Stefan-Boltzmann Law of Radiation

Statement

In 1879 Josef Stefan proposed, from experimental observations, an empirical relation between the rate at which heat is radiated from unit area of a perfect blackbody — the radiant power $R$ — and its absolute temperature $T$. Five years later Boltzmann derived the relation using thermodynamics, and it is therefore known as the Stefan-Boltzmann law.

Statement: the rate of emission of radiant energy per unit area, or the power radiated per unit area, of a perfect blackbody is directly proportional to the fourth power of its absolute temperature.

$$R \propto T^{4} \qquad \Rightarrow \qquad R = \sigma T^{4}$$

where $\sigma$ is Stefan's constant.

• $\sigma = 5.67 \times 10^{-8}$ J m$^{-2}$ s$^{-1}$ K$^{-4}$, that is W m$^{-2}$ K$^{-4}$. Some numericals take $\sigma = 5.7 \times 10^{-8}$.

• Dimensions of $\sigma$: $\left[L^{0}M^{1}T^{-3}K^{-4}\right]$.

Thus the power radiated by a perfect blackbody depends only on its temperature and not on any other characteristic such as its colour, material or nature of surface.

Energy radiated in a given time

If $Q$ is the amount of radiant energy emitted in time $t$ by a perfect blackbody of surface area $A$ at temperature $T$, then

$$\dfrac{Q}{At} = \sigma T^{4} \qquad \Rightarrow \qquad Q = \sigma A t T^{4}$$

Ordinary body

For a body which is not a blackbody, the energy radiated per unit area per unit time is still proportional to the fourth power of the temperature, but is less than that for a blackbody:

$$R = e\sigma T^{4}$$

where $e$ is the emissivity of the surface, and correspondingly $Q = e\sigma A t T^{4}$.

Body in surroundings at a different temperature

If a perfect blackbody at absolute temperature $T$ is kept in surroundings at a lower absolute temperature $T_{0}$, then

• energy radiated per unit area per unit time $= \sigma T^{4}$;

• energy absorbed from the surroundings per unit area per unit time $= \sigma T_{0}^{4}$;

• net loss of energy by the perfect blackbody per unit area per unit time $= \sigma T^{4} - \sigma T_{0}^{4} = \sigma\left(T^{4} - T_{0}^{4}\right)$.

For an ordinary body at temperature $T$, the net loss of energy per unit area per unit time is $e\sigma\left(T^{4} - T_{0}^{4}\right)$, and the net rate of loss of heat from the whole surface is

$$\dfrac{dQ}{dt} = e\sigma A\left(T^{4} - T_{0}^{4}\right)$$

On the other hand, if the body is at a temperature lower than that of its surroundings, $T < T_{0}$, then $e\sigma\left(T_{0}^{4} - T^{4}\right)$ is the net gain in thermal energy of the body per unit area per unit time.

Significance

Because the loss or gain of energy per unit area per unit time is proportional to the fourth power of the absolute temperature, this law is very significant in deciding the thermal equilibrium of physical systems. If the absolute temperature of a body is doubled, the power radiated increases by a factor of $2^{4} = 16$. A body radiating at some rate at room temperature (300 K) will radiate at double that rate if its temperature is increased by only $57\,^{\circ}$C.

Standard forms used in numericals

$$R = \sigma T^{4}, \qquad R = e\sigma T^{4}, \qquad Q = e\sigma A t T^{4}, \qquad \dfrac{dQ}{dt} = e\sigma A\left(T^{4} - T_{0}^{4}\right)$$

$$\dfrac{R_{1}}{R_{2}} = \dfrac{T_{1}^{4} - T_{0}^{4}}{T_{2}^{4} - T_{0}^{4}} \qquad \text{(equal } e, \sigma, A \text{ cancel)}$$

$$\dfrac{P_{1}}{P_{2}} = \dfrac{r_{1}^{2}T_{1}^{4}}{r_{2}^{2}T_{2}^{4}} \qquad \text{for two spheres, since } A = 4\pi r^{2}$$

Worked patterns from the textbook:

• Energy radiated in one minute by a blackbody of area 200 cm$^{2}$ at $127\,^{\circ}$C: $Q = \sigma A t T^{4}$ with $A = 200 \times 10^{-4}$ m$^{2}$, $t = 60$ s, $T = 400$ K, giving about $1742$ J.

• A 60 W filament of emissivity 0.5 and area $5 \times 10^{-5}$ m$^{2}$: from $\dfrac{dQ}{dt} = e\sigma A T^{4}$, $T^{4} = 4.23 \times 10^{13}$, so $T = 2.55 \times 10^{3} = 2550$ K.

• Rate of loss of heat from a sphere at $827\,^{\circ}$C compared with the same sphere at $427\,^{\circ}$C, surroundings at $27\,^{\circ}$C: $\dfrac{R_{1}}{R_{2}} = \dfrac{1100^{4} - 300^{4}}{700^{4} - 300^{4}} = \dfrac{14560}{2320} = \dfrac{182}{29}$, that is $182 : 29$.

• A virtual star at 3000 K radiating 25 times the power of the Sun at 6000 K: from $\dfrac{P_{\text{star}}}{P_{\text{Sun}}} = \dfrac{r_{\text{star}}^{2}T_{\text{star}}^{4}}{r_{\text{Sun}}^{2}T_{\text{Sun}}^{4}}$, one gets $r_{\text{star}} = 20\,r_{\text{Sun}}$.

Standard results

$$R = \sigma T^{4} \ \text{(blackbody)}, \qquad R = e\sigma T^{4} \ \text{(ordinary body)}, \qquad \sigma = 5.67 \times 10^{-8}\ \text{W m}^{-2}\text{K}^{-4}$$

$$\text{Net rate of loss of heat} = e\sigma A\left(T^{4} - T_{0}^{4}\right)$$

• Convert all temperatures to kelvin, and never write $\left(T - T_{0}\right)^{4}$ in place of $\left(T^{4} - T_{0}^{4}\right)$.