Light and Heat
Why does the balloon look dark?
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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”.
Image from National Geographic
Part of the explanation for the color is that the balls are coated with a carbonbased 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
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?
Photo by Chelsea Carter, © Burt Carter
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?
Back to the shirt (a recapitulation):
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.)
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???
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!
Image from ABC News
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
4m
25m
3m
25m
51m
107 m
131 m
NOTE: At about 200m you would probably barely have
enough light to see your hand in front of your face.
254 m
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.
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.
Photo by Chelsea Carter, © Burt Carter
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?