Black Body Radiation: Why Does the Black Spot Appear Brighter?

AI Thread Summary
A polished metal plate with a rough black spot, when heated to around 1400 Kelvin and placed in a dark room, shows the black spot appearing brighter than the plate. This phenomenon occurs because the black spot acts as a near-perfect black body, absorbing more heat than the polished plate, which has lower emissivity and reflects some radiation. The black spot's higher emissivity leads to greater power emission according to the black-body radiation equation. Consequently, the black spot emits more thermal radiation, making it appear brighter. This highlights the relationship between emissivity, absorption, and thermal radiation in black body physics.
Amith2006
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Sir,
A polished metal plate with a rough black spot on it is heated to about 1400 Kelvin and quickly taken into a dark room. It is said that the black spot will appear brighter than the plate. Is it because the black spot is a perfect black body and hence is a better absorber of heat than the plate?
 
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Amith2006 said:
Sir,
A polished metal plate with a rough black spot on it is heated to about 1400 Kelvin and quickly taken into a dark room. It is said that the black spot will appear brighter than the plate. Is it because the black spot is a perfect black body and hence is a better absorber of heat than the plate?
The plate has a lower emissivity than the black spot (ie it is not a perfect absorber of radiation - it reflects radiation whereas the black spot absorbs almost all incident radiation). So in the black-body radiation equation:

P = \sigma Ae(T^4)

the power emitted by the black spot is higher for the black spot (higher e value).

AM
 
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