# Using light from the sun to heat something above temperature of the sun? **URL:** \ **Category:** Factual Questions\ **Created:** [February 20, 2015, 12:04am UTC](https://boards.straightdope.com/t/using-light-from-the-sun-to-heat-something-above-temperature-of-the-sun/713140 "2015-02-20T00:04:17Z")\ **Posts on this page:** 20\ **Page:** 1
**Author:** ![DaveBfd](https://avatars.discourse-cdn.com/v4/letter/d/c6cbf5/32.png) [@DaveBfd](https://boards.straightdope.com/u/DaveBfd)\ **Post date:** [February 20, 2015, 12:04am UTC](https://boards.straightdope.com/t/using-light-from-the-sun-to-heat-something-above-temperature-of-the-sun/713140/1 "2015-02-20T00:04:17Z")
This is a question I stole from another source. The answer given was that it was impossible. This seems counter-intuitive to me, could somebody explain what the answer is, and why? Lets use ideal circumstances of an unrealistically large magnifying glass (or many?), on a small object which absorbs most light that has no atmosphere around it, as well as any other factors which I have neglected. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [February 20, 2015, 12:22am UTC](https://boards.straightdope.com/t/using-light-from-the-sun-to-heat-something-above-temperature-of-the-sun/713140/2 "2015-02-20T00:22:42Z")
Think of it from the ant’s point of view. When the ant looks up, every direction it looks, it sees the surface of the Sun. It’s like it’s completely surrounded by the surface of a star. And hence, it heats up to the temperature of the surface of a star. ---
**Author:** ![Arkcon](https://avatars.discourse-cdn.com/v4/letter/a/96bed5/32.png) [@Arkcon](https://boards.straightdope.com/u/Arkcon)\ **Post date:** [February 20, 2015, 12:28am UTC](https://boards.straightdope.com/t/using-light-from-the-sun-to-heat-something-above-temperature-of-the-sun/713140/3 "2015-02-20T00:28:46Z")
Well, the reason you’re having a problem with this question is that you’re mixing up some of the terms. And like you’ve already said, leaving some out. Consider. The sun give up some heat. You take a portion of it, and concentrate it on a small area. How can you have more, when you took a portion? You mean you have more temperature per surface area? Well, sure, but not more heat, if you follow. There was this old children’s book of experiments that had a caper where you could “Generate 100,000 pound per square inch of force.” What you did was push a pin through a cloth. The force your hand generated, per the tiny surface area of a pin point, generated the result. I always though that experiment was lame. It still doesn’t seem to me to teach anything. Is it any wonder I get just as confused in this older thread as you are in this thread? > **[Winter Garbage](https://boards.straightdope.com/sdmb/showthread.php?t=737634)** > > I read, on the usual source, that that sun thing, in its core, produces about as much heat per cubic foot as an active compost heap, so I got to wondering how big a compost heap would I need to keep my 1500sqft, fairly well insulated, single story... ---
**Author:** ![breezman](https://avatars.discourse-cdn.com/v4/letter/b/4491bb/32.png) [@breezman](https://boards.straightdope.com/u/breezman)\ **Post date:** [February 20, 2015, 12:30am UTC](https://boards.straightdope.com/t/using-light-from-the-sun-to-heat-something-above-temperature-of-the-sun/713140/4 "2015-02-20T00:30:36Z")
You might want to look at [http://en.wikipedia.org/wiki/Heliostat](http://en.wikipedia.org/wiki/Heliostat). ---
**Author:** ![DaveBfd](https://avatars.discourse-cdn.com/v4/letter/d/c6cbf5/32.png) [@DaveBfd](https://boards.straightdope.com/u/DaveBfd)\ **Post date:** [February 20, 2015, 12:57am UTC](https://boards.straightdope.com/t/using-light-from-the-sun-to-heat-something-above-temperature-of-the-sun/713140/5 "2015-02-20T00:57:48Z")
> [@Arkcon](#): > > There was this old children’s book of experiments that had a caper where you could “Generate 100,000 pound per square inch of force.” What you did was push a pin through a cloth. The force your hand generated, per the tiny surface area of a pin point, generated the result. > > I always though that experiment was lame. It still doesn’t seem to me to teach anything. Is it any wonder I get just as confused in this older thread as you are in this thread? > > [Winter Garbage - Factual Questions - Straight Dope Message Board](http://boards.straightdope.com/sdmb/showthread.php?t=737634) This analogy was very useful, and it answers my question. Thank you for stating it in this way. ---
**Author:** ![Napier](https://avatars.discourse-cdn.com/v4/letter/n/ce73a5/32.png) [@Napier](https://boards.straightdope.com/u/Napier)\ **Post date:** [February 20, 2015, 1:42am UTC](https://boards.straightdope.com/t/using-light-from-the-sun-to-heat-something-above-temperature-of-the-sun/713140/6 "2015-02-20T01:42:55Z")
You can certainly do what the OP asks, by creating electricity through solar cells and then using it to drive the right kind of heating, perhaps an arc. Doing it by heating a surface through absorption of the light, that’s impossible. I think. ---
**Author:** ![Jeff\_Lichtman](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/jeff_lichtman/32/1531_2.png) [@Jeff\_Lichtman](https://boards.straightdope.com/u/Jeff_Lichtman)\ **Post date:** [February 20, 2015, 8:00am UTC](https://boards.straightdope.com/t/using-light-from-the-sun-to-heat-something-above-temperature-of-the-sun/713140/7 "2015-02-20T08:00:47Z")
It certainly is possible. All you have to do is collect solar energy over a wide area (or over a long period of time) and concentrate it in a small amount of matter. There is no law of conservation of temperature. ---
**Author:** ![CalMeacham](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/calmeacham/32/35_2.png) [@CalMeacham](https://boards.straightdope.com/u/CalMeacham)\ **Post date:** [February 20, 2015, 2:28pm UTC](https://boards.straightdope.com/t/using-light-from-the-sun-to-heat-something-above-temperature-of-the-sun/713140/8 "2015-02-20T14:28:45Z")
There is something related to the OP’s question, though. It used to be called the _Brightness Theorem_, but the People In Charge of Confusing Names have now decreed that it is the law of _Conservation of Radiance_. It states that the Radiance (which used to be called brightness) divided by the square of the refractive index of its medium is always the same: [http://www.its.bldrdoc.gov/fs-1037/dir-009/\_1241.htm](http://www.its.bldrdoc.gov/fs-1037/dir-009/_1241.htm) > **[Radiance](https://en.wikipedia.org/wiki/Radiance)** > > In radiometry, radiance is the radiant flux emitted, reflected, transmitted or received by a given surface, per unit solid angle per unit projected area. Radiance is used to characterize diffuse emission and reflection of electromagnetic radiation, and to quantify emission of neutrinos and other particles. The SI unit of radiance is the watt per steradian per square metre (W·sr−1·m−2). It is a directional quantity: the radiance of a surface depends on the direction from which it is being observed... As the Wikipedia page (or any text on radiometry) tells you, the Radiance is the Radiant flux (power emitted in photons) per unit area of the emitter per unit solid angle it’s emitted into divided by the angle between the surface normal and the direction. See also Etendue: > **[Etendue](https://en.wikipedia.org/wiki/Etendue)** > > Etendue or étendue (/ˌeɪtɒnˈduː/; French pronunciation: ​\[etɑ̃dy\]) is a property of light in an optical system, which characterizes how "spread out" the light is in area and angle. It corresponds to the beam parameter product (BPP) in Gaussian beam optics. Other names for etendue include acceptance, throughput, light grasp, light-gathering power, optical extent, and the AΩ product. Throughput and AΩ product are especially used in radiometry and radiative transfer where it is related to the vie Fr... What this means is that, do what you will in a passive ootics system (one that can’t add energy to the system), if the index of the emitter and receiver are the same, you can’t get more than the radiuance of the emitter (You can always get _less_, by having elements that don’t perfectly transmit all the light). So your magnifying glass can’t make the radiance at the any exceed that of the sun. What you _can_ do is to change the solid angle, or the area under consideration. Suggestions of storing the power given above in the thread violate the requirement of passivity. ---
**Author:** ![iamthewalrus\_3](https://avatars.discourse-cdn.com/v4/letter/i/258eb7/32.png) [@iamthewalrus\_3](https://boards.straightdope.com/u/iamthewalrus_3)\ **Post date:** [February 20, 2015, 5:11pm UTC](https://boards.straightdope.com/t/using-light-from-the-sun-to-heat-something-above-temperature-of-the-sun/713140/9 "2015-02-20T17:11:43Z")
> [@Chronos](#): > > Think of it from the ant’s point of view. When the ant looks up, every direction it looks, it sees the surface of the Sun. It’s like it’s completely surrounded by the surface of a star. And hence, it heats up to the temperature of the surface of a star. I’m pretty sure this is the wrong way to think about it. If you collected all the light that was output from, say, one square foot of the sun’s surface and concentrated it on a well-insulated square inch of the Earth’s surface, that square inch would heat up to much hotter than the surface of the sun. The energy radiated from the surface of the sun is a function of its temperature. For the square inch to reach an equilibrium temperature, it would have to radiate as much energy as is being put into it (144 times as much as the sun). If it’s outputting energy the same way the sun is (black body radiation), it would have to get much hotter. ---
**Author:** ![74westy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/74westy/32/3950_2.png) [@74westy](https://boards.straightdope.com/u/74westy)\ **Post date:** [February 20, 2015, 6:39pm UTC](https://boards.straightdope.com/t/using-light-from-the-sun-to-heat-something-above-temperature-of-the-sun/713140/10 "2015-02-20T18:39:01Z")
> [@Jeff\_Lichtman](#): > > It certainly is possible. All you have to do is collect solar energy over a wide area (or over a long period of time) and concentrate it in a small amount of matter. There is no law of conservation of temperature. > [@iamthewalrus\_3](#): > > I’m pretty sure this is the wrong way to think about it. > > If you collected all the light that was output from, say, one square foot of the sun’s surface and concentrated it on a well-insulated square inch of the Earth’s surface, that square inch would heat up to much hotter than the surface of the sun. The energy radiated from the surface of the sun is a function of its temperature. For the square inch to reach an equilibrium temperature, it would have to radiate as much energy as is being put into it (144 times as much as the sun). If it’s outputting energy the same way the sun is (black body radiation), it would have to get much hotter. You both refer to collecting and concentrating. I thing the OP is disallowing any process that introduces energy into the system. So you can collect and concentrate with lenses, mirrors and any other passive system you want but the second law of thermodynamics will not allow you to use the sun to heat an object to a temperature hotter than the sun that way. If the object in question was hotter, it would cool itself by heating up the sun. ---
**Author:** ![dracoi](https://avatars.discourse-cdn.com/v4/letter/d/90db22/32.png) [@dracoi](https://boards.straightdope.com/u/dracoi)\ **Post date:** [February 20, 2015, 6:52pm UTC](https://boards.straightdope.com/t/using-light-from-the-sun-to-heat-something-above-temperature-of-the-sun/713140/11 "2015-02-20T18:52:14Z")
No… I’m with the walrus on this one. Black body radiation alone, without any magnifying glasses or lenses, would never do more than bring the objects to equilibrium because at a certain point, they’re both radiating the same amount of energy. But black body radiation is a function of surface area as well as temperature. (From [Wikipedia](https://en.wikipedia.org/wiki/Black-body_radiation#Stefan.E2.80.93Boltzmann_law): “The Stefan–Boltzmann law states that the power emitted per unit area of the surface of a black body is directly proportional to the fourth power of its absolute temperature”). If you can use a lens to take a square foot of the sun and focus it on a square inch of the Earth, the square inch will have to be much hotter before it can reach an equilibrium between incoming and outgoing energy. ---
**Author:** ![CalMeacham](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/calmeacham/32/35_2.png) [@CalMeacham](https://boards.straightdope.com/u/CalMeacham)\ **Post date:** [February 20, 2015, 6:59pm UTC](https://boards.straightdope.com/t/using-light-from-the-sun-to-heat-something-above-temperature-of-the-sun/713140/12 "2015-02-20T18:59:39Z")
> [@dracoi](#): > > No… I’m with the walrus on this one. > > Black body radiation alone, without any magnifying glasses or lenses, would never do more than bring the objects to equilibrium because at a certain point, they’re both radiating the same amount of energy. > > But black body radiation is a function of surface area as well as temperature… See my post # 8 “Radiance” is radiant flux per unit time per unit area per unit solid angle and divided by the cosine of the angle. ---
**Author:** ![GusNSpot](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/gusnspot/32/436_2.png) [@GusNSpot](https://boards.straightdope.com/u/GusNSpot)\ **Post date:** [February 20, 2015, 7:31pm UTC](https://boards.straightdope.com/t/using-light-from-the-sun-to-heat-something-above-temperature-of-the-sun/713140/13 "2015-02-20T19:31:42Z")
I am with the walrus also. I do have a question. With the best lens humans can make today, what is the percentage of energy of the square foot of sun that is actually being concentrated? As a kid, I used a fair dinkium sized magnifying glass to burn/blow up stuff and I know I was not using the total surface of the sun. Sun rays are parallel as far as is practical for us so, with loss caused by the atmosphere, is not what I am using just a percentage of the power in that small diameter? :dubious: With a 10 sq in glass, I am getting only a percentage of the power that the glass can see/utilize and then the losses of the imperfect glass get added, so do I have enough power on a spot 1/10th of a square inch, to get the target over the point source temperature of the sun? 😕 Not what the OP really asked but I was wondering how hot the spot I made as a kid in the instant of maximum temperature? 😃 ---
**Author:** ![74westy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/74westy/32/3950_2.png) [@74westy](https://boards.straightdope.com/u/74westy)\ **Post date:** [February 20, 2015, 8:56pm UTC](https://boards.straightdope.com/t/using-light-from-the-sun-to-heat-something-above-temperature-of-the-sun/713140/14 "2015-02-20T20:56:39Z")
I’m clearly out of my depth so I’ll just ask. How is the Stefan–Boltzmann law reconcilable with [the second law of thermodynamics](http://en.wikipedia.org/wiki/Second_law_of_thermodynamics#Clausius_statement)? ---
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [February 20, 2015, 9:51pm UTC](https://boards.straightdope.com/t/using-light-from-the-sun-to-heat-something-above-temperature-of-the-sun/713140/15 "2015-02-20T21:51:51Z")
> [@74westy](#): > > I’m clearly out of my depth so I’ll just ask. How is the Stefan–Boltzmann law reconcilable with [the second law of thermodynamics](http://en.wikipedia.org/wiki/Second_law_of_thermodynamics#Clausius_statement)? Why would you think it wouldn’t be? The Clausius statement that you link to is true for the net exchange of energy. ---
**Author:** ![74westy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/74westy/32/3950_2.png) [@74westy](https://boards.straightdope.com/u/74westy)\ **Post date:** [February 20, 2015, 9:58pm UTC](https://boards.straightdope.com/t/using-light-from-the-sun-to-heat-something-above-temperature-of-the-sun/713140/16 "2015-02-20T21:58:56Z")
> [@naita](#): > > Why would you think it wouldn’t be? The Clausius statement that you link to is true for the net exchange of energy. So can a body with a large surface area transfer heat to a hotter body with a small surface area? ---
**Author:** ![iamthewalrus\_3](https://avatars.discourse-cdn.com/v4/letter/i/258eb7/32.png) [@iamthewalrus\_3](https://boards.straightdope.com/u/iamthewalrus_3)\ **Post date:** [February 20, 2015, 11:32pm UTC](https://boards.straightdope.com/t/using-light-from-the-sun-to-heat-something-above-temperature-of-the-sun/713140/17 "2015-02-20T23:32:01Z")
> [@74westy](#): > > So you can collect and concentrate with lenses, mirrors and any other passive system you want but the second law of thermodynamics will not allow you to use the sun to heat an object to a temperature hotter than the sun that way. If the object in question was hotter, it would cool itself by heating up the sun. Hmm. If you set up mirrors/lenses/whatever to channel radiation from the sun to your object, then as soon as your object got up to the temperature of the sun, there’s no more net heat transfer, since the object would be radiating as much energy _back_ through those same mirrors/lenses/whatever to the sun as it was receiving. Maybe? I’ll have to think about this some more. ---
**Author:** ![dracoi](https://avatars.discourse-cdn.com/v4/letter/d/90db22/32.png) [@dracoi](https://boards.straightdope.com/u/dracoi)\ **Post date:** [February 20, 2015, 11:49pm UTC](https://boards.straightdope.com/t/using-light-from-the-sun-to-heat-something-above-temperature-of-the-sun/713140/18 "2015-02-20T23:49:49Z")
> [@iamthewalrus\_3](#): > > Hmm. > > If you set up mirrors/lenses/whatever to channel radiation from the sun to your object, then as soon as your object got up to the temperature of the sun, there’s no more net heat transfer, since the object would be radiating as much energy _back_ through those same mirrors/lenses/whatever to the sun as it was receiving. > > Maybe? I’ll have to think about this some more. No, because the smaller patch would have to radiate more energy per unit of area. You can only radiate more energy per unit of area if you at a higher temperature. There certainly is an equilibrium point between the two objects, but in the given example (using lenses to focus light), the smaller object must have a higher temperature in order to reach equilibrium. ---
**Author:** ![rsa](https://avatars.discourse-cdn.com/v4/letter/r/839c29/32.png) [@rsa](https://boards.straightdope.com/u/rsa)\ **Post date:** [February 21, 2015, 2:52am UTC](https://boards.straightdope.com/t/using-light-from-the-sun-to-heat-something-above-temperature-of-the-sun/713140/19 "2015-02-21T02:52:50Z")
Put your object inside a greenhouse. Wouldn’t the object absorb the incoming energy as well as retaining the infrared heat that is trapped within the greenhouse? The object would heat up faster than heat could be radiated out from the greenhouse glass, no? ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [February 21, 2015, 4:06am UTC](https://boards.straightdope.com/t/using-light-from-the-sun-to-heat-something-above-temperature-of-the-sun/713140/20 "2015-02-21T04:06:09Z")
If the glass can trap heat, then the glass itself can also radiate. And the Sun’s rays are _not_ parallel: They vary by as much as half a degree. [Next page](https://boards.straightdope.com/t/using-light-from-the-sun-to-heat-something-above-temperature-of-the-sun/713140.md?page=2)