Emission of infra red from hot bodies

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SUMMARY

The discussion centers on the differences in infrared emission between matte black and shiny silver surfaces when heated to approximately 100°C. It is established that matte black surfaces are superior emitters of infrared radiation due to their higher emissivity, which is inversely related to reflectivity as per Kirchhoff's law. While Wien's law pertains to the peak wavelength of emitted radiation based on temperature, it does not account for the efficiency of emission, which is significantly influenced by surface properties. The conservation of energy principle further clarifies that a surface's ability to emit heat correlates directly with its absorptivity.

PREREQUISITES
  • Understanding of Kirchhoff's law regarding emissivity and absorptivity
  • Familiarity with Wien's law and its implications on blackbody radiation
  • Basic knowledge of thermal radiation and heat transfer principles
  • Concept of reflectivity and its relationship to emissivity
NEXT STEPS
  • Research Kirchhoff's law in detail to understand its applications in thermal radiation
  • Explore the principles of thermal emissivity and its measurement techniques
  • Study the effects of surface texture on thermal properties in materials science
  • Investigate the quantum mechanical models of radiation emission from surfaces
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Physicists, materials scientists, engineers, and anyone interested in the principles of thermal radiation and heat transfer efficiency.

DGriffiths
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I've seen much mention in websites of matt black surfaces being better emitters of infra red than shiny silver ones so if a matt black and shiny silver surface are heated to about 100 deg C the black one will cool down more quickly as a result of this! Doesn't Wien's law state that the amount of I.R. emitted by a hot body is only dependent on the temperature??

My question is then why is a shiny silver surface a much poorer emitter than a matt black one?
 
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DGriffiths said:
<snip>

My question is then why is a shiny silver surface a much poorer emitter than a matt black one?

Because of the conservation of energy: absorptivity + reflectivity + transmissivity = 1. For an opaque material, transmissivity = 0. Also, absorptivity = emissivity (Kirchoff's law), so emissivity = 1 - reflectivity. More reflectivity = less emissivity.
 
Yep. The key to understanding this is to understand Kirchhoff's law equating absorptivity to emissivity. Imagine two people of different size sitting by a camp fire. Who heats up the quickest, and who stays hot the longest?
 
DGriffiths said:
I'Doesn't Wien's law state that the amount of I.R. emitted by a hot body is only dependent on the temperature??
No Wien's law says that the peak position of the blackbody curve only depends on temperature
 
Thanks very much for all replies - I'd only heard of Kirchoff in respect to electrical circuits.

I get the "emissivity = 1 - reflectivity. More reflectivity = less emissivity." if that's what the law says but I still don't get why? Is there any mental picture (even if not totally accurate) that may help. I can see that the bending of molecules (due to the fact that charged particles are being accelerated) can result in the emission of e.m radiation in the infra red region but I still have no model of what is going on at the surface of a shiny metal that results in the emission of i.r. and also why it is so poor at doing so.

I appreciate I may be wanting a handwavy classical model that is not appropriate for a quantum mechanical concept but if anyone has any ideas I'd really appreciate it.
 
Andy was right in pointing you to conservation of energy. Consider a silver ball and a black ball illuminated by the same heat source. They should come to the same temperature. But the silver ball reflects most of the light falling on it while the black ball absorbs the light. Therefore to shed the same amount of heat at the same temperature, the black ball must have a much higher emissivity.
 

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