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Please do provide assurance of what formula you used and that you did the calculations for the answer. Thanks.

My back of letter calculation says that CMBR was never in visible range since it became free from the opaque universe and that the point when it could be used to boil water is still far in the future.

Please can you also tell me how far into the future it is, and whether there's enough energy in CMBR to actually boil water, or will like the heat from CMBR will leak off into environment before water boil?

Formula used: T = 2.73 × (1 + z) K, Wien's Displacement Law assuming CMBR is a perfect blackbody. But http://hyperphysics.phy-astr.gsu.edu/hbase/wien.html doesn't fit in with given values of CMBR in other sources. Using wavelenght=12.24 cm for water heating in microwave and z=1100 for T=3000K for the recombination. That that temperature, it's still infrared, the universe never had visible spectrum as background radiation.

Can someone confirm if I am right?

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    $\begingroup$ The peak is not everything, a black body at 3000K has a spectrum that peaks in the infrared, but still emits a large part of his radiation in the visible spectrum. So when the CMB was at 3000k some of its photons were visible. $\endgroup$
    – AnOrAn
    Sep 28 '20 at 7:55
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    $\begingroup$ What do you think is the condition for it to boil water? $\endgroup$
    – nasu
    Sep 28 '20 at 13:21
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Yes, it was. Spectral radiance of black body with temperature of $3000\,K$ in a visible EM frequency range is not zero, as can be seen by Plank radiation law :

enter image description here

If you want to find total power emitted per solid angle $\Omega$ in visible wavelength range- you need to integrate Plank law's spectral radiance over that wavelength range like :

$$ P_{~\Omega} = \int_{0.3\mu m}^{0.8\mu m} {\frac {2hc^{2}}{\lambda ^{5}}}{\frac {1}{e^{\frac {hc}{\lambda k_{\mathrm {B} }T}}-1}} ~~d\lambda$$

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