How Does Absorptivity Affect Heat Radiation in Small Temperature Differences?

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SUMMARY

The discussion centers on the derivation of the heat loss rate due to radiation for a small body in an evacuated enclosure, specifically when the temperature difference between the body (θ) and the walls (θ0) is minimal. The formula derived is dQ/dt = 4(θ0)^3 * Aaσ(θ - θ0), where A represents the area, a is the absorptivity of the body, and σ is the Stefan-Boltzmann constant. The participants emphasize the application of Stefan's law, Q = σ[θ^4 - (θ0)^4], to arrive at the final equation without altering the constant σ.

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  • Understanding of Stefan-Boltzmann law
  • Knowledge of thermal radiation concepts
  • Familiarity with calculus for differentiation
  • Basic principles of heat transfer
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Students in thermodynamics, physicists studying heat transfer, and engineers working on thermal management systems will benefit from this discussion.

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Homework Statement



A small body at temperature θ is suspended in an evacuated enclosure with walls at uniform temperature θ0.If (θ-θ0) is small show that the the rate at which the body loses heat bt radiation is given by

dQ/dt=4(θ0)^3*Aaσ(θ-θ0)
where A is the area and a is the sbsortivity of the body.

Homework Equations


The Attempt at a Solution



We should use the Stefan's law:Q=σ[θ^4-(θ0)^4]
the differentiation may be carried out...But I do not understand how to replace σ
 
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[tex]\sigma[/tex] is a constant. You don't replace it with anything. :confused:
 

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