Question

State and prove Kirchhoff's law of heat radiation.

Hint:

Remember the simple concept: "Good absorbers are good emitters." A black T-shirt gets hotter in the sun (good absorber) and also cools down faster in the shade (good emitter) compared to a white one.

Answer 1

Statement: Kirchhoff's law of heat radiation states that at a given temperature, the ratio of the emissive power (\(E\)) to the coefficient of absorption (\(a\)) is the same for all bodies and is equal to the emissive power of a perfectly black body (\(E_b\)) at that same temperature. \[ \frac{E}{a} = E_b \] This implies that a good absorber of radiation is also a good emitter of radiation at a given temperature.
Proof (Thought Experiment):
Consider an ordinary body 'O' and a perfectly black body 'B' of the same surface area, suspended in a constant temperature enclosure. After some time, both bodies will reach thermal equilibrium with the enclosure, meaning they will all be at the same temperature.
Let \(E\) be the emissive power of the ordinary body and \(a\) be its coefficient of absorption.
Let \(E_b\) be the emissive power of the black body. By definition, its coefficient of absorption is \(a_b = 1\).
Let \(Q\) be the total radiant energy incident per unit area per unit time on each body from the enclosure.
For the ordinary body 'O' at thermal equilibrium:
Energy absorbed per unit area per unit time = \(aQ\).
Energy emitted per unit area per unit time = \(E\).
Since the temperature is constant, Energy absorbed = Energy emitted. \[ aQ = E \quad \cdots (1) \] For the perfectly black body 'B' at thermal equilibrium:
Energy absorbed per unit area per unit time = \(a_b Q = 1 \cdot Q = Q\).
Energy emitted per unit area per unit time = \(E_b\).
Since the temperature is constant, Energy absorbed = Energy emitted. \[ Q = E_b \quad \cdots (2) \] Now, substitute the value of \(Q\) from equation (2) into equation (1): \[ a(E_b) = E \] \[ \frac{E}{a} = E_b \] This proves Kirchhoff's law of heat radiation.