Droplets collect near the rim of a mug filled with hot water.
During my mug cooling experiment last week, I was surprised to find that taking the lid off a vacuum insulated mug increased its initial cooling rate by a factor 7.5.
Removing the lid allowed air from the room to flow across the surface of the water, cooling it in two ways.
- Firstly, the air would warm up when it contacted the hot water, and then carry heat away in a convective flow.
- Secondly, some hot water would evaporate into the moving air and carry away so – called ‘latent heat’.
I wondered which of these two effects was more important?
I decided to work out the answer by calculating how much evaporation would be required to explain ALL the cooling. I could then check my calculation against the measured mass of water that was lost to evaporation.
Where to start?
I started with the cooling curve from the previous blog.
Graph#1: Temperature (°C) versus time (minutes) for water cooling in an insulated mug with and without a lid. Without a lid, the water cools more than 7 times faster.
Because I knew the mass of water (g) and its heat capacity (joule per gram per °C), I could calculate the rate of heat loss in watts required to cool the water at the observed rate.
In Graph#2 below I have plotted this versus the difference in temperature between the water and the room temperature, which was around 20 °C.
Graph#2: The rate of heat flow (in watts) calculated from the cooling curve versus the temperature difference (°C) from the ambient environment. The raw estimates are very noisy so the dotted lines are ‘best fit lines’ which approximately capture the trend of the data.
I was struck by two things:
- Firstly, without the lid, the rate of heat loss was initially 40 watts – which seemed very high.
- Secondly:
- When the lid was on, the rate of heat loss was almost a perfect straight line This is broadly what one expects in a wide range of heat flow problems – the rate of heat flow is proportional to the temperature difference. But…
- When the lid was off, the heat flow varied non-linearly with temperature difference.
To find out the effect of the lid, I subtracted the two curves from each other to get the difference in heat flow versus the temperature of the water above ambient (Graph#3).
[Technical Note: Because the data in Graph#2 is very noisy and irregularly spaced, I used Excel™ to work out a ‘trend line’ that describes the underlying ‘trend’ of the data. I then subtracted the two trend lines from each other.]
Graph#3: The dotted line shows the difference in power (watts) between the two curves in the previous graph. This should be a fair estimate for the heat loss across the liquid surface.
This curve now told me the extra rate of cooling caused by removing the lid.
If this was ALL due to evaporative cooling, then I could work out the expected loss of mass by dividing by the latent heat of vaporisation of water (approximately 2260 joules per gram) (Graph#4).
Graph#4. The calculated rate of evaporation (in milligrams per second) that would be required to explain the increased cooling rate caused by removing the lid.
Graph#4 told me the rate at which water would need to evaporate to explain ALL the cooling caused by removing the lid.
Combining that result with the data in Graph#1, I worked out the cumulative amount of water that would need to evaporate to explain ALL the observed extra cooling (Graph#5)
Graph#5: The red dashed line shows the cumulative mass loss (g) required to explain all the extra cooling caused by removing the lid. The green dashed lines show the amount of water that actually evaporated in each of the two ‘lid off’ experiments. The green data shows additional measurements of mass loss versus time from a third experiment.
In Lid-Off Experiments#1 and #2, I had weighed the water before and after the cooling experiment and so I knew that in each experiment with the lid off I had lost respectively 25 g and 31 g of water – just under 10% of the water.
But Graph #5 really needed some data on the rate of mass loss, so I did an additional experiment where I didn’t measure the temperature, but instead just weighed the mug every few minutes. This is the data plotted on Graph#5 as discrete points.
Conclusions#1
In Graph#5, it’s clear that the measured rate of evaporation can’t explain all the increased cooling rate loss, but it can explain ‘about a third of it‘.
So evaporation is responsible for about a third of the extra cooling, with two thirds being driven by heat transfer to the flowing air above the cup.
It is also interesting that even though the cooling curves in Graph#1 are very similar, the amount of evaporation in Graph#5 is quite variable.
The video below is backlit to show the ‘steam’ rising above the mug, and it is clear that the particular patterns of air flow are very variable.
The actual amount of evaporation depends on the rate of air flow across the water surface, and that is driven both by
- natural convection – driven by the hot low-density air rising, but also by…
- forced convection – draughts flowing above the cup.
I don’t know, but I suspect it is this variability in air flow that caused the variability in the amount of evaporation.
Conclusions#2
I have wasted spent a several hours on these calculations. And I don’t really know why.
Partly, I was just curious about the answer.
Partly, I wanted to share my view that it is simply amazing how much subtle physics is taking place around us all the time.
And partly, I am still trying to catch my breath after deciding to go ‘part-time’ from next year. Writing blog articles such as this is part of just keeping on keeping on until something about the future becomes clearer.
P.S. Expensive Mugs
Finally, on the off-chance that (a) anybody is still reading and (b) they actually care passionately about the temperature of their beverages, and (c) they are prepared to spend £80 on a mug, then the Ember temperature-controlled Ceramic mug may be just thing for you. Enjoy 🙂
Protons for Breakfast 