enter image description here

Look at the brightest parts of the photo and use a radiance calculator to find the blackbody radiation curve (show the plot of the curve) and the wavelength of maximum emission. How do these compare with the Earth’s?

Could someone please explain how this would be solved? The calculator I can use is: https://astrogeology.usgs.gov/tools/thermal-radiance-calculator/

Thank you

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    $\begingroup$ That is impossible unless you know how to convert pixel brightness to radiation intensity at a specific wavelength. Thermal cameras often produce an output in which pixel brightness is related to temperature. If you know the temperature, then the backbody radiation curve is easy to get. But your question can't be answered unless you provide more details about the image. $\endgroup$ Feb 15 '18 at 20:45
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    $\begingroup$ That is what I thought, but my TA said to think of the temperature of Earth being something like 288K and use that information to solve. Basically plug in the known value into the only radiance calculator. The brightest part of the photo is the face, where the heat is. So I guess I would use the average temperature of the human body to plot the black body curve? $\endgroup$
    – yre
    Feb 15 '18 at 20:53
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    $\begingroup$ @CamiloRada okay so I found out that the average temp of the human body is37.0*C(310.15K) and the average temp of Earth is 0.9*C(274.05K). When I plugged both of those into Nasa's radiance calculator I got this as an answer. imgur.com/a/ZTAp4 $\endgroup$
    – yre
    Feb 15 '18 at 20:59
  • $\begingroup$ Though it's a somewhat interesting question, suggest this might be more applicable elsewhere, perhaps in the broader Physics Stack Exchange than here, as it only very tangentially involves much of anything Earth Science covers. If you wish to have it moved, flag it for moderator attention and indicate where you'd like to try it at :-) $\endgroup$ Feb 16 '18 at 9:19

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