# Question:

If...

• I have to identical vessels containing an equal volume of water.
• I place them in a room with an ambient temperature of 20 C, at normal pressure, in a normal atmosphere.
• I heat one volume to 30 C and cool one volume to 10 C, then remove the cooling/heating at exactly the same time...

... which will reach ambient temperature the soonest? Why?

• If I change the vessels does this increase the relative speed of cooling/heating? I.e., does exposed surface area have a greater impact on cooling rather than heating?
• Is the rate change constant?
• If the ambient temperature is changed, say to 80 C but the difference is kept the same, i.e., 90 C and 70 C, does this affect the outcome?

## Use case/origin of this question

I was getting water from the break room and I have the option of either hot or cold water... So if I had to pick, and not mix, which I do, then ... that's how I started wondering about this. :)

• What's the relative humidity in the room? Evaporation will affect the answer, and relative humidity and temperature will affect the evaporation rate. Feb 26, 2019 at 3:10
• @DavidWhite good point. I think the summary of the answers provided is that; evaporation and airflow affect this, which is affected many things, including humidity.
– Tim
Feb 26, 2019 at 18:11

For the original case, I would expect the warmer one to reach equilibrium quicker.

This for two (probably quite small in practice) reasons. First, the warmer one will evaporate more of it's water than the cooler one. It may not be a significant amount of evaporation; but any evaporating would reduce the amount of mass to be cooled. Depending on how you set up the scenario, getting the water to 10 degrees hotter would cause evaporation before you even start the timers.

The second reason is due to thermal convection. Heat rises, so when you have water warmer than the surroundings, this would cause more airflow upwards from the container, compared to cool water where the air would be more prone to stagnate on top, reducing the convection; thus reducing the heat transfer.

Increasing surface area of the water will only make these two effects more pronounced for the warm water, by increasing the surface area for evaporation and convection.

Increasing the ambient to 80 degrees shouldn't really change this.

The rate of change will not be completely constant. For starters, as both approach equilibrium, the heating and cooling will slow down. Also, the rate may be affected by stagnation in the water itself. The cooler bucket may decrease it's heat transfer rate as a layer of warmed water begins to develop on top. This would greatly reduce the heat transfer rate on top, only getting worse as that layer gets closer to equilibrium. This doesn't happen with the warmer bucket, because the cooler room cools the water surface, and that water sinks below the warmer water, leading to natural convection.

The only way I can think of that you might get different results is if you kept the water in a sealed container, and kept it closed with an insulated lid, while having the sides of it conductive. I'm not even sure if that would work though, it would just eliminate evaporation and reduce the effects of stagnant convection.

• Thanks @Jmac, very good explanation. If we were able to ignore the air movement around the warmer vessel would the rate of temperature change be the same?
– Tim
Feb 25, 2019 at 19:09
• Evaporation not only decreases the mass being heated/cooled, but also has a cooling effect on its own. The water molecules with the highest kinetic energy are the ones that evaporate, so when they leave, the remainder have a collectively lower temperature. Water will evaporate for both the warm and cool vessels, but the evaporation will accelerate the cooling and slow the warming. Feb 25, 2019 at 19:11
• @varlogtim No, evaporation would still be significant enough to make the warmer one cool quickest.
– JMac
Feb 25, 2019 at 19:22

Very interesting questions. I’ll take a stab at them.

... which will reach ambient temperature the soonest? Why?

I’m thinking the volume at 30 C will reach ambient (20 C) before the volume at 10 C. My reasoning is I think that the higher temperature water will cool primarily based on a combination of convection, conduction, and evaporative cooling. The lower temperature water will heat up primarily based on convection and conduction from the surrounding air to the water. The evaporative cooling favors cooling of the warmer water.

• If I change the vessels does this increase the relative speed of cooling/heating? I.e., does exposed surface area have a greater impact on cooling rather than heating?

Increasing exposed surface area of the water. That increases the evaporative cooling effect. It will favor increasing the speed of cooling of the warmer water.

• Is the rate change constant?

Since the rate of heat transfer is proportional to the temperature difference, all other things being equal, as each vessel approaches room temperature the rate of heating or cooling decreases.

• If the ambient temperature is changed, say to 80 C but the difference is kept the same, i.e., 90 C and 70 C, does this affect the outcome?

Increasing the temperature differences between the vessels and ambient air increases the initial rate of cooling and heating. But I believe the warmer vessel will still reach ambient before the cooler vessel.

Hope this helps.

• ... and still, 30 C water has to become 20 C water first, before cooling down further... Feb 25, 2019 at 19:29
• @AtmosphericPrisonEscape I could be wrong, but I think once the warmer vessel reaches the ambient air temperature, the average temperature of the water will remain the same and equal the ambient air temperature. Evaporative "cooling" is restricted to the surface where water molecules having higher kinetic energy than the average of all the water molecules escape. Then heat transfers to the surface from water just below the surface, and from air to the surface and through the vessel walls to make up the difference. End result, average water temperature remains constant and equal to the air. Feb 25, 2019 at 19:49
• Sorry, I've misread the question. Late-night physics... Feb 25, 2019 at 21:17
• @AtmosphericPrisonEscape Late night physics, Ahh I remember it well (or not so well) Feb 25, 2019 at 21:28