I am not sure about my answer. I think that the cup of tea cools down by the following modes:

  1. Radiation between the surface of the tea and the air molecules.
  2. Conduction between the tea and the cup itself and between the cup and the table or whatever the surface is.

Do you think it is correct?

  • $\begingroup$ Evaporation. The faster, hotter molecules at the surface are moving fast enough to escape from the liquid into the air. This leaves slower, cooler molecules behind. $\endgroup$ – mmesser314 Aug 22 '15 at 20:44
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    $\begingroup$ You will find answers to your question at this earlier question even if the question posed is slightly different... $\endgroup$ – Floris Aug 22 '15 at 20:47
  • $\begingroup$ While a cup of tea certainly does radiate photons, it mostly interacts with the air through conduction, not radiation. Also evaporation, like mmesser314 said. $\endgroup$ – Jahan Claes Aug 22 '15 at 20:49
  • $\begingroup$ Also, evaporation is essentially convective heat transfer to the environment. $\endgroup$ – David White Aug 24 '15 at 2:52

An open cup of hot water loses a lot of its heat through evaporation. You can easily test this: prepare two cups of hot water, and then float a thin layer of cooking oil on top of one. Come back in a half hour: you'll find that the cup without oil is a lot cooler, and the water level has also significantly gone down.

Edit: I did the experiment:

  1. I poured boiling water into two ceramic mugs
  2. I put a layer of canola oil on top of one mug
  3. I took a picture:Two mugs of hot water, left one with layer of oil
  4. I waited an hour, feeling the temperatures of the two mugs
  5. I took another picture, and compared the temperatures:Same mugs after waiting an hour

Nope: didn't have a thermometer, didn't weigh them. We're talking basic. But here are the results:

  1. Starting after the first five minutes, the mug with oil on top was consistently warmer
  2. The mug without oil lost about 4% of its water to evaporation
  3. Final temperature of right mug was a bit over room temperature

Water has a heat of vaporization of 540 calories per gram. So, losing 4% of the water means that the right mug's contents lost about 22 calories per gram to evaporation heat, meaning that evaporation alone lowered the temperature by 22°C. If the water temperature went from 90°C to 30°C, then about 1/3 of the lost temperature was due to evaporation.

That's actually less than I expected. However, note that this was a full hour of cooling, down to close to room temperature. The rate of evaporation is strongly temperature-dependent, so I would expect that initially the heat loss was mostly evaporation, but that ratio quickly fell with the temperature. (I edited the initial sentence of this answer to be closer to reality.)

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    $\begingroup$ Never trust the internet - do your own experiment. Like it. Did you actually ever do this experiment, or do you just expect it to be true? $\endgroup$ – Floris Aug 23 '15 at 2:56
  • $\begingroup$ Ask me tomorrow. $\endgroup$ – Daniel Griscom Aug 23 '15 at 5:24

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