Can you use sun light to heat an objects surface to hotter than the surface of the Sun? [closed]

I was reading this question: Concentrating Sunlight to initiate fusion reaction and some of the comments, as well as an answer, suggest that thermodynamics second law prevents what I ask in the title. I was wondering if that is really the case because: 1) The sun and a small target (eg deuterium pellets) are not equal amounts of gas in equilibrium. 2) Even if equal amounts of gas aren't required, it's still more like focusing light from one side of the box onto a smaller region on the other side. 3) There is mass being converted into energy in the center of the sun which can flow from place to place and drive an engine.. in other words this set up wouldn't make perpetual motion or free lunch.

And here is my attempt at a calculation: The largest parabolic reflector I know about is 500 meters wide (area ~ 500*500*3.14 = 785000 m^2). The earths surface receives 1000 watts per square meter, so 785000*1000=785,000,000 J/s it was also pointed out in the other question that you can't focus onto an arbitrarily small point. So lets dump the energy into a cup of water no? Specific heat of water: 4.18 joules per gram per degrees c. grams, one cup of water weighs about 236 grams

So if we throw all those together we get: (785,000,000 J/s)/[(4.18 J/gc)(236 g)] = 795759 c/s

If the surface of the sun is 5505 degrees then ignoring losses how long would it take to heat the water that high from room temperature or 25 degrees c? (x seconds)(rate)=5505-25=5480

x~6.9*10-3 seconds

So in order to prevent the cup of water from exceeding the temperature at the surface of the sun it'd have to cool off quicker that that fraction of a second, of course it's going to boil and plasmafy and rapidly expand, but keep in mind that I'm using a relatively small telescope parabola we could do the same calculation but with a much larger diameter and get a much smaller x seconds. Also the more sunlight you capture the higher pressure it's going to put on the target which prevents cooling. I can't see why exceeding the surface temperature wouldn't be possible.

Edit:

I'm confident that the assertion in the above linked question (that you can't make the target hotter than the surface of the sun) is wrong. So basically my question is what is the theoretical problem, and how is the real problem different from the theoretical? Alternatively, if I'm wrong, and the real problem doesn't deviate significantly from the theoretical, then why is the above calculation misleading? In the above calculation I show that there is a relation between the square meters of cross section of the focusing optic (e.g. reflective parabola) and the number of seconds it would take to raise the target to the suns surface temperature. Some number of orders of magnitude more area, and it becomes inconceivable to me that the target would be able to cool off fast enough to not heat up higher than the specified temp. So is there some missing piece of physics that would forbid that?

As I understand it the theoretical problem stems from second law of thermodynamics. One statement of that is as follows - you can't use a colder body to spontaneously heat a hotter body. In other words you can't make a perfect refrigerator. Or refrigeration requires work to be done on the system. Perhaps the optics are doing the work? Which gives me an idea, if the system was just the reflector, target, and star, then perhaps the solar radiation pressure would push everything apart fast enough.

• Once the target gets hotter than the sun - what stops the photons leaving the hot target and flowing back through the same optics to heat the sun? Jul 7 '17 at 4:32
• There is no theoretical limit on the temperature. There is only a limit on the energy flux of the concentrated sunlight. In thermodynamics we distinguish between work and heat, and work is energy that you can put in a single degree of freedom. There are limits from the second law about how much work you can extract from a given system. If instead of work you just want to pump hat at some temperature T, then you'll find that the efficiency is always larger than in case of extracting work and it becomes more the lower T is and you recover the result for work in the limit of infinite T. Jul 7 '17 at 4:44
• Martin Beckett, If the photons created in the sun only came out along those optics and to the target and back along the same optics then the target and the sun itself would both continue to heat up, but there is still no free lunch because you have a fusion core continually dumping energy into the two surfaces, and in reality the light will go every which way! Why would it be confined to just those optical paths? Jul 7 '17 at 4:51
• @user27382 You're missing a key fact in your above comment - when an object emits photons, it cools down. Also, fusion doesn't start until well above the sun's surface temperature, so you don't have any additional photons entering the picture. Another key fact that might not be obvious to you (it wasn't at first to me) - optical systems made of lenses and mirrors are always reversible. Jul 7 '17 at 8:22
• Jul 7 '17 at 15:48

Here's an argument I like, which is paraphrased from user chrisitan.wolff on the XKCD forums (http://forums.xkcd.com/viewtopic.php?t=113444), combined with some of the logic of Randall Munroe on the corresponding XKCD What-If (https://what-if.xkcd.com/145/):

The temperature of any object at equilibrium with the electromagnetic field is the average over angular size of the temperatures of the sources it sees, weighted by their emissivities.* This means that bodies that appear larger to the object in question contribute more to the object's temperature.

An optical system that converges light onto a target makes the light source bigger in the object's field of view. It does not make any individual point on the source appear hotter, because that would violate the reversibility of the optical system.

Here's why: if it made any point on the source appear brighter (which here is synonymous with hotter), then the system would have to gather light from at least two points on the source and combine it into one point on the image. Reversibility means that a single light ray that enters the system has precisely one path out of the lens, regardless of the direction in which it travels. With this source-brightening system, if we take rays of light leaving the target and entering the system, these rays now have two or more possible paths to take back to the source. Thus the system is not reversible, which is impossible. Therefore, it can't make any point on the source appear brighter/hotter.

So now that we've established that your system can't increase the perceived temperature of the source, but can only increase its perceived angular size, the rest of the argument follows easily. Take the best-case scenario, where the system is so cleverly designed that the Sun takes up the entire field of view of the object. Then, taking the average of a bunch of identical points, the temperature of the object is the temperature of the Sun. Any other configuration has less of the Sun in the target's field of view, so the target's temperature would be lower. Thus, the target's temperature cannot exceed the temperature of the Sun.

*The emissivity of the target itself does not affect its temperature, surprising as it may seem (see Is the equilibrium temperature of a black body higher than other objects?).

• Hmm.. I don't know about this "perceived angular size." But wouldn't increasing the size of the lens increase the amount of flux captured? I.e. increase the number of rays in whatever diagram and thus increase the temperature it's possible to achieve? It's not splitting any photons, larger lenses literally capture more photons and thus allow higher temperatures. Jul 7 '17 at 20:55
• @user273872 Yes, a larger lens captures more photons from the source. These photons make the source appear larger, while keeping the photon flux of the source the same at every point. It doesn't increase the flux itself, just the area over which the object sees a higher flux. Do you see a flaw with the logic I've demonstrated? So far you haven't refuted any of the logic in the answer; if you're right, then you should be able to refute my answer. Jul 8 '17 at 4:57
• Perhaps the flaw has to do with treating things as points where in reality the two photons from the source combine on an atom? Jul 8 '17 at 22:39