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So, I have some problems understanding Kirchoff's Radiation law.

My textbook, Transport Processes and Separation Process Principles, by Geankoplis, states that at the same temperature T1 the emissivity and absorptivity of a surface is equal, which holds for any black or non black solid surface.

In a problem from my professor it is given that : The sun radiate a flat surface with 1000 W/m2. The absorptivity of the plate is 0.9 and the emissivity is 0.1. The air temperature is 20 C and the heat transfer coefficient is 15 W/Km2. Calculate the surface temperature at equilibrium if the bottom is isolated.

My question is: how is it possible that the emissivity and absorptivity in this case is not equal, which contradicts Kirchoff's law?

Jmei
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2 Answers2

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To add some detail to Pieter's answer:

The absorptivity quoted is an average over all the wavelengths of light incident on the body from the Sun. This light may have an average wavelength of $\lambda \approx 0.5 \mu \text{m}$ or so which presumably has an absorbtivity of 0.9 in your case.

The emissivity is also averaged, but is given as an average relative to the blackbody radiation of the plate itself. Since its not glowing hot we conclude this average wavelength is in the infrared and according to your data the emissivity for that case must be 0.1.

Kirchoff's law only states that the two equal each other at a given wavelength. The data we are comparing are at two different wavelengths (visible and IR).

jacob1729
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Emissivity and absoprtivity are both functions of wavelength. The plate may absorb 90 % of sunlight ($\lambda \approx 0.5 \mu$m) and have an emissivity and absorptivity of 0.1 in the thermal infrared $(\lambda \approx 10 \mu$m).

These numbers are a bit unlikely, it is usually the other way around.