As I understand it, during boiling the input of heat destroys or re-arranges the hydrogen bonds. It is used, in other words, against the potential energy of the intermolecular bonds.

But if some hydrogen bonds between molecules are destroyed then why is the kinetic energy of these particular molecules not increased and, consequently, temperature?


4 Answers 4


This is because the external pressure is constant (at one atmosphere). If you increase the pressure, e.g. by using a pressure cooker, then the temperature goes up, or likewise if you reduce the pressure the temperature goes down.

Water boils when the chemical potential of the water is the same as the chemical potential of the steam. If we consider steam as an ideal gas then the chemical potential is controlled by the pressure and temperature.

If you start with just water at below 100ºC then the water evaporates, and the partial pressure of the water vapour increases until the chemical potential of the vapour and water match. At that point there is no net evaporation of the water.

However at 100ºC the partial pressure of the steam in equilibrium with the water rises to one atmosphere and it can't get any higher. So if you raise the temperature above 100ºC the water and vapour cannot be in equilibrium so the water boils continuously in a desperate but hopeless attempt to raise the steam pressure.

This is how pressure cookers raise the boiling point. At 100ºC the water boils, but in a pressure cooker it can raise the steam pressure to above an atmosphere so the water can remain in equilibrium with the steam above 100ºC.

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    $\begingroup$ Doesn't boiling in kettles and saucepan occur mainly at the bottom of the head of water and thus at a slightly higher pressure? Looking at my kettle during the annoyingly long time it takes for it to turn itself off, I get the impression that strikingly violent formation of steam bubbles at the base might dominate over surface evaporation at the top as a means for carrying energy out of the fluid - am I wrong? (no surprise if so) $\endgroup$ Oct 16, 2017 at 15:00
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    $\begingroup$ You're not wrong, but the effect of the increase in pressure from the depth of the boiling vessel is usually small. Considering that 1atm of pressure is like 10m of seawater (or something like that), the depth of your pot is going to have a negligible effect on boiling temperature. $\endgroup$ Oct 16, 2017 at 18:20
  • $\begingroup$ but it doesn't clearly explain why water can't get any higher than 100 degrees at atmospheric pressure because suppose its more than 100 degrees the formation of vapors would increase,,,,,,,I still don't get why the relation that "SVP=atmospheric pressure" is necessary for boiling to occur $\endgroup$ Jul 15, 2022 at 10:56
  • $\begingroup$ @ArunBhardwaj we can discuss it in the Physics SE chat room if you want. $\endgroup$ Jul 15, 2022 at 11:15
  • $\begingroup$ yeah I am ready $\endgroup$ Jul 15, 2022 at 11:47

The kinetic energy and temperature do increase, but these molecules are no longer liquid water; they detach and leave as vapor. If you measure the temperature of actively boiling water, the thermometer is affected by the hot bubbles around it and shows a temperature slightly hotter than the boiling temperature (it can be off by a degree or so, depending on the setup).


The energy is increased. However below the surface, water molecules are densly packed. So unless they can form a steam bubble the energy below the surface is immediately passed on to other molecules. Only when a steam bubble forms (or the molecule is on the surface) the energy is not immediately transfered on but the molecule can escape.

The energy necessary to become steam can be interpreted as the energy necessary for the water molecule to leave the liquid and not be bounced back in by the gas around the liquid. It is an equilibrium depending on the molecules mean movement distance without hitting another molecule or atom and transfering too much energy to that molecule so that it would become liquid again. The molecules mean free movement distance however is nothing else than the gas pressure. So with rising pressure it gets exceedingly difficult for the molecules to leave the liquid body of water because the chance of hitting another molecule in the air around it is getting higher, thus the boiling temperature rises.

When that molecule then becomes steam, it is no longer part of the liquid. When the molecule left the liquid it however took the excess energy with it. The excess energy is no longer in the liquid which remains at a constant temperature. The steam however does not necessarily do the same, depending on the conditions of the other gas around the liquid.

Also the reason why water boils with bubbles rising up from the bottom is mostly due to uneven energy distribution within the liquid. The heat source heats water molecules directly where it has contact to the liquid. Some molecules have so much more energy that even with the pressure the water exerts on them, they become steam while below the surface.

On a macroscopic level however, the liquid water temperature as a whole stays mostly constant, as the steam (having the higher energy) escapes the liquid leaving only the liquid behind at constant temperature.


But if some hydrogen bonds between molecules are destroyed then why is the kinetic energy of these particular molecules not increased and, consequently, temperature?

The idea that temperature is proportional to the average kinetic energy of molecules is exactly true for classical ideal gases, and an approximation for other systems.

The fundamental property of thermodynamic temperature is that when two systems are placed in intimate contact, there will be a spontaneous net transfer of heat from the warmer system to the cooler one, while net flow of heat in the opposite direction is not spontaneous. The reason for this spontaneity is that the cooler system will gain more entropy from receiving a small increment $\delta q$ of heat than the warner system will lose from giving up the same amount of heat. This transfer of heat will continue until the two systems reach thermal equilibrium. At that point, their temperatures are the same.

For example, a litre of water at 50 °C is seen to be hotter than a litre of water at 40 °C because heat would spontaneously flow from the former to the latter and not the other way around. This would eventually result in both systems settling down at the intermediate temperature of 45 °C.

But when water is boiling, there is no state in between water and steam. Steam that is right at the boiling point does not spontaneously transfer net heat to water that is right at the boiling point. Any small increment $\delta q$ of heat transferred from the steam to the water would simply result in a small amount of steam $\frac{\delta q}{\Delta H_{vap}}$ turning into water and exactly the same amount of water turning into steam, so the entropy of the system as a whole wouldn't change.

For this reason, we must admit that the temperature of steam at the boiling point is the same as the temperature of water at the boiling point, even though the average kinetic energy might be higher in the gaseous state.

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    $\begingroup$ The vapor bubbles form at the heater with the temperature higher than the boiling point. Then they come up warming up water above the heater and at first shrink and disappear turning back into water. This shrinking causes typical noise heard before all water starts boiling. Once all water boils the noise disappears and the hot bubbles come all the way up to the surface. At this time, when they flow around the thermometer, they warm it up above the boiling temperature. I've tried this, it is true. $\endgroup$
    – safesphere
    Oct 17, 2017 at 0:28

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