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If a hot cup of hot chocolate is just standing there, can it cool itself down by transferring the kinetic/thermal energy that the liquid has into the mug/cup?

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  • $\begingroup$ Also interesting: physics.stackexchange.com/q/5265/26969 $\endgroup$ – Floris Dec 10 '15 at 6:33
  • $\begingroup$ There's a strong dependence of this phenomenon on the substance. Some hot chocolates take longer to cool than others. This one's still very much hot after nearly 50 years, even notwithstanding the loss of one of its chief constituents. $\endgroup$ – WetSavannaAnimal May 23 '18 at 3:53
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Absolutely, although transfer into the mug is typically a small component of heat loss. If the mug is initially cold, some heat will be transferred to get it to the same temperature as the liquid; after that, the mug will cool due to contact with air (and a tiny bit due to thermal radiation), and this will set up an equal flow of heat from the liquid to the mug in the steady state.

A much larger component of heat loss is evaporation. You can slow this down by putting whipped cream on the chocolate (acts like an insulator), or you can speed it up by blowing on the liquid - removing the water vapor from the surface of the liquid reduces the rate at which water vapor returns to the liquid; and liquid evaporating takes a lot of heat with it, almost 540 calories per gram. This means that losing one gram of liquid due to evaporation will cool 540 ml of remaining liquid by 1 degree C.

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Yes, and into the atmosphere, which will dissipate the thermal energy into the atmosphere around that the chocolate

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Why not? Heat always flows from a higher level(from high temperature) to a lower level(to low temperature). So if there is a temperature difference then heat will surely flow.

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