My friends and I were discussing whether cold water in a single use plastic bottle would heat up to room temperature faster than if the same volume of water was placed in an uncovered ceramic mug (better insulated vessel but open top).

I know for cooling hot liquids, there is evaporative cooling which plays a big role in heat transfer rate so a hot liquid in a mug would heat up faster, but is there a similar reverse mechanism like "condensative heating" which would apply to cold liquids heating up to room temperature situation?

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    $\begingroup$ You're really comparing apples and oranges here. How fast the liquid heats up depends on a lot of factors. $\endgroup$
    – Gert
    Dec 4, 2019 at 14:16
  • $\begingroup$ What makes you think the ceramic mug is a better insulator?. The thermal conductivity of glass is nearly 10 times that of PET (plastic used in water bottles). $\endgroup$
    – Bob D
    Dec 4, 2019 at 14:57
  • $\begingroup$ @BobD the mug is a lot thicker and fourier's law says it's inversely proportional q=kAdT/dx. Also if you touch a hot mug vs hot water bottle the surface temperature of the water bottle is probably hotter (the same applies for cold). But my question was really more about the influence of having a cover so you can take two ceramic mugs, one with a ceramic cover one without, but I'd like it quantified in that case. $\endgroup$
    – Curtis
    Dec 4, 2019 at 17:58

2 Answers 2


Initial Thoughts

Your question initially seems to focus on whether a counter process exists to heat a cold liquid that is the inverse of the process of natural convection + evaporation that cools liquids.

The short answer is ... No, such an inverse process does not exist entirely as stated.

First, the presumption we will make is that the liquid is below its boiling temperature at the given pressure. Any heat transfer to the liquid only raises the temperature of the liquid, it does not cause the liquid to boil.

Let's take the case when the air has no vapor for the liquid component. This addresses evaporation by making it present in both cases, hot or cold fluid. Let's also set the air temperature as a constant $T_a$. This allows a direct comparison by vapor pressures only of the liquid.

An open mug (open at the top) with a liquid at $T_h$ that is hotter than the surrounding air at $T_a$ will experience natural convection and evaporation to the surrounding air.

An open mug (open at the top) with a liquid at $T_c$ that is colder than the surrounding air at $T_a$ will not experience natural convection processes. Since the air has no component of the liquid, the liquid will still evaporate. Since the vapor pressure of the cold fluid is lower than that for the hot fluid. The evaporation rate of the cold liquid to hot air will therefore be lower than that for a hot liquid to cold air.

In conclusion, all else being equal for the same temperature difference ($T_h - T_a$ = $T_a - T_c$) for a liquid in an open mug, a hot liquid will cool faster than a cold liquid will heat up. Natural convection occurs in the case for the hot liquid. The evaporation rate is faster for the hot liquid.

The relative ratio of heating/cooling will depend on the temperature difference ($T_h - T_a$ versus $T_a - T_c$), the relative saturation of the air with the liquid component (essentially a vapor pressure to partial pressure difference), and the absolute temperature of the liquid (to set the magnitude of the vapor pressure of the liquid itself).

Also, when both mugs are closed off entirely and all else is exactly equal, both systems will have the same rate of temperature change. All else in this case is emissivity (as it affects radiation to/from the mug and surrounding air), thermal conductivity (as it affects conduction through the mug), and convection coefficient (as it affects convection to/from the mug and surrounding air).

Finally, when the liquid is at its boiling temperature for the given pressure, heat transfer into the liquid will cause the liquid to boil. In an open container, a film of cooler vapor will form above the liquid as a boundary to the hotter gas (air) in the surroundings. This will impeded the conduction of heat from the hotter air back to the liquid. Heat instead will go first from the surroundings to heat up the vapor above the liquid, and then the vapor will rise out of the mug. So, all else being equal, a boiling liquid can in practice experience lower heat transfer rates from the gas above it.

Additional Thoughts

The comment added clarifies the question further: Does a cold liquid heat up faster when it is a covered mug or an uncovered (at the top) mug.

Fill the mug to the brim or to the top (no extra air space). Set the air temperature to be above the liquid temperature and keep it stagnate (not flowing). Allow that we are talking about heat flow through a thermal conductor (not a thermal insulator) in the direction from the (hotter) surroundings to the (colder) liquid. Normalize the heating rate to the rate of heat transfer per unit volume of liquid for both cases.

Opening the top of the mug changes heat transfer modes from the top only.

In a closed mug, heat conducts through the top cap into the water. In an open mug, heat flows from the hot air to the cold liquid through a boundary layer vapor film above the cold liquid. Conduction rates of heat through a gas boundary layer are lower than conduction rates through the same thickness of a (thermally conductive) solid. As the solid cap is thicker, the conduction rate across it becomes smaller. As an order of magnitude, the thermal conductivity of stagnate air is perhaps a factor of 10x lower than glass. For a 100 micron air boundary layer above the liquid, a 10 micron ceramic plate has the same conduction rate, and a 1 mm thick ceramic plate as 100x smaller conduction rate.

--> The jury is out here. When the solid cap is thicker than the boundary layer using the ratio of thermal conductivities of the air to the cap material, the open mug will have more heat conduction from the top air.

Evaporation occurs from the open mug (and not at all from the closed mug). The evaporating liquid builds a colder vapor layer above the liquid. This slows down the conduction rate to the liquid below because heat that would go to heat the liquid is instead taken to heat the cold vapor.

Finally, in the open mug, the surrounding gas radiates heat to the colder liquid. In the case that the cap is only as thick as the boundary layer on the open mug and when we neglect evaporation as a process to limit heat transfer to the open mug liquid, radiation can be a deciding factor to make the open mug heat faster.


The jury is out in practice. You can set the restrictions on the system configuration, apply first principles, and reach a conclusion either way. Otherwise, the arguments become "what-if" cases that go on forever.

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    $\begingroup$ This addresses most of it, but OP does potentially have a point about "condensative heating" that it might be helpful to address. This could be especially important when the mugs are closed off. Assuming the mugs aren't ideal insulators, you might be able to make the case that there is a slight heating effect when the ambient air condenses on the cold mug; and that effect would still exist when the mugs are sealed; while it would prevent the reverse (evaporation) in the hot mug. $\endgroup$
    – JMac
    Dec 4, 2019 at 15:22
  • $\begingroup$ Doesn't this really depend on the humidity of the air, though? If the partial pressure of vapor in the warm air is higher than the vapor pressure of the cold liquid, wouldn't there be a net absorption of vapor into the liquid? $\endgroup$ Dec 4, 2019 at 15:28
  • $\begingroup$ In fact, isn't the statement "an open mug of liquid colder than the air will not evaporate" disproved by everyday experience? If what you're saying was true, puddles wouldn't evaporate after a rainstorm, right? $\endgroup$ Dec 4, 2019 at 15:30
  • $\begingroup$ Puddles are generally warmer than the surrounding air. Yes humidity is to be considered and I will address it. $\endgroup$ Dec 4, 2019 at 16:59
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    $\begingroup$ @JeffreyJWeimer Actually, here's a really obvious counterexample: liquid nitrogen boils at 77 K under 1 atm of pressure. It cannot be higher than this temperature in the liquid state at that pressure. This is much, much colder than the temperature of the surrounding air (which is at 270-300 K). And yet an open container of liquid nitrogen eventually evaporates away, despite being colder than the air it's surrounded by. $\endgroup$ Dec 4, 2019 at 18:02

It's quite in reverse. If you want a colder liquid than it's surroundings (room temperature) to heat-up faster, then you need to spit liquid onto some vessel which would have biggest uncovered area, so to maximize liquid's direct contact area with air molecules. So that air molecules could transfer their kinetic energy to liquid faster. Otherwise air molecules have to overcome vessel walls material thermal conductivity. Of course, I assume that thermal conductivity of vessel material is lower than air's, otherwise we will get the reverse situation (there may exists special materials so to maximize heat transfer). Another way to understand this phenomenon, just think why you need to keep Vacuum flask closed if you want to keep longer it's content temperature different (colder/hotter) than it's surroundings. (Hint: vacuum flask has very low thermal conductivity)


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