The question asked by a website is as such:
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.
According to the explanation (emboldened sentence), the water in the cup does reach a 100 degrees. Doesn't that mean that the water in the cup does indeed boil, because the boiling point of water happens to be a 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?
Best Answer
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.