Light and Heat
Why does the balloon look dark?
Unless otherwise noted the artwork and photographs in this slide show are original and © by Burt Carter. Permission is granted to use them for non-commercial, non-profit educational purposes provided that credit is given for their origin. Permission is not granted for any commercial or for-profit use, including use at for-profit educational facilities. Other copyrighted material is used under the fair use clause of the copyright law of the United States.
Image from National Geographic
The answer will help us understand why Los Angeles County, CA decided to dump tons of hollow black plastic “shade balls” into their reservoir to save water. All the internet sites that reported this had comments in the comments sections (where the “trolls” operate) wondering why the balls are black. Black objects get hotter in the sun than lighter colored objects. Wouldn’t that defeat the purpose? As we think about this, think about the implications of the name of these balls: “shade balls”.
Part of the explanation for the color is that the balls are coated with a carbon-based material to keep the plastic from breaking down into dangerous organic compounds in the city’s water supply. The black color keeps the light that breaks plastic down off the balls, extending their life-spans. But this is only part of the answer. Black would be a good choice of color anyway. It is clear from reading the comments sections of the webpages that most of the participants don’t really understand the way heat is made and budgeted on Earth. Of all the electromagnetic energy that leaves the sun only about a quarter ever reaches the surface. Of that energy a tiny bit is UV (ultraviolet), almost half is visible light, and a little over half is infrared (IR, or heat). All those wavelengths are absorbed by the Earth – rocks, soil, water, vegetation, etc. Then something important for understanding the Earth’s heat budget happens.
Remember that a red balloon submerged in water looks darker than it does in air, and a pink hand looks whiter. We left this as a question to be answered later and now is later – it’s time to think about it. Why do we see these differences?
The nice fleshy pink tan of the hand that is obvious at the surface is missing at depth. The hand looks
very pale. (In fact, in general almost everything looks more “bleached out” deeper in the water.
Furthermore the balloon is a much darker red at
depth than at the surface.
Why?
Here’s another related experiment This is my brand new bright red shirt. Notice that it is roughly the same color as the balloon.
Photo by Chelsea Carter, © Burt Carter
Here it is wet. This picture is just to calibrate the color when the shirt is wet. Notice that there isn’t a lot of difference in color above the water.
Photo by Chelsea Carter, © Burt Carter
If I dive to ~2m (~6 feet) deep things change. Notice that the shirt now looks significantly darker in color. It can’t just be the fact that it’s wet – we’ve calibrated for that already.
Photo by Chelsea Carter, © Burt Carter
In fact, the shirt at 2m looks about like the balloon did at 3m, maybe a little darker.
Photo by Chelsea Carter, © Burt Carter
At ~3m (10 feet) deep the shirt looks even darker – almost black at this depth. What is happening? Why does the shirt get darker with depth? (And why, at the same depth, isn’t the balloon as dark as the shirt?)
Photo by Chelsea Carter, © Burt Carter
Photo by Chelsea Carter, © Burt Carter
Notice that the bottom of the pool looks about the same in both the shallow end and at the deep end in 1m and 3.2m respectively. Why doesn’t the blue get darker like the red does?
The photographs below show a light meter under the same lights in a lighted room. They were taken within seconds of each other. The only physical difference in the setup is the red filter covering the light meter in the right-hand picture.
There is clearly a difference in the readings – the red light is recorded at around 1/3 the level as the unfiltered light – 235 lux vs 661 lux.
What happened to the missing 426 lux that didn’t reach the sensor through the red filter?
(And can it be the same thing that happens to the red light in water?)
Let’s remind ourselves of something. Why does the balloon look red? Sunlight is white, not red. When we look at the balloon all we see is red. What happens to the
other colors in the sunlight?
Well?
What does it mean to say that the red filter or the water “absorbs“ the red light, or that the red ball absorbs the non-red wavelengths? When something absorbs light, what happens to it? Does it simply cease to exist? If so, where does it go? (Here’s a related question that has nothing to do with the present topic of climate, but everything to do with the nature of science. When a meteor enters the atmosphere people say that it is “vaporized”. What does that mean? Again, does it cease to exist? If so, where does it go?) (Here’s another. When you flush the toiled or put the garbage in the roadside bin for pickup, does it cease to exist? Where does it go?)
One of the most basic rules in the universe is that matter and energy can neither be created nor destroyed. Nothing of matter, nothing of energy can simply cease to exist – it must “go” somewhere”. Only time and life go away for good and cease to exist. Don’t waste them. The light that is missing in our examples, that absorbed by the balloon or the water or the red filter, is a case in point. It must still exist. It clearly doesn’t exist as light any more, but it can exist in a different form. What might that be?
1) The red shirt darkens to near black with increasing depth when seen from the surface. It looks almost black when photographed through 3m of water.
2) The blue pool bottom does not. It looks pale blue at all depths. (If anything it is lighter in deeper water.)
3) The red balloon also appears to darken with depth, even when photographed at depth rather than at the surface.
4) In fact, the red balloon looks only slightly darker photographed at 3m than the shirt does photographed through 2m of water. -- (Notice how I worded that.)
5) Remember that the hand apparently lost it’s pinkness when photographed at 3m.
There is something about redness and water that we need to consider!
(Blueness and water don’t act the same way, at least within 3m of depth.)
Back to the shirt (a recapitulation):
The reason that most solar radiation doesn’t reach Earth’s surface is that it is partially reflected back to space (a small amount) and partly absorbed by atmospheric gasses. What does this mean? When an atom or molecule “absorbs” light, the energy carried by that light excites the atom (or atoms, in a molecule) causing the electrons to “jump” to higher orbital levels. In some cases, in a part of the atmosphere called the ionosphere the valence electrons are actually knocked off the atoms and the atoms are thereby ionized. It takes high-energy cosmic and x-rays to do this and it happens in the upper part of the atmosphere. Of course this can’t last forever. Electrons don’t stay in the higher orbital shells. If they did it wouldn’t take long for all of them to be as high as possible and energy absorbance would cease. Neither can the energy simply disappear or cease to exist. Instead, the energy is quickly re-radiated when the electron falls back into its proper position. It is not, however, radiated as light, but as heat – IR radiation. If the energy originally was absorbed was IR it is still radiated as IR, only at a longer wavelength/lower energy. Air molecules and atoms convert light to heat.
This is the point that all the commentators are trying to make. The black balls on the water’s surface absorb all the solar radiation that hits them and convert it to heat. They are black because
they absorb every wavelength and preferentially reflect none. In LA in the summertime this can be a lot of energy. The balls get really hot, like the street does or like a black steering wheel or
bucket seat. Wouldn’t all that heat make the evaporation rate go up???
Image from ABC News
There are a couple of reasons why not, and they both have to do with what is (and isn’t) going on beneath the balls.
1) There is not a single layer of balls, but rather multiple layers. Sure the top layer of balls gets very hot, and sure that layer radiates all of that heat away. However, it is far easier, for a variety of reasons, for that heat to be radiated into the atmosphere rather than into the next deeper layer of balls. That next layer does get some radiant heat from above, and probably is also partly exposed to the sun, thereby directly absorbing some energy. But it will be a smaller amount than the top layer and when it radiates it will also preferentially radiate upward rather than downward. Each layer of balls shades the underlying layer and protects it from absorbance.
2) More importantly, the balls shade the water surface. No light or IR or UV reaches the surface of the water, therefore none can penetrate it!
Water is much better at absorbing light than air. Whereas it takes hundreds of kilometers of air (most of it admittedly very thin) to absorb half the light and IR coming from the sun, the rest is gone within about 200m of water, even in perfectly clear water. All that energy concentrated in the top thin layer of water is converted to heat, which is therefore much more densely produced in the water than it is in the overlying air. This is why it’s important to shade the water’s surface from the sun – to keep the light out of it.
Different wavelengths are absorbed more or less readily by water, meaning that the penetrate to different depths. This diagram shows the depths at which 99% of the energy of various wavelengths have been absorbed by seawater.
UV IR
WATER SURFACE 3m 4m
25m
51m
131 m
254 m
107 m
25m
NOTE: At about 200m you would probably barely have enough light to see your hand in front of your face.
Modified from data in Garrison and Ellis, Oceanography, 9th ed., Cengage, p. 187.
Before we go back to our original question there is one more to deal with. Why is shirt red? That is, why does it look red to us?
It looks red because it is reflecting the red wavelengths of sunlight toward us. What about the other wavelengths? What happens to them?
The shirt absorbs the other wavelengths (including IR and UV) and converts them to heat.
Photo by Chelsea Carter, © Burt Carter
In the pool it does the same thing: absorbs everything but red and reflects that toward our eyes. But since the red light has been mostly absorbed by the time it passes through almost 6m of water (down and back, remember) we see very little red light and interpret that as darker.
The balloon at ~3m does not look so dark because it was photographed with a camera at 3m. The red was not absorbed through two directions of travel (3m down and 3m back, for ~6m total) but only through one direction – 3m total. Thus less red light had been absorbed. When the camera “saw” the balloon.
Similarly, that’s why, at the same depth, The balloon doesn’t look as dark as the shirt – the red light is absorbed through only one trip through the water if the camera is also at 3m with the balloon and through two trips if it has to go down, reflect off the shirt and come back to the camera.
Photo by Chelsea Carter, © Burt Carter
The heat in Earth’s atmosphere that drives weather and controls climate is, for the most part, not solar heat absorbed directly by the air. Instead, it is heat radiated to the air from below – from the Earth beneath that absorbed the solar radiation of all wavelengths and converted it all to heat. The equatorial climate is warmer than the polar climate, and in general, climates get cooler toward the poles. As we’ll see, this is because there is more heat for the air to absorb at the equator, very little at the poles, and a gradient of available heat in between. Why should this be so?