Using light from the sun to heat something above temperature of the sun?

Factual Questions
21

Yes, but the glass is much more transparent to the incoming visible range light than the resulting infrared radiation from our “object”. Once the temperature of the glass reached equilibrium with the interior temperature there would be no additional heating within the greenhouse but the temperature of the object within the greenhouse should be higher than one in the open.

I know this is probably wrong, but I don’t quite see why. Perhaps because infrared radiation has less energy that visible light?

22

You’re assuming that the resulting radiation from the object would be infrared. At low temperatures, this is true, but as the object approaches the temperature of the surface of the Sun, the re-radiated radiation would approach visible.

23

:confused:

If you get far enough away from the sun and are measuring perpendicular, at what distance does it get to ½°?

For my practical application across a 6" lens, at the sun to earth distance, what would the angle between one side to the other of that lens and how much would it change the practical application of the experiment in my question? The math?

24

The apparent size of the sun in the sky is about 1/5°. If the sun’s rays were parallel, the sun would look like a point.

25

1/2, not 1/5.

26

Yes, thank you.

27

Yes, that is true but you have not told me how many rays are actually going through the magnifying glass I am holding in my hand.

I was always taught that for all practical purposes, I can assume they are parallel when hitting my magnifying glass.

So what conversion factor do you use to determine the amount of rays that are hitting such a small area because they are not actually parallel?

You all have the knowledge to answer my original question but won’t do it. Why? Why this hang up because I stated something wrong. So tell me how to do it, don’t just ignore me, at least tell me you will not answer my question so I don’t ask it here anymore. ::: sheesh ::::

28

I’m going to guess that the sun’s rays are not parallel because the light could be emitted from anywhere on the surface of the sun. I can’t imagine how they could be parallel unless they are somehow aimed at the earth through some unknown mechanism.

29

Why does it have to be passive? Why does it have to be optical at all?

Millions of years ago plants collected energy from the sun through photosynthesis. They died and became coal, oil, and natural gas. Is energy from coal, oil, and natural gas not ultimately energy from sunlight? Very old sunlight, sure, but sunlight.

How about hydroelectric? Water evaporates from a lower elevation and precipitates on a higher elevation. That elevation difference stores potential kinetic energy from sunlight which is converted to electricity through the use of turbines. Sure, it’s electrical energy, but it comes from sunlight.

Photovoltaic? Wind?

What energy do we have access to that isn’t ultimately from sunlight? Nuclear? Tidal? Geothermal?

30

The photosphere, the visible surface of the sun is in the 6000°K range, but it is also less than one percent the density of earth air at sea level. So, how do you measure that heat as compared to a patch of dirt? To have an comparable temperature, the dirt would have to be much, much hotter than the sun, because it is denser.

31

I think you mean the dirt would have to contain more heat energy because it is denser than the surface of the sun. That depends on how much dirt you’re talking about. The photosphere of the sun isn’t dense, but there’s a huge amount of it. The amount of heat energy in an object is its specific heat times its temperature times its mass.

According to this, the sun’s diameter is about 1,392,000 kilometers, and the photosphere is 500 kilometers thick. This means the volume of the photosphere is about 3.04 * 10^24 cubic meters. Wikipedia says the density of the photosphere is about 2 * 10 ^ -4 kilograms per cubic meter, which means the total mass of the photosphere is about 6 * 10 ^ 20 kilograms.

The photosphere is mostly made up of hydrogen and helium plasma. I can look up the specific heats of these elements, but I don’t know if those numbers can be used to figure the specific heat of a plasma. Maybe someone here knows how to do that.

Now let’s consider the patch of dirt. Soil varies in density, but here’s a source that says a cubic meter of rammed soil has a mass of about 1600 kilograms. The mass of the photosphere is about 3.75 * 10^17 times the mass of a cubic meter of densely packed soil. If the temperature of the soil were 6000 K (the same as the photosphere), the specific heat of the photosphere would have to be ridiculously low for it to have less heat energy that the soil.

To summarize, the photosphere may not be dense, but it’s really big. Its size means it contains a huge amount of heat energy.

32

Or if the sun was a point source, infinitely distant - neither of which can actually happen of course - but in school physics, it’s not uncommon to find that simplification being prescribed to remove confounding factors in some example or other.

33

One of the reasons folks aren’t answering this directly is it’s a more complex question than it at first appears. IOW, easy to ask, hard to answer. And part of the challenge is the common habit many folks have of confusing power, energy, heat, and temperature. They’re four very different, albeit related things. To work this out, we need to start with energy, work our way through heat, then finally arrive at temperature. It goes something like this:
The sun applies energy to the top of the atmosphere. The intensity per unit area depends on the grazing angle; it’s more square-on in the tropics and more grazing at the poles. So where you’re standing are on Earth is a big factor.

The atmosphere then absorbs / attenuates some percentage of the energy until it gets to the surface where you’re standing. That also depends on the grazing angle and hence your location. And the clarity or obscured-ness of the atmosphere, e.g. clouds, dust, water vapor, particulates.

Then we have to account for what percentage of the solar energy is within the frequency band the magnifying glass refracts. IOW, it doesn’t do much with UV or IR, and doesn’t focus insanely high frequency solar X-rays or low frequency RF noise even a little bit.

Then we have to account for the losses in the glass itself.

Then we have to account for how large the lens is versus how large is the focus spot, and how accurately you keep the focus spot targeted on your aim point over time.

Finally, that gives us the effective average rate of energy deposited on the target area. With *energy *established, now we’re ready to start talking about heat.
First we need to consider the energy capacity of the target material . Some materials can hold little heat, and injecting only a small amount of energy produces a big gain in temperature. Others can hold a lot of heat; this means injecting a lot of energy causes relatively little temperature change. The technical term is “specific heat”.

Once we know how susceptible our target is to energy, we can begin to introduce temperature as the result of energy over time >> heat stored >> temperature.
In the temperature range we’re talking about, heat is the net result of energy injected versus energy emitted over time.
So the first factor is the reflectivity of the target. Energy reflected isn’t adding energy to the target.

Next we need to consider conduction & convection. At the target warms up, it’ll try to conduct heat away into the Earth, and convect heat away by warming the adjacent air which then rises and is replaced by cool air. So some of the incident energy is spent warming things near the target even if 100% of the energy of your spot is originally incident on the target. For round numbers, the quantity of heat loss is proportional to the diference between the temperature of the target and the surroundings.

Your target also radiates. That is, energy is lost directly as EMR. The rate of radiation also increases as the target absorbs energy, contains more heat, and the temperature increases. e.g. A white-hot target is radiating a lot more energy than a red-hot target.
Ultimately, as the target absorbs heat and temperature increases, the rate of all three forms of heat loss increases too. Eventually you reach an equilibrium where the rate of energy deposition equals that of energy loss. At that point net heat change is zero and the temperature stops increasing.
There are also non-linear interactive effects …

You can see that as the target gets hot, it might change reflectivity. Nothing is 100% reflective and once warmed enough, the surface may start to char which reduces reflectivity, increases absorption, and will move the equilibrium point higher, resulting in greater temperature.

But … once your target starts smoking, the smoke partly obscures the incident energy. Energy spent heating smoke isn’t heating the target. So this will move the equilibrium point lower, resulting in lower temperature. This is one of the hard problems with laser weapons against slow-moving targets or targets in a vacuum.
All in all, as you can see there’s a lot going on. The early folks’ comments that the temperature of the Sun forms an upper limit to the temperature of your target are spot on. But that’s just an upper limit; the actual temperature achieved is going to be less.

34

See? Now that was not so hard was it? :smiley: :::: flee :::::

Thank you very much for the trouble, time, & effort you had to go to to give me a very understandable answer.

( to go to to :smack: )

BUT, don’t you just hate that word?

The question that I really want answered that I failed to explain in such a way as for anyone to understand , is this… ( if it makes no sense to you I will attempt to draw it out. )

At aprox 93,000,000 miles or what ever unit of measure makes the math the easiest for you ( when it gets to exponents, I fall off the wagon anyway ) from the outer most point of the Sun’s fuzzy gas , plasma, what ever to the outermost point of the earths atmosphere is XX,XXX,XXX distance.

We now build a 1mm ( actual number to make the math easiest is up to you) perfectly straight tube made of the least reflective material there is for the visible light spectrum that goes up to that altitude above the earth. ( perfect material if that simplifys the math ) which is closed at the bottom so no light can get in at that end.

Place the tube with target at the bottom of the tube in the exact position on the equator, aimed perfectly straight at that spot we determined on the sun and make our observation at the instant the sun is perfectly over the tube.

Question one. Since we are looking only at the visible light spectrum and we know the exact size of the target getting hit at that perfect time, how many rays are we talking about?

Question two. The one I am most curious about. How big is the patch of the Sun that we are seeing? Actual, or % or % compared to our target area?

I am not really concerned about power, temperature, rate of transfer or anything except the size of the part of the Sun that the target sees.

So, the spot on the sun is 1 millions times bigger, one thousand times bigger, 93 million times bigger, 93 times bigger, only 1,2,3,4,5 rays total? How big is a ray or a photon?

Do what ever is the easiest math, geometry, or whatever for you.

If it is only a %, or a ratio I will be fine with that.

Do you understand what I am asking?

Can it be calculated?

Just thought of question #3.
How small would the hole have to be, even if we can not actually build it, but technically be still open that light can not get in because the rays or photons are too big? Just a hypothetical I understand. Does a photon or a ray actually have a cross section?

My last question on the subject in this thread, promise.

Or you can just tell me to take a flying leap… :smiley:

35

I’ve got jury duty on Monday, oh joy. It was supposed to be one of my days off, so I’m not even skipping work for it. C’est la guerre.

I’ll tackle this when I get done. You’re asking good questions that deserve decent answers.

36

If the tube is perfectly straight and 1 mm across, then you’re looking at a 1 mm across piece of the solar surface. Needless to say, this is far less than the total surface of the Sun, or even than the fraction of the surface we see without the tube.

And I don’t know what you mean when you ask “how many rays”. One can mathematically draw an infinite number of rays through any finite-sized tube.

37

Cool, I will be watching… Thanks. :slight_smile:

38

“then you’re looking at a 1 mm across piece of the solar surface.”

If I look at a horse through a straw and can see the whole head, am I not getting more than a straw sized look at the horse & if I walk up to the horse and draw a ring the size of the straw, that will be much smaller than what I can see of the horse from the original distance I was. Say it is 15% of the shape of the horse.

Now I do the same math for the size of the piece of the Sun I would be seeing through my long tube as I asked up thread. What % of the Sun am I seeing?

“One can mathematically draw an infinite number of rays through any finite-sized tube.”

So Sun light is not a wave and a stream of photons? Or particles? or a ray?

IIRC, I was taught that is acted in both ways. Am I using the wrong term?

Photons can be placed side by side in infinite numbers and they will all fit through the smallest hole we can mathematically come up with at the same time? If photons can do this, then what is a photon, something with no substance?

:confused:

Are they smaller than an electron? Smaller than a proton? Or do photons not have a physical dimension? If they have no size, then light should be able to go through anything? Well, that ain’t right I was led to believe.

So, is there not a mathematically sized tube with an hole that light can’t get through even if it hits the tube perfectly? This is a side question.

All I really want to know is what % of the Sun I can see with a very small tube that from here to there that I can see through mathematically under the conditions I postulated in the question LSL Guy & I were talking about just above.

See post #33 & #34 in this thread.

I know, I know but I would really like an answer to this question and what better place to ask then the SDMB?

Sun light, particle or wave

39

Have fun on jury duty. I always have a blast or at least a good game of bridge if I don’t get picked. :smiley:

40

If you’re looking through a straw, then you have to take into consideration the length of the straw. Since you gave no length it was reasonable for Chronos to assume you ment a tube of sufficient length compared to width to make the length irrelevant. And in that case you are looking at a spot 1 mm wide.

How long is your tube?

A ray is not the same term as a photon, in fact they are pretty much incompatible and showing light behaves the same for either is really complicated.