I think this just further confuses the issue.
Let's take heat, temperature, even energy out of the equation and use the following terms, "Emitted power per unit area", and "Delivered power per unit area".
The sun or an LED will "emit" a set amount of power per unit area in a somewhat defined distribution pattern. For the sun, that is a about 67 megawatts per per square meter in a roughly spherical distribution. For an LED, that is up to say 2-3 megawatts per square meter in a roughly lambertian distribution.
1) Using purely optical concentration/reflection/refraction, and no conversion to other energy forms, whether it be electrical, chemical, or otherwise, you cannot, using the sun as a source deliver onto a surface, more than 67megawatts per square meter. It does not matter how big the reflector, or how many mirrors, it cannot be done.
2) IF you could create a Perfect optical system, and IF you could create a Perfect Single Sided black body, then the maximum temperature of that black body would be limited to the surface temperature of the sun, or 5500K approximately. A perfect black body will absorb everything and turn it into heat. AND everything that is hot radiates! Once that black body reached the same temperature as the sun, or 5500K, it will radiate as much power as it is being bombarded with and stop increasing in temperature!
Semiman ..... moonlighting as junior physics man.
This is sorta where i was today thinking thru this issue. I recall reading years ago that the sun puts out about 10,000 horsepower per square yard of surface area. However, the sun is 93 million miles away, and the inverse square law is going to kill you here. Standing on the surface of the earth, no collector of any size is going to allow you to exceed what is emitted from the sun's surface at the bottom of the photosphere. Not even if you built a Dyson sphere and captured all the energy that way. Again, this just seems common sense to me, notwithstanding the fact that there are some things about Etendue I'm still digesting.
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