Kirchhoff's Law of Heat Radiation and its Theoretical Proof

Statement

Kirchhoff's law of thermal radiation deals with wavelength-specific radiative emission and absorption by a body in thermal equilibrium.

Statement: at a given temperature, the ratio of the emissive power to the coefficient of absorption of a body is equal to the emissive power of a perfect blackbody at the same temperature, for all wavelengths.

$$\dfrac{R}{a} = R_{B}$$

Since the emissive power of an ordinary body can be described in comparison with that of a perfect blackbody through its emissivity, the law may also be stated as: for a body emitting and absorbing thermal radiation in thermal equilibrium, the emissivity is equal to its absorptivity.

$$a = e \qquad \text{or, more specifically,} \qquad a(\lambda) = e(\lambda)$$

Thus if a body has high emissive power it also has high absorptive power, and if a body has low emissive power it also has low absorptive power.

Theoretical proof

The law can be proved by the following thought experiment.

Diagram in words: an enclosure at a fixed temperature, containing two bodies of identical geometric shape side by side — an ordinary body A and a perfect blackbody B — with radiation from the walls of the enclosure falling on both.

i. Consider an ordinary body A and a perfect blackbody B of identical geometric shapes placed in an enclosure. In thermal equilibrium both bodies will be at the same temperature as that of the enclosure.

ii. Let $R$ be the emissive power of body A, $R_{B}$ the emissive power of blackbody B, and $a$ the coefficient of absorption of body A. Let $Q$ be the quantity of radiant heat incident on each body in unit time, and $Q_{a}$ the quantity of radiant heat absorbed by body A, so that

$$Q_{a} = aQ$$

iii. As the temperatures of body A and blackbody B remain the same, each must emit in unit time the same amount of heat as it absorbs. Since emissive power is the quantity of heat radiated from unit area in unit time,

quantity of radiant heat absorbed by body A = quantity of heat emitted by body A

$$\therefore \ aQ = R \qquad ....(1)$$

iv. For the perfect blackbody B, all the incident radiation is absorbed, so

$$Q = R_{B} \qquad ....(2)$$

v. Dividing equation (1) by equation (2),

$$a = \dfrac{R}{R_{B}} \qquad \text{or} \qquad R = a R_{B}$$

vi. But by the definition of emissivity, $e = \dfrac{R}{R_{B}}$.

$$\therefore \ a = e$$

Hence Kirchhoff's law is theoretically proved.

Applications in daily life

• The bottom of a cooking utensil is blackened so that it is a good absorber of the heat radiated by the flame, while the top is polished so that it is a poor emitter and loses little heat.

• The walls of a thermos flask are silvered: a good reflector is a poor absorber and hence, by Kirchhoff's law, a poor emitter, so the flask loses very little heat by radiation.

Standard results

$$\dfrac{R}{a} = R_{B}, \qquad a = e, \qquad a(\lambda) = e(\lambda)$$

• Good absorbers are good emitters; good reflectors are poor emitters.