Does the colour of a container affect how fast water in it cools?

[phantomvommand]

Containers are painted with the 7 colours of the rainbow; each colour corresponds to a different wavelength. Pour boiling water into each container.
Which container would cool the fastest?
Does the result change depending on whether the exterior or the interior or both are painted?
The material of the can is tin. Google ‘tin can’ for an image of the can used.

Based on experiment, red is just slightly better than orange, orange > green > yellow > blue > purple. (Purple and blue used were quite dark)

 

[Dale]

Yes, it will affect it. A black color will make radiative heat transfer faster.

As to what approach is best, that is a little complicated. What material is the unpainted container?

 

[Baluncore]

Does the result change depending on whether the exterior or the interior or both are painted?

Yes, any surface that is involved in heat transfer is important.
A thick layer of paint will insulate the container so it will take longer to cool.
If you paint the surface with a layer that for IR has a thickness of λ/4, with a RI = √(wall material) then you will maximise the cooling rate.

 

[phantomvommand]

Yes, it will affect it. A black color will make radiative heat transfer faster.

As to what approach is best, that is a little complicated. What material is the unpainted container?

The material is tin. Thanks for this, i forgot to mention that black is not allowed. The colours used are the 7 colours of the rainbow. Based on experiment, red was just slightly more effective than orange, orange > green > yellow > blue > purple.

 

[phantomvommand]

Yes, any surface that is involved in heat transfer is important.
A thick layer of paint will insulate the container so it will take longer to cool.
If you paint the surface with a layer that for IR has a thickness of λ/4, with a RI = √(wall material) then you will maximise the cooling rate.

Thanks for your reply, may I know what IR is?

 

[Baluncore]

IR = Infra Red = heat.
λ/4 = https://en.wikipedia.org/wiki/Anti-reflective_coating

 

[hutchphd]

Were these experiments carried out in bright light?. The black body energy emitted in the visible by a body at 400K is very very small fraction, so the visible color of them will matter little. They may well have different thermal transfer characteristics in the far IR as well as different thermal conductivities but the visible radiative transfer outward is not the controlling factor.
If you do the experiment in direct sunlight the incoming visible light can, and likely will, make a difference.
Also can you make an estimate of whether your results are statistically significant ?

 

[russ_watters]

The material is tin. Thanks for this, i forgot to mention that black is not allowed. The colours used are the 7 colours of the rainbow. Based on experiment, red was just slightly more effective than orange, orange > green > yellow > blue > purple.

Please provide the experimental data/setup. We can't do better than guessing about an experiment if we know virtually nothing about it.

One key issue/problem here is that the blackbody wavelength at the temperature of the experiment is far longer than the colors of the paint. So they tell us near nothing about the blackbody emissivity of the paint. It's likely that to within a few percent your paint is in fact "black" at the relevant range of wavelengths. Note, the color you see is the color it reflects; all other colors it is absorbing (and emitting; ie, "black") and it is likely "black" to infrared as well. An infrared thermometer can confirm that.

 

[phantomvommand]

Please provide the experimental data/setup. We can't do better than guessing about an experiment if we know virtually nothing about it.

One key issue/problem here is that the blackbody wavelength at the temperature of the experiment is far longer than the colors of the paint. So they tell us near nothing about the blackbody emissivity of the paint. It's likely that to within a few percent your paint is in fact "black" at the relevant range of wavelengths. Note, the color you see is the color it reflects; all other colors it is absorbing (and emitting; ie, "black") and it is likely "black" to infrared as well. An infrared thermometer can confirm that.

I did this experiment 2 years ago when I was in 10th grade. It was done under white light. I realize my recollection was not perfect, resulting in errors in my description. Attached below are the results and the setup.

@hutchphd , yes the results seem statistically significant.

 

[russ_watters]

What is the container sitting on? Is it under normal room lighting? What kind of lights? What time of day? How well is room temperature controlled? Was it the same can painted multiple times or different cans? Sitting next to each other or tested separately? Was the entire apparatus heated to 100C or was boiling water poured in at the start? Was any attempt made to predict or explain the results quantitatively? I notice the can is uncovered...

The results do seem statistically relevant, but there is a lot of room for error. There's also a few simple variants that could be tried to account for the errors. Like insulating and/or covering it. And trying bare tin, black paint and silver paint.

Edit: wait, the graph is presented rotated 90 degrees from what one would typically see, which makes it hard to read, but aren't the results backwards from what theory would predict? What exactly was the prediction and conclusion here?

 

[phantomvommand]

What is the container sitting on? Is it under normal room lighting? What kind of lights? What time of day? How well is room temperature controlled? Was it the same can painted multiple times or different cans? Sitting next to each other or tested separately? Was the entire apparatus heated to 100C or was boiling water poured in at the start? Was any attempt made to predict or explain the results quantitatively? I notice the can is uncovered...

The results do seem statistically relevant, but there is a lot of room for error. There's also a few simple variants that could be tried to account for the errors. Like insulating and/or covering it. And trying bare tin, black paint and silver paint.

Edit: wait, the graph is presented rotated 90 degrees from what one would typically see, which makes it hard to read, but aren't the results backwards from what theory would predict? What exactly was the prediction and conclusion here?

The container is sitting on a kitchen top. I think it is marble. Yes, it is under normal room lighting, white light. I did this at night. Room temperature was around 28 degrees Celsius. There were 7 different cans, painted with 7 different colours. Tested separately, 1 by 1. The 20 degree drop, for all 7 cans, was from 70 to 50. Recording did not start until the water was 70 degrees Celsius. The can is uncovered, but the water is held in the covered calorimeter. If you are wondering why water surrounds the calorimeter, that is because the assignment was to design a can that cools water in the calorimeter by 20 degrees Celsius in the shortest possible time. Colour of can was just 1 of the 4-5 factors investigated, other factors include medium surrounding the calorimeter, starting temp of water, volume of medium surrounding water etc. Because I was testing colour, I kept the material surrounding the calorimeter (water) constant. My group was able to come up with reasonable explanations for all the other factors; colour was the only factor We couldn’t explain. The explanation we ended up giving was using Wien’s displacement law, but I myself am not convinced given the fact that the law only applies at very high temperature. (A few thousand Kelvins). Unfortunately, we did not investigate black, as black was explicitly prohibited. We thought silver was reflective and would reflect heat back into the calorimeter, so we eliminated silver too.
The conclusion is that red was the most effective colour. This conclusion is contentious. This project was a school wide science performance task that was graded, and most groups were split between purple and red, although I think more groups found red to be the most effective colour.

for the entire project, to make the most effective cooling device, we used a red can, water in calorimeter with starting temperature 90 degrees (maximum allowed starting temperature, temperatures > 90 were explicitly banned), the medium surrounding the calorimeter was vinegar, and vinegar was filled to the maximum height of the calorimeter.

 

[russ_watters]

Thanks, that's much better. Was there any prediction or conclusion explaining why red gave the best result? Odd assignment. Odd constraints. Odd that there doesn't seem to be an agreed upon conclusion.

 

[phantomvommand]

Thanks, that's much better. Was there any prediction or conclusion explaining why red gave the best result? Odd assignment. Odd constraints. Odd that there doesn't seem to be an agreed upon conclusion.

My group used Wien’s Displacement law, but I doubt we knew much about it. We just saw that at lower temperatures, the peak wavelength emitted was red, so red would be the most effective colour. Yes, lots of logic leaps, and I’m not even sure if Wien’s Displacement’s Law applies here, and that is why I am asking this question now (though 2 years late). Do you have any theoretical explanation for why either red or purple would be the most effective?

for the other factors investigated, the conclusions were similar. Groups that tested vinegar all found it highly effective.

 

[russ_watters]

My group used Wien’s Displacement law, but I doubt we knew much about it. We just saw that at lower temperatures, the peak wavelength emitted was red, so red would be the most effective colour. Yes, lots of logic leaps, and I’m not even sure if Wien’s Displacement’s Law applies here, and that is why I am asking this question now (though 2 years late). Do you have any theoretical explanation for why either red or purple would be the most effective?

Well, yeah, you can use Wein's displacement law here, but:
1. It's difficult to apply since you don't know the black-body spectrum of the paint. (Still not easy if you do)
2. You used it backward/got a wrong conclusion. The red paint doesn't emit red, it reflects red.

 

[phantomvommand]

Well, yeah, you can use Wein's displacement law here, but:
1. It's difficult to apply since you don't know the black-body spectrum of the paint. (Still not easy if you do)
2. You used it backward/got a wrong conclusion. The red paint doesn't emit red, it reflects red.

I am not sure if Wien’s law even applies here. Isn’t Wien’s law about the wavelength emitted at certain temperatures? Even if the calorimeter were at 6000K, Wien’s Law just tells us the calorimeter is burning blue, but does not say anything about the heat loss rate? Of course, the temperature is just 300K, so I really don’t think Wien’s Law applies. Is there a more plausible theory for why colour affects, or does colour even affect?

 

[russ_watters]

I am not sure if Wien’s law even applies here. Isn’t Wien’s law about the wavelength emitted at certain temperatures? Even if the calorimeter were at 6000K, Wien’s Law just tells us the calorimeter is burning blue, but does not say anything about the heat loss rate?

There's a next step or more direct method; the Stefan-Boltzmann law/equation. What you'll want to do for a first "gut check" is calculate the heat loss rate due to radiation and compare it to the observed rate. That will tell you how big of a factor the radiation was altogether and how much it could have contributed to the observed discrepancy.

You might be able to Google the or find an app that can generate the curve. It might be instructive on how much of the curve is actually missing if you remove a slice far from the peak. It is likely tiny.

Of course, the temperature is just 300K, so I really don’t think Wien’s Law applies. Is there a more plausible theory for why colour affects, or does colour even affect?

Not color itself, but polished bare metal vs paint should have a noticeable impact. If radiation has a noticeable impact, then paint should noticeably impact it as well. But the exact color shouldn't matter.

 

[Richard R Richard]

Interesting debate ...
Do you also have the results of the experiment with the lights off available?
Does it have the spectrum of intensities of each color of the light with which the experiment was done?
I am inclined that the difference in results is the percentage of each color in the total intensity of the light.

It was not clear to me which side of the container was painted, inside, outside or both?

from wikipedia https://en.wikipedia.org/wiki/Thermal_radiation#Surface_effects

Lighter colors and also whites and metallic substances absorb less of the illuminating light, and as a result heat up less; but otherwise color makes little difference as regards heat transfer between an object at everyday temperatures and its surroundings, since the dominant emitted wavelengths are nowhere near the visible spectrum, but rather in the far infrared. Emissivities at those wavelengths are largely unrelated to visual emissivities (visible colors); in the far infra-red, most objects have high emissivities.

Yes, it is under normal room lighting, white light.

The environment can alter the interior temperature, not by the emission, but by the absorption of light.
what seems strange to me (not to say impossible) is that the difference is so great in such a short time.

 

[Tom.G]

Wow! Lots of unknowns there.

The graph shows total experiment time of 7.25 minutes. The wall-clock from start to finish must have been much longer, leaving time for environmental changes.

Was the can placed on the same spot on the thermal mass (marble countertop) for each test? (If so, the cooling time would increase with each test.)

Was there forced air heating/cooling active in the laboratory room this was in?

Was the hot water always at the same temperature when placed in the can? (If no, then different temperatures of the thermal mass (marble) when measurements started.)

Were the outer surfaces of each of the cans dry? (Both paint and water.)

Those are the things that popped up anyhow, there are probably more.

Cheers,
Tom

 

[256bits]

I am not sure if Wien’s law even applies here. Isn’t Wien’s law about the wavelength emitted at certain temperatures? Even if the calorimeter were at 6000K, Wien’s Law just tells us the calorimeter is burning blue, but does not say anything about the heat loss rate? Of course, the temperature is just 300K, so I really don’t think Wien’s Law applies. Is there a more plausible theory for why colour affects, or does colour even affect?

The 'photograph' is from Wiki Wien displacement law - for black body radiation.
And if radiative power is related to the color of the object, or temperature, then we see that a blue object ( of the emmissive wavelength blue - not reflected blue ) radiates more power than a similar object of green, yellow, or red color. ( ~ T⁴ )

Since your object is only painted a certain color of the visible spectrum, it does not follow that Wien law would not apply to the color seen.

Your object I would suspect is selective - it reflects ( as stated by Russ ) the colour red of the visible spectrum, and it might not exhibit a black body radiation even at the temperature of it is at ( 300 K or so ).

Do you have any theoretical explanation for why either red or purple would be the most effective?

The conclusion is that red was the most effective colour. This conclusion is contentious. This project was a school wide science performance task that was graded, and most groups were split between purple and red, although I think more groups found red to be the most effective colour.

Pigment.

Each colour of paint applied to the can would be using certain particles of pigment to make the colour.
For example, some particles to make a pigment, might be larger in size than the particles of a pigment to make another colour - from the reflected and absorbed light - it is not an emissive colour.
We cannot be sure if the pigments are selective in the other wavelengths of light, at least at the longer emissive wavelengths associated with the temperature of the object.

If purple had the same red pigment it would be expected that purple should exhibit somewhat a similar curve as the red. This may have been borne out, as noticed by the data of the several experimenters data as mentioned. By the way, your graph has red and purple showing the extremes and not side by side, contrary to the several experimenters conclusion.

I tend to feel that the experiment studied certain aspects of pigments not related to their reflected colour, but inherent to the pigments themselves, and their ability of reflection, absorption, and emission wgt wavelength.

 

[rbelli1]

A thick layer of paint will insulate the container so it will take longer to cool.

Was it the same can painted multiple times or different cans?

If the same can did you remove the paint for each test run? Or did you apply orange paint then test. Next apply red paint over the orange paint etc. in the order of fastest to slowest cooling time?

BoB

 

[DrClaude]

Have a look at
R. A. Bartels, Do darker objects really cool faster?, Am. J. Phys. 58, 244 (1990)
https://www.physlab.org/wp-content/uploads/2016/03/Cooling_paint.pdf

 

[hutchphd]

I once built a wood stove from a shiny stainless beer keg. I knew that it would need to be blackened but fired it up before painting to test . The output was amazingly small (even to protophysicist me), until the stove got to very high temperature ( maybe some plasma frequency cutoff for the metal ?). Anyhow its behavior exactly comports with this article.

With black stove paint the stove was fine.

 

[Tom.G]

A demonstration of heat transfer in high school Science class (Many yrs ago!) made an impression.

The experiment was two tin cans filled with water and a thermometer in each. One can was painted flat black and the other was the bare, shiny metal.

There was a Heat Lamp aimed at each can identical distances. The purpose was to demonstrate the higher absorbance of the Black paint.

The result was a bit different. The Black can was cooler than the shiny one.

The instructor had made an obvious (in retrospect) mistake. The Heat Lamps were placed rather close to the two cans. As a consequence the radiant energy was hitting less than half the surface area, resulting in an increase of cans radiating surface... which cooled them fairly well!

Oh well. At least I remembered the lesson.

Cheers,
Tom