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

Click For Summary
SUMMARY

The discussion centers on the phenomenon of black body radiation, specifically regarding a polished metal plate with a rough black spot heated to approximately 1400 Kelvin. The black spot appears brighter than the plate due to its higher emissivity, making it a better absorber of radiation. The black spot's ability to absorb nearly all incident radiation results in a higher power emission, as described by the black-body radiation equation P = σAe(T^4), where the emissivity (e) of the black spot is significantly greater than that of the polished plate.

PREREQUISITES
  • Understanding of black body radiation principles
  • Familiarity with the Stefan-Boltzmann law
  • Knowledge of emissivity and its impact on thermal radiation
  • Basic concepts of thermal equilibrium and heat transfer
NEXT STEPS
  • Research the Stefan-Boltzmann law and its applications in thermal physics
  • Explore the concept of emissivity and its measurement techniques
  • Study the properties of black bodies and their significance in thermodynamics
  • Investigate real-world applications of black body radiation in engineering and materials science
USEFUL FOR

Physicists, engineers, and students studying thermodynamics or heat transfer, particularly those interested in the principles of black body radiation and emissivity effects.

Amith2006
Messages
416
Reaction score
2
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?
 
Physics news on Phys.org
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
 

Similar threads

Black Body Radiation and ambient temperature
  • · Replies 7 ·
Replies
7
Views
4K
Radiation of a Spherical Body Inside a Black Body
  • · Replies 2 ·
Replies
2
Views
2K
How Does Sunlight Affect Black Body Radiation?
  • · Replies 3 ·
Replies
3
Views
6K
How does a black body emit as much energy as it absorbs?
  • · Replies 5 ·
Replies
5
Views
2K
Second Law of Thermodynamics - Radiation
  • · Replies 20 ·
Replies
20
Views
3K
Graduate Understanding Perfectly Black Bodies
  • · Replies 5 ·
Replies
5
Views
2K
Undergrad Blackbody radiation and the cosmic microwave background
  • · Replies 29 ·
Replies
29
Views
5K
Undergrad Questions about Jean-Rayleigh's derivation of Ultraviolet Catastrophe
  • · Replies 1 ·
Replies
1
Views
2K
Application Of Kirchoff Law (Desert)
  • · Replies 6 ·
Replies
6
Views
2K
Undergrad Trouble understanding the idea of a cavity radiator being a Black Body
  • · Replies 9 ·
Replies
9
Views
3K