the flame could heat the pot up more, and so on.

Eventually there would be another pseudo-equilibrium point where the flame just isn't putting out enough energy to raise the temperature of the pot (and indirectly the water) any higher, because of the cooling effect of the air (which we're presuming can be infinitely replaced and so never actually heat up). But the exact state would depend on how much energy is coming from the flame; what setting we have the gas at. With water boiling at a fixed temperature of 100°, that puts the contents of the pot in a known state (although the temperature of the pot itself will be at whatever temperature balances the energy it transfers to 100° water with the energy it receives from the flame; we don't know what that is). If the water is receiving more or less energy it will evaporate faster or slower, but the energy in the evaporated water is carried away by the air. The evaporation makes the system self-adjusting to a wide range of energy input rates from the flame.

This is exactly why cooking things in boiling water is such an easy and important technique humans have been using for many thousands of years! As long as you can get your cooking heat source hot enough to boil water (and within a reasonable upper limit that's hard to reach accidentally), it doesn't matter very much exactly how hot it is. The boiling water will have a consistent effect on the food you're cooking every time! And because the water is receiving "too much" energy to remain at 100° but having the excess carried away as steam, the system can often self-adjust when you drop some colder food into the water to start cooking; as it's now losing more energy heating up the food it simply has less excess going to evaporation (unless you put in so much food that is so much colder that the water loses heat to it faster than it can pull heat from the pot; I see this when I cook frozen vegetables in boiling water). If the 100° temperature was being maintained by a very careful balancing of input and output energy rater than the boiling point effect, then any change at all would affect the balance and change the cooking temperature quite a bit, requiring a lot more fine tuning and control from the cook.


My thought was that the steam bubbles and therefore the ladle might be the absolute tiniest fraction of a degree hotter than the water. A miniscule fraction of a degree that is. My reasoning is that the steam could be hotter than the 'water' but the water couldn't be as hot or it would be steam.

But how do you measure? Dipping a thermometer into the water will have the steam bubbles touching it as well. Temperature is a measure of averages after all, that's why you get water evaporating slowly at room temperature because a portion of the molecules have enough energy to break away from the other water molecules. This causes cooling of the water because they carry that excess energy away.

I would think the part of the ladle above the water would be affected more by the air temperature above the pot and convection in the room, or by the heat from the flames heating air above the pot. So what do you mean by the 'temperature of the ladle'?

So discounting the part of the ladle above the water, and comparing the temperature of the ladle below the water to the temperature of the liquid water, I think the ladle would be a tiny fraction of a degree warmer because the steam bubbles would be warmer. I don't know how to measure it, but the steam must be at least as hot as the water and can be hotter and the average temperature of atoms interacting with the ladle would be hotter then.

There is dissolved water vapor in the water too, do you therefore count that? The bubbles form on the bottom both because that is where the heat is applied and because the atoms of the pot serve as a nucleation point for bubbles to form. Since that is where the heat is applied and the metal of the pan is hotter than the boiling point of the water, I can't think of a reason for the bubbles to not increase in temperature by a small amount as they form beyond the boiling point.


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