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


*

*Radiation between the surface of the tea and the air molecules.

*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? 
 A: 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:


*

*I poured boiling water into two ceramic mugs

*I put a layer of canola oil on top of one mug

*I took a picture:

*I waited an hour, feeling the temperatures of the two mugs

*I took another picture, and compared the temperatures:
Nope: didn't have a thermometer, didn't weigh them. We're talking basic. But here are the results:


*

*Starting after the first five minutes, the mug with oil on top was consistently warmer

*The mug without oil lost about 4% of its water to evaporation

*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.)
