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When boiling water on a stove, will the temperature of the steam vary significantly with the temperature of the burner?

Person A's argument: So, once individual water molecules reach 100C/212F, they become vapor. The water molecules in the pot are <100C; the water molecules in the air are >100C. Generally, the only way to heat up the water vapor to significantly more than 100C would be to trap the water vapor. In a big kitchen, the water vapor rises rather quickly and gets sufficiently far away from the burner. Within the first couple seconds that the molecule becomes vapor, the vapor may still be close enough to the burner to become slightly more than 100C (101C?), but generally, no matter what the temperature of the burner, the water molecules will escape at 100C and won't reach a temperature significantly above 100C, given a large room.

Person B's argument: With a hotter burner, the water in the pot is hotter and as a result the water molecules that become steam - and bubble up from the bottom of the pot - transfer less heat to the surrounding water on their way to the top of the pot and leave as hotter steam.

Or do persons A and B just have a poor grasp of physics?

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  • $\begingroup$ Person A is correct. Under normal conditions when water reaches its boiling point its temperature will not increase further and any extra heat transferred to it goes into boiling the water, overcome the latent heat. $\endgroup$ – By Symmetry Feb 11 '17 at 17:59
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    $\begingroup$ Are you the Charles Woodson who played for the U of M, won the Heisman Trophy, played for the Oakland Raiders, and is now an ESPN analyst? $\endgroup$ – Chet Miller Feb 11 '17 at 21:05
  • $\begingroup$ When you say "the water molecules in the air are >100C", you mean "in the air in general", or "in the air above the boiling water"? I interpreted as being the first option, what is reflected in my current answer and should be corrected if necessary. $\endgroup$ – stafusa Aug 16 '17 at 23:30
  • $\begingroup$ The steam leaving the water is at 100 degC at sea level. At a different elevation/pressure it will differ. Also if the liquid in pot is not pure water and contains other volatile substances a possible azeotrope boiling point will be the temperature of the vapour. If you had beer in the pot the water-ethanol azeotrope would be about 78 degC. $\endgroup$ – KalleMP Jan 21 '18 at 13:41
  • $\begingroup$ @KalleMP, people don't commonly boil multi-component mixtures in the open air on their kitchen stove. Also, beer has approximately 5% ethanol, and the water-ethanol azeotrope is 95% ethanol-5% water. You wouldn't detect such an azeotrope for long because the ethanol would quickly boil off. $\endgroup$ – David White Feb 7 '18 at 2:36
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Both A and B are slightly wrong. The 'boiling point' of water is the temperature at which steam and liquid exist at equilibrium, and the roiling boil of a pot of water on the stove indicates a lack of equilibrium. Each steam bubble, expanding as it rises from the bottom of the pot, is accumulating vapor from the surrounding liquid (not staying a constant volume).

So, A is wrong to think that there is an equilibrium-temperature indication in the boiling pot. A single molecule can become vapor only at the water surface, or by doing work against surface tension and water pressure by expanding the diameter of a bubble. If the work is done leaving uncondensed water vapor, it must have been hotter than 'the boiling point'.

And, if B is naiive in thinking that the temperature outside the pot is important in determining the temperature inside.
The evaporation of water is a heat sink more than capable of cooling the metal, it might just be that higher outer temperature turns a boil with four streams of bubbles into a similar boil with eight streams of bubbles. More heat doesn't guarantee higher temperature, just higher heat flow.

As for 'significantly higher' temperature of the bubbles, that calls for judgment. The observation of small bubbles expanding as they rise, means there is significance, because it's observable.

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  • $\begingroup$ Boiling is even more complicated than this. Good summary. $\endgroup$ – Dan Z Apr 30 at 2:56
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In order for there to be a phase change from liquid to vapor, water must release latent heat of 100 degrees centigrade. A hotter burner will not raise the latent heat temperature of water vapor at phase change (the boiling point) of the water.

However, if the burner is large enough to heat the entire room and the room is closed, it could raise the temperature of water vapor already in the air by transferring sensible heat from the burner to the air in the room.

As the temperature of the room rises, the saturated vapor pressure inside the room would also rise, and molecules of water vapor in the air would move with greater kinetic energy, raising their temperature. In order for the burner to raise the temperature of water vapor in the ambient air, the room would have to be closed. Otherwise, the saturated vapor pressure would not rise and neither would the temperature of the water vapor.

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The boiling pot of water acts as a temperature regulator once the water begins to boil. As long as the burner doesn't have any additional means of transferring heat into the steam above the water, it does not increase in temperature but actually decreases rapidly as you'll often note condensation on the upper part of the pot.

If you wanted to create steam above $100^o C$ then you need to apply additional heating above the water with another burner/heater.

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Person A is essentially right, with the exception that the water molecules in the air are not at 100°C or higher, but at room temperature, since they're (supposedly) in equilibrium with the other components of air. Remember that even at room temperature water turns into vapor: this evaporation is what makes puddles on the sidewalk eventually disappear after the rain.

Supporting person A are measurements made in kitchen experiments that attest the water temperature won't exceed 100°C even very close to the bottom surface.

If you preheat the pot to at least some 200°C and throw a bit of water in, you'll get Leidenfrost effect, i.e., the drops will float for a while over a cushion of vapor. With more water, it'll splatter around with vapor that might be in part above 100°C close enough to the hot surface.

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  • $\begingroup$ How can you say that the water molecules in the air are not 100°C when the steam rising from the pot is clearly not yet in equilibrium with the surrounding air? $\endgroup$ – noah Aug 16 '17 at 23:23
  • $\begingroup$ It's a matter of interpretation of what he means by "in the air". My understanding is that he meant "the air" in general", but even above the surface of the boiling water it won't be above 100°C. $\endgroup$ – stafusa Aug 16 '17 at 23:26
  • $\begingroup$ It will also not at be room temperature, as you could easily verify by holding your hand right above a boiling pot. $\endgroup$ – noah Aug 16 '17 at 23:27
  • $\begingroup$ Repeat: my understanding is that OP meant "the air in general", not above the boiling water; only they can tell what they meant. $\endgroup$ – stafusa Aug 16 '17 at 23:28
  • $\begingroup$ @stafusa It's actually quite clear what he meant, because he used the term "steam", not just vapour. Steam in this context is pretty much universally referring to the visible wet (supersaturated) steam you get over a boiling pot. $\endgroup$ – JMac Aug 16 '17 at 23:33

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