1
$\begingroup$

If all non-UV light was filtered from sunlight, does this approximate a different type of black body radiation? Regular sunlight has a black-body temperature of 5777 K.

This is in relation to the problem of concentrating sunlight to a temperature higher than 5777 K. Thermodynamics rules this out. However, if only the UV portion of the solar spectrum is used in some hypothetical solar concentrator, could a temperature higher than 5777 K be achieved?

Improve this question
$\endgroup$
1
  • $\begingroup$ I don’t think so because it’s the total energy absorbed by the target would heats it up. If you just use the UV component, overall you will absorb less energy, than if you used the full spectrum, so temperature will be less. But it’s an interesting question in terms of spectral shifting. $\endgroup$
    – boyfarrell
    Commented Jan 24, 2021 at 9:33

2 Answers 2

Reset to default
1
$\begingroup$

You are quite right that it is thermodynamically impossible to heat something hotter than the surface of the sun using only sunlight and passive optical elements (such as a filter that would block some wavelengths). Heat would then be spontaneously flowing from somewhere less hot to somewhere hotter, and that doesn’t happen.

The roadblocks you will encounter when trying to do this don’t really have anything to do with the spectrum of light coming from the sun—you can heat something to pretty high temperature in a microwave oven, even though the wavelength of the radiation that does the heating corresponds to the peak emission of a very, very cold blackbody. It’s all about the intensity of the radiation.

It is impossible to passively concentrate light to an intensity higher than the source. Having used a magnifying glass to start a fire, you might think that you just need a big enough magnifying glass. But as you use a bigger and bigger magnifying glass to focus the sunlight, the image you form of the sun gets bigger and bigger too, spreading the increased power over a larger area. And the hottest it can ever get is the temperature of the sun. If you filter out some of the light, it’s less hot than that.

Improve this answer
$\endgroup$
1
  • $\begingroup$ But as you use a bigger and bigger magnifying glass to focus the sunlight, the image you form of the sun gets bigger and bigger too, . . . . . The size of the image produced by a lens does not depend on its diameter but rather its focal length. so the statement needs altering. $\endgroup$
    – Farcher
    Commented Jul 2, 2022 at 7:40
-1
$\begingroup$

  1. No, the truncated spectra is not a different type of Black-body radiation. (Please look at the spectras of different temperatures.)

  2. As @boyfarrell suggests, if you truncate the incoming spectra, the overall energy absorbtion will be lover. (The integrated radiance will be lower, because you integrate over a limited frequency range.)

Note:
At first glance it might seems that theorically you can crate temperatures higher than 5777 K with "concentrated" sunlight. Let's say we collect sunlight (without any absorbtion) on 10 m$^2$ and focus that on an ideal black body (that absorbes all the radiation) with a surface of 1 m$^2$. In radiative equilibrium of the cube the absorbed energy equals with the emitted energy: $$10L_s=1L_c$$ $$10\frac{\sigma}{\pi}T^4_s=\frac{\sigma}{\pi}T^4_c$$ $$^4\sqrt{10}T_s=T_c\approx1.8T_s$$ Where $L$ is the (integrated) radiance, $\sigma$ is the Stefan–Boltzmann constant and $T$ is the temperature. The lower index '$s$' stands for the sun and '$c$' stands for the cube.

But (as suggested by @shaihorowitz) it turns out that you can't just focus the sunlight on an "infinitesimally small" point due to the conservation of etendue. (It's impossible to concentrate non-parallel rays from an extent source into an infinitesimally small point.) Practically you can't achive a higher (integrated) radiance (with passive elements) than the radiance of the original source. (In accordance with thermodynamics. Further discussion can be found here.)

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ Heating something up using black body radiation to a temperature hotter then the surge that emitted that radiation isn't so simple. The situation is much more nuanced. In general it is impossible to focus the black body radiation into a smaller surface as you say $\endgroup$
    – shai horowitz
    Commented Feb 17, 2021 at 0:51
  • $\begingroup$ Theorically I don't see why it's impossible to focus (part of) the radiation on a surface small enough. $\endgroup$
    – fanyul
    Commented Feb 17, 2021 at 10:05
  • 1
    $\begingroup$ there's the conservation of etendue for one thing. en.wikipedia.org/wiki/Etendue#Conservation_of_etendue. See here for more details as well physics.stackexchange.com/questions/338865/… $\endgroup$
    – shai horowitz
    Commented Feb 17, 2021 at 10:15
  • 1
    $\begingroup$ Hmm... OK, my bad. What I didn't think about, that it's impossible to concentrate non-parallel rays from an extent source into an infinitesimally small point. Because that's what the conservation of etendue tells, isn't it? (And it would be necessary, as the parallel part of the black body radiation is infinitesimally small.) $\endgroup$
    – fanyul
    Commented Feb 17, 2021 at 12:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.