Cup of warm water suspended in a pot of water held at a steady boil

Question

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.

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.

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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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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?

Answer 5

I love this question since, thermodynamics states that as time approaches infinity, all objects will reach the same temperature.

But in reality, that's not the case.

Take the flame or stove for example, does the pot ever get as hot as the flame, or as hot as 450C?

Answer to that is no.

Similar concept with boiling water, but this is due to phase change.

When water boils, the energy from the rise in temperature is being used to break the water bonds to become vapor. So the vapor is actually now "lower" than right when water boils. As vapor floats to the top and touches the cup, it transfers that temperature to the cup, but is no longer at boiling temp.

Therefore, in a realistic situation, you can never get the cup of water to boil the same way the water boils in the pot.

Answer 6

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.