What is black-body equivalent of UV part of solar spectrum? 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?
 A: 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.
A: *

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


*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.)
