State and prove Kirchhoff’s law of heat radiation.

Answer: At a given temperature, the ratio of the emissive power to the coefficient of absorption is the same for all bodies and is equal to the emissive power of a perfectly black body at that temperature: \(\dfrac{E}{a} = E_b\).

Equivalently, at a given temperature the coefficient of absorption of a body equals its emissivity, \(a = e\); good absorbers are good emitters.

Proof: Place an ordinary body A and a perfectly black body B, of equal surface area, inside an enclosure maintained at a constant temperature \(T\). After some time both reach thermal equilibrium at \(T\), so each emits as much energy as it absorbs.

Let \(Q\) be the radiant energy incident per unit area per unit time on each body. Let \(E\) and \(a\) be the emissive power and coefficient of absorption of A, and \(E_b\) the emissive power of B (for which \(a = 1\)).

• For A: energy absorbed per unit area per second \(= aQ\), energy emitted \(= E\). In equilibrium, \(E = aQ\).
• For B: energy absorbed \(= Q\), energy emitted \(= E_b\). In equilibrium, \(E_b = Q\).

Dividing,

Since the emissivity \(e = E/E_b\), this also gives \(a = e\). This proves Kirchhoff’s law.