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Could you check me on this: if I'm interested in total power (summed over all frequency) of CMBR absorbed by a black body in empty space, would that be proportional to the surface of BB?

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Yes, for a black body with emissivity of one, it is proportional to the area of the collector.

For any other object it is also proportional to the emissivity of the object.

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  • $\begingroup$ Heh ... I was just about to ask about that division :) Thx, man! $\endgroup$ May 29, 2011 at 18:50
  • $\begingroup$ I gotta stop sipping that Drano. $\endgroup$
    – Bill Slugg
    May 29, 2011 at 18:52
  • $\begingroup$ Didn't you get the def of emissivity backwards (1 equal perfectly efficient, 0 equals no radiation absorbed/emitted)? You were probably thinking of albedo, which runs the other way, but is typically used in relation to non thermal radiation. $\endgroup$ May 31, 2011 at 0:59
  • $\begingroup$ You are correct. A perfect emitter, is 1.0 and would be dark black sooty looking. A perfectly shiny aluminum telescope mirror would have an emissivity of .04. $\endgroup$
    – Bill Slugg
    May 31, 2011 at 2:44
  • $\begingroup$ Thx. I have revised and extended my original answer. BTW - Some bikers paint their chrome flat black. The exhaust cools down a lot faster that way. $\endgroup$
    – Bill Slugg
    May 31, 2011 at 2:46

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