Emissivity as a function of wavelength

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 6K views
yellowputty
Messages
8
Reaction score
0

Homework Statement



Hello, I'm a little confused. I did an experiment today with a Leslie Cube. One of the faces was a black paint, and another a white paint. I know that white reflects well and has a low emissivity, at least in the visual wavelengths. In the Leslie's cube, was hot water, so I was measuring the Irradiance of thermal radiation. From the results I concluded that the white and black faces had similar emissivities (in fact the white face was slightly higher). I understand that emissivity depends on several factors, temperature, angle and also wavelength. How come in the infrared spectrum, both faces have similar emissivities? I have a feeling the answer lies in a blackbody diagram. Can someone point me in the right direction?

Thanks in advance :)
 
Physics news on Phys.org
I think I just worked it out. Colour has nothing to do with a materials ability to absorb and hence emit thermal radiation. Colour is important in reflecting and absorbing wavelengths in the visible spectrum only. A white material will reflect all visible wavelengths, but a black surface would absorb all, and re-emit as thermal radiation.

I think oberlooked the way that visible light is then turned into thermal radiation. Thank you!
 
You may be correct. I thought of another problem. Maybe you are measuring reflected radiation that the cube reflects from external sources? How have you isolated this?