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Comment by grog454 | Hacker News Reader

Why would you not consider that caveat? The cooling rate should be generally proportional to both the surface area and the current temperature delta between the liquid and whatever it contacts. Taken to the extreme, if you added the creamer and then poured the coffee out onto the floor, the creamer's mass and temperature would have relatively little impact compared to the surface area change.

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For a non-sealed cup of hot liquid, the significant majority of cooling happens at the exposed surface with the air, through a mixture of convection and evaporation. Modeling the cooling as being proportional to total surface area would be pretty inaccurate, although there is definitely still some conduction through the cup/mug worth considering as well.

… my work here is done.

I think it's just beginning...

> For a non-sealed cup of hot liquid, the significant majority of cooling happens at the exposed surface with the air, through a mixture of convection and evaporation.

I don't know enough to argue with any confidence but this is very surprising to me. Ignoring for a moment the thermal conductivity of different mug materials, it seems like a large amount of energy would go toward heating the mug up to near liquid temperature rather quickly. Then you'd have at least as much heat loss between the mug and air (compared to exposed liquid and air).

The mug I imagine we're talking about is a ceramic mug, which I believe to have high thermal conductivity just based on what processor covers are made of. It also has plenty of mass.

If you're talking about an insulated mug obviously this changes. But just the fact that insulated mugs exist proves my point that a large amount of heat is lost through the mug...

Assume for simplicity that we mix half coffee and half creamer. This halves delta t. However the area will not double, since the top and bottom areas are the same. Furthermore, as another commenter pointed out, the top area is where most of the action happen.

However, if we really want to overcomplicate things we could consider the possibility of an insulating air pocket in a half-empty cup, leading to less convective losses. Consider a vacuum flask half full of hot coffee outside in a strong wind. If you fill it to the brim with creamer it might cool faster. Evaporation might become important too.

If you don't fill the thermos to the top, we can also add the small temperature impact of Helmholtz resonance from the wind blowing across the lip to make this more complicated. (Now I wonder how loud a sound needs to be to boil water...)

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