So what is the temperature?‏

I am trying to figure out what the temperature would be if I directed the sun light 500 times in one pin-point spot (with 500 mirrors of course) when the outside temperature is 80 degree F. What it the temperature at the pin-point spot?

Without getting too technical, is it:

500 * 80 = 40,000 degrees F

This can’t be and seems unlikely, am I missing something? Iron melts at around 875 C. to 900 C. At 40,000 degree C, the iron should turn into a gas.
Is there a different temp to calculate from the temp. outside to the temp. of a single sun-ray over the surface area of the mirrors being used?
What is the temp. of a single ray of sunshine?

The mirrors are 1/2" x 1/2" sq. mounted on a old satellite dish (like an inverted disco-ball). At about 5 feet out from the dish, centers the very hot spot. At the hot-spot another mirror 12" x12" sq. is angled downwards and directs the heat to my backyard pool and heats the water really fast. I have always wanted to know what the generated heat temperature is. Anybody’s help would be gladly appreciated?

Thank you for your interest, and time.

Temperature doesn’t really work like that. It isn’t a measurement of heat, but rather a measure of properties typically associated with common notions of heat.

As to the mirrors, I recommend this Mythbusters episode:

You can’t simply just multiply air temperature like that (try converting to Celcius first, multiplying, and converting back to Fahrenheit and see if you get the same answer). If you want to get a better answer, you’re going to have to multiply your local solar insolation (Watts/m^2) value by the total area of mirrors you have directed on your point source in order to find out how much power you’re putting into the object, and then work out a heat-transfer calculation to see at what temperature everything stabilizes at. Not exactly an easy task.

Thanks, I’ll check it out

The air temperature on the day is largely irrelevant. The material and heat transfer properties of the substance you are heating are relevant.

Let’s say the sunshine is delivering 2 kW per square metre to your backyard (actual figure depends on where you live and time of day). You take a perfectly reflective dish of one square metre and concentrate the sunlight by a factor of 500 (to an area of 20 square centimetres) onto some surface. That 20 square centimetres of surface is now receiving 2 kW of heat, or 1000 kW/m[sup]2[/sup].

What temperature will it reach? That depends on all sorts of unknowns - its thickness, its heat capacity, its colour, the temperature and heat capacity of its environment, etc. Basically it will heat up until it is at a temperature where it is losing 2 kW of heat (by conduction, convection and radiation) to balance the amount of heat coming in. The iron could well become too hot to touch, but it won’t vaporise!

Now, instead of reflecting the heat onto a surface, you are reflecting it into your pool. Effectively, in that case, you are adding 2 kW of heat to the pool, over and above the 2 kW per square metre of water surface that was being added to the pool anyway.

Thanks Yamato, but I’m a little slow, already started drinking, my 500 mirror squares are 1/2"x1/2" sq. each (I don’t know what you mean by “Local Solar insolation” ) that I can figure out, but where do I get the value of watts to /m of the power of 2? If this is really not an easy task, just tell me if I need to start drinking heavily. Thanks again.


How do I find out the wattage per 1/2"x1/2" sq.silver backed tile? I know you were just making an example in you’re post, but is there a cheap way to find out?

Use 500W/square meter. That’s probably a little low, but close enough. If you were to concentrate 1 square meter of sunlight 500x, you would get 10,000W/square centimeter.

Someone linked to a site where someone had done this, and was melting/burning a bunch of stuff. Is that your site? That was pretty cool.

The Sun’s surface temperature is about 5800K, or about 5500 C or 10,000 F, so those are your upper limits. I tried a simple average based on the Sun’s solid angle, but got 315 K or 107 F, which is obviously way too low.

On preview: to use the 2K Watts/square meter, multiply that by the area of the reflecting part of the dish in square meters. 500 1/2 inch square tiles would have an area close to 1 [del]square meter[/del] square foot, so you’d get around 200 Watts.

If you are using mirrors and/or lens to concentrate sunlight, you are limited by thermodynamics to the temperature of the sun’s surface. So, no hotter than about 5500C. Inefficiencies will probably keep you from getting that hot.

My example of 2 kW per square metre was too high. Your figure of 500W per square metre is probably closer to reality.

However something went awry with your calculation. 500 x 500 / 10000 is only 25 W per square centimetre, not 10,000.

That’s what I get for doing it in my head…

The guy at cockeyed put mirrors on an old satellite dish.

Sunlight falling on the earth is approximately 700watts/sq m; meaning a sheet of whatever a meter square, pointed so it’s center is pointing directly at the sun, is receiving 0.7KW; at northern US latitudes, that sheet would probably be tilted about 45 degrees to get maximum sunlight.

As you know if you ever walked around barefoot, some items will heat more than others. If you concentrate 10,000 times sunlight radiant heat in one spot, that spot will start heating up.

As mentioned above, anything being heated will get hotter and hotter until it reaches a steady state, where heat is lost (radiance, convection, etc.) at the same rate it is being absorbed; or a chemical state change alters the situation (it burns up or melts.) In the atmosphere, an extremely hot item will cool quite a lot just from heating the air around it. In space, with vaccuum, it may instead heat until it vaporizes.

Realistically, conduction and convection will create a steady state by about 1000 to 1500 degrees; in oxygen, almost anything will probably burn before that. Also, consider the simple incandescent lightbulb, which is giving off 100 watts, a significant amount by radiation…

As YamatoTwinkie said, you can’t multiply temperatures like that. Is 80F twice as hot as 40F? If so that means 27C is twice as hot as 4C, as those are the Celsius equivalents.

Now that we have learned this truth. We are smarter than about 90% of science reporters and TV weather men. Similar multiplying of temperatures is pretty common in what should be informed media reports.

What if the outside temperature was below zero? Multiply that by 500, and you get a really good cooling system! Just concentrate the sunlight on a cold day onto a small area and watch it get really cold. If the outside temp was -10, then we could get to -5000 degrees.

If you’re trying to get it really hot, the thermal tranport will be dominated by radiation. Then things are not too complicated. The important thing to know is that the radiant flux density per unit area goes like the fourth power of the temperature (T), measured from absolute zero. For a perfectly black object, the constant of proportionality is called the Stefan-Boltzmann constant (sigma). If the body isn’t black, you have to multiply by the emissivity (e). Thus:

Radiant flux = esigmaT^4

Now, if you could concentrate the flux density by some factor, say 500 times, the object will heat up until it radiates the same amount of energy as it recieves. Emissivity works both ways, if the body isn’t black it only absorbs a fraction e of the radiant power, so the emissivity actually cancels out of the analysis (in this simplified picture). Since the temperature of the Earth is the equilibrium temperature reached by the sunlight hitting the Earth, very very roughly, we would expect the temperature of the object to be (Earth’s temperature)*Fourth Root(500) in your favorite absolute temperature scale. Since the Earth’s temperature is ~300 Kelvin, we get that the object’s temperature is (300 K)*4.7, which is about 1400 K or roughly 1100 degrees Celsius.

If you want to do the problem more accurately, you need to calculate the actual power absorbed by your object, then equate it to esigmaT^4*(the effective area of your object). You can’t arbirtrarily increase this by using more and more mirrors. If you think about using a really big lens or mirror to grab a large amount of light and focus it to a tiny object, you know that the absorbed power is best when the object is at the focal point of the lens, but you also know that the image of the sun will have a finite size. The best way to concentrate the light is with non-imaging optics, but suffice to say, if you consider all the geometrical factors, the very best you could possibly do is to heat the object to the surface temperature of the Sun (about 5800 K). To a physicist, it is obvious that you can’t exceed this, since otherwise you could shine light back and forth and get energy for free.

Another consequence of this, by the way, is that you can’t use a lens or mirror system as a long-range solar-powered death ray. Think of it this way: With no mirror at all, you see the Sun filling some portion of the sky, and cold space filling the rest of it. This brings the Earth to a temperature somewhere in between the temperature of the Sun and of cold space. Since the Sun fills a very small portion of the sky, the temperature of the Earth is closer to the temperature of cold space than it is to the temperature of the Sun.

Now, consider the point of view of an ant that’s getting burned by a magnifying glass. If the optics are good and the ant is right at the focal point, then the ant will see the Sun filling the entire magnifying glass. If the glass is large enough and close enough, this will fill a significantly larger part of the sky than it does without the glass, so the temperature the ant reaches will be closer to the temperature of the Sun than it is without the glass. In other words, the ant will get hotter, possibly hot enough to burn up.

But now imagine trying to burn something a great distance away (much further than the size of your mirror or lens). Now, your target will still see the Sun filling up the area of your optics, but if your optics are far away, that won’t be a very large part of the sky, and might well be only as much as you’d get from the natural Sun, anyway. So your mirror isn’t going to make much difference. In order to affect something a significant distance away, you need to cover a very large area with mirrors (like, the entire hillside facing the ocean, for Archimedes’ death ray).

Wow! This has been quite an education. I truly appreciate everyone’s comments. I just wanted a rough guesstimate of the temp. at the center of the hot-spot, instead I got a lot more to think about I didn’t know before. So about 1100 degrees C.

One last question:

Not long ago I wanted to see what would happen if a put a telescope backwards at the focal point of the heat (hot-spot) and passed it through to see if it would stretch the hot-spot into a thinner hot-spot to control the stream better…however the telescope melted in seconds. I should have put it on a base instead of trying to hold it suspended with pliers. Does anyone have any ideas or theories on what will happen. I don’t have another telescope…yet, I’m still searching garage sales to find a cheap one I might be able to melt again…any thoughts?

Thanks again for everyone’s comments.