<p><a href="https://i.sstatic.net/xHQgm.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/xHQgm.png" alt="enter image description here"></a></p>

<p>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?</p>

<p>Could someone please explain how this would be solved?
The calculator I can use is:
<a href="https://astrogeology.usgs.gov/tools/thermal-radiance-calculator/" rel="nofollow noreferrer">https://astrogeology.usgs.gov/tools/thermal-radiance-calculator/</a></p>

<p>Thank you</p>

CC BY-SA 3.0
2018-02-15T20:45:16.790Z
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.
Camilo Rada
CC BY-SA 3.0
2018-02-15T20:53:26.100Z
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?
yre
CC BY-SA 3.0
2018-02-15T20:59:31.507Z
@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. https://imgur.com/a/ZTAp4
yre
CC BY-SA 3.0
2018-02-16T09:19:28.357Z
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](https://earthscience.stackexchange.com/help/on-topic).  If you wish to have it moved, flag it for moderator attention and indicate where you'd like to try it at :-)
JeopardyTempest
<waves><atmospheric-radiation>
Looking at the brightest parts of the image how would one find the blackbody curve and wavelength maximum using a radiance calculator?
yre
# Looking at the brightest parts of the image how would one find the blackbody curve and wavelength maximum using a radiance calculator?

(Asked by: username_0 on 2018-02-15)

[![enter image description here](https://i.sstatic.net/xHQgm.png)](https://i.sstatic.net/xHQgm.png)

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

**Tags:** waves, atmospheric-radiation

> **Comments:**
> **username_1**: 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.  
> **username_0**: 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?  
> **username_0**: @username_1 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. https://imgur.com/a/ZTAp4  
> **username_2**: 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](https://earthscience.stackexchange.com/help/on-topic).  If you wish to have it moved, flag it for moderator attention and indicate where you'd like to try it at :-)  

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