The question asked by a website is as such (possibly behind a paywall):

A cup of warm water is suspended in a large pot of water held at a steady boil. Will the water in the cup ever boil? Assume that the pot never runs out of water.enter image description here

The provided answer & explanation:

No. When the cup of water is placed in the boiling water, the cup is cooler than the surrounding water and heat will flow into it. Eventually, the water in the cup will increase to $100 ^{\,\circ} $ C — if any more energy goes into the cup, then the cup will begin boiling. But at this point, the boiling water in the pot and the $100 ^{\,\circ}$ C water in the cup are at the same temperature. Since there is no temperature difference, there will be no more heat flow into the cup, so it will never boil.

According to the explanation (emboldened sentence), the water in the cup does reach 100 degrees. Doesn't that mean that the water in the cup does indeed boil, because the boiling point of water happens to be at 100 degrees Celsius? Additionally, I don't really get the difference (temp or state wise) between the cup-water & the water in the pot, because the heat source is the same and it ought to always flow into the cup to reach thermal equilibrium?

  • $\begingroup$ We do have steam and water both at 100ºC. $\endgroup$ Jul 2, 2021 at 13:49
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    $\begingroup$ This looks like a perfect candidate for doing the actual experiment yourself. This just requires household equipment and should be easy to do. $\endgroup$ Jul 2, 2021 at 14:27
  • $\begingroup$ @silverrahul, I agree. The stated answer looks like an assumption based on theoretical considerations, those theoretical considerations are based on an assumed physical and mathematical model, and all mathematical models are simplifications of real world phenomena. I wouldn't be surprised if the water in the cup did boil, and I may decide to do this experiment myself to find out. $\endgroup$ Jul 2, 2021 at 16:11
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    $\begingroup$ @DavidWhite I actually did try this. And i can confirm that the water in the cup did not boil WHEN the water outside started to boil. But i cannot state if the water in the cup would NEVER boil. I ran out of patience and did not want to waste fuel waiting to see if the water in the cup does boil after some time or not. $\endgroup$ Jul 2, 2021 at 18:25

5 Answers 5


The illustration here is enlightening. The vapor bubbles form at the bottom of the pot, where the temperature of the heat source exceeds 100°C, as is required in practice for boiling*. As they float to the top of the water, they condense somewhat, indicating that the water is at 100°C or cooler. This water is thus not capable of making the water in the cup boil.

*A summary of the linked discussion: the formation of a vapor bubble costs energy beyond the latent heat, as we also need to create a new gas–liquid interface. Surfaces cost energy because the bonds are comparatively unsatisfied relative to the bulk. Consequently, some degree of overheating is always required for boiling, even in the heterogeneous nucleation case.


Just reaching the boiling temperature is not enough for boiling. Boiling happens when the the vapour pressure becomes equal to atmospheric pressure. When the temperature of $100ºC$ is reached then too we need more heat to vaporise it. The term is Latent Heat.

As the two systems (cup and pot) reach thermal, the heat flow do not literally stop, but yes, there is no net heat gain in the cup system. So, no extra heat is acquired by cup water, to set the bonds free and to change the state from liquid to vapour.


I don't have the rep to add to the above answer. But FYI I did attempt this experiment IRL and can say that the temp in the cup never got much above 90 deg C. I did leave it at a rolling boil for 10 mins even. I also tried with a lid over top, I could not get the water in the cup to boil that way either. It was not possible to fit the thermometer under the lid.

pic of experiment

  • $\begingroup$ This experiment reminds us that part of the problem is that the water in the cup loses energy to the environment. Water at the bottom is in contact with the metal pot at a temperature well above 100C, so it boils. The water in the cup is only ever in contact with water just below 100C or steam slightly above 100C (as soon as it goes above 100 it turns into a gas bubble and shoots upward). So when the water in the cup is at, lets say 99C, it's barely getting any energy from the tiny temperature difference, and it's radiating a lot of energy. The equilibrium will be well below, like at 90C. $\endgroup$
    – AXensen
    Sep 27 at 10:09

Indeed, the boiling point of water is 100°C, thus to clarify, the reason isn't that the water inside the cup is salt water.

Firstly, you have to understand that extra energy is needed for water to change from liquid to gas (i.e. boiling) even when it reaches the boiling point. Water doesn't instantly become water vapour when it reaches 100°C. Instead, when water reaches boiling point, it needs to absorb some extra amount of energy (latent heat of vaporisation) and increase their potential energy before "escaping" the cup of liquid water and becoming water vapour.

Now back to square one, how does the pot water normally transfer heat or thermal energy to the cup water? That is by a temperature difference (difference in average kinetic energy of the molecules). If there is no temperature difference, then there is no thermal energy transfer. Consider the case when you put some ice cubes at 0°C in a cup of water which is also at 0°C. Ignoring heat loss to the surroundings, neither would the ice suddenly melt, nor would the water suddenly turn to ice. Why is it the case? It is because there is no temperature difference between the two, thus there is no energy transfer and neither of the substance can undergo change of state.

Apply the above concept to the question. Even though the pot water is boiling, it will not exceed 100°C. Thus, there is no temperature difference between the two, and thus there is no thermal energy transfer. Therefore, there is no extra energy for the water to undergo change of state, and the water wouldn't boil and would just stay at 100°C.

Extra question: What would happen if the cup water is replaced by a substance of lower and higher boiling point?


I agree with you, the water in the cup is boiling at 100°C, and after reaching the 100°C there is no difference of the two separate waters the inflowing energy will evaporate same of the water outside and inside the cup.

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    $\begingroup$ Both liquid and gaseous phases of water can exist at 100C - the fact that the water is at 100C does not necessarily mean it's turning into vapor (i.e. boiling). The difference between the water in the pot and cup is that the pot is heated by the stove which is more than 100C, while the cup is heated by the water in the pot, which is never more than 100C. Thermal energy only flows when there is a temperature difference, so once the liquid water in the cup and pot both reach 100C, there is no heat exchange between them. Water in the cup stays at 100C, but does not boil. $\endgroup$ Jul 2, 2021 at 14:08
  • $\begingroup$ +1 If it was otherwise, water would be boiling only close pot boundaries. Let's remove the cup and we could treat water in the center of the pot the same as a water inside the cup. Without a cup it is moving more freely and energy flow is better, but still, maybe to lesser degree, water in a cup should boil. The same as the water in the center of the pot without a cup. $\endgroup$
    – dankal444
    Jan 21, 2022 at 7:30
  • $\begingroup$ @dankal444 When water at the bottom of the pot turns to vapor, it can climb above 100 degrees, and as the steam bubble rises through the pot, it can transfer heat to not-boiling water and cause it to boil. Unlike water in the middle of the pot, water in the cup can never come in contact with anything above 100 degrees. $\endgroup$ Sep 19 at 17:26

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