Solar Furnace

I have two answers, a practical and a theoretical. First, the solar furnaces that are currently in operation do not reach temperatures higher than 3,000-4,000K, although they concentrate a huge amount of energy to a small spot.

Second, imagine that the Sun and the target are at the two focal points of a HUGE, closed elliptical mirror. Due to the properties of the elliptical mirror, all of the Sun's energy will be focused to the target. That is a closed system, and the second law applies in the from I have stated, and the target cannot be at a higher temperature than the Sun (of course, the temperature will start to raise as soon as the system is closed). If you remove some of mirror, the energy that gets to the target will be less, so that the temperature will be less, too. In a realistic situation the Sun's radiation is always coming to the target from a solid angle that is less than $2\pi$ and the thermal radiation of the target always goes to a solid angle of $4\pi$ , therefore that the target cannot reach the temperature of the Sun.

You wrote "If your claim was true, then it would be valid to say that nothing on earth that directly or indirectly gets energy from the sun (almost everything) can ever be more than 6000K." Here on Earth we have a high temperature heat source (the part of the sky covered by the Sun, 6000K) and a low temperature heat sink (the part of the sky not covered by the Sun, 2.8K). Naturally, we can use that (and the life that evolved on Earth used it for millions of years) the decrease entropy locally. To achieve that you need clever mechanisms, that are not present in the simple optical system we are discussing here.

Here is another way of looking at this, that does not use the general entropy argument. Consider a parabolic mirror of focal length $f$ and radius $R$ . The intensity of radiation (energy per unit area per unit time) coming from the Sun is $I$ . (This is also called "solar constant" or "solar flux" and its value is 1.36kW/m $^2$ ). The angular diameter of the Sun is $\alpha=0.53^{\circ}= 9.25 \times 10^{-3} rad$ . The question is: what is the intensity of radiation in the focal plane of the mirror?

Simple optics tells us that the size of the Sun's image is an $r=f\alpha/2$ circle. The area of that circle is $A'=(\alpha f/2)^2$ . The power hitting the mirror is $P = AI=R^2\pi I$ . This power is transferred to the image circle with the intensity of $I'= P/A'= (4R/f)^2 (1/\alpha^2) I$ .

Here is the first interesting point: scaling up the size of the mirror does not make the intensity at the focal plane larger. As long as we keep a reasonable relationship between the radius of the mirror and the focal length (let us say, $R=f$ ) the intensity is independent of the size.

We can conclude that the intensity in the focal plane is $I'\approx 4I/\alpha^2$ . Is this "strong" or "weak"? To answer this question, let us consider what would happen if all the sky is covered with the Sun and we do not use focusing mirror. The larger the solid angle is covered by the radiation source the larger is the intensity $I''=I \frac{\text{solid angle covered by the new source}}{\text{solid angle covered by the Sun}}=I\frac{\text{solid angle covered by the new source}}{(\alpha/2)^2\pi}$ . When the whole $4\pi$ solid angle is covered by the sun the intensity is $I''= I\frac{4\pi}{(\alpha/2)^2\pi} = 8I/\alpha^2$ .

Surprisingly, this is about the same intensity as the intensity we obtained by focusing. The ratio of the two intensities is $I'/I''=1/2$ ; in other worlds in the focal plane we can get half of the intensity of the radiation relative to the intensity when the whole sky is covered by the Sun. Of course, if all the sky is covered with the Sun, the temperature on Earth would be equal to the temperature of the Sun. We cannot do better by focusing the light.