State Stefan - Boltzmann Law and write its express – Class 11 Physics

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Q. State Stefan - Boltzmann Law and write its expression.

Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2019Subjective· 2mImportance★★★★★

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Concept understanding — Stefan Boltzmann Law

From a Hot Plate to a Law of Nature

Imagine you hold your hand near a hot iron. You feel warmth — that's radiant energy leaving the metal and reaching your skin. Now imagine the iron is twice as hot. Does it feel twice as warm? No — it feels much more intense. That jump is not linear; it's dramatic. The Stefan–Boltzmann law captures exactly how dramatic.

Every object above absolute zero emits electromagnetic radiation. The hotter it is, the more energy it radiates per second from each square metre of its surface. For a perfect emitter — called a black body — the relationship is astonishingly simple.

E = \sigma T^{4}

Here $E$ is the energy radiated per unit area per unit time (power per area, in $\text{W/m}^2$ ), $T$ is the absolute temperature in kelvin, and $\sigma$ is the Stefan–Boltzmann constant:

\sigma = 5.67 \times 10^{-8} \; \text{W m}^{-2} \text{K}^{-4}

The key point: $T$ is raised to the fourth power. Double the temperature, and the radiated power increases by a factor of $2^4 = 16$ . That is why a red-hot object (say 1000 K) radiates far more intensely than a warm object at 300 K — about $(1000/300)^4 \approx 123$ times more.

Why the Fourth Power?

The $T^4$ dependence is not arbitrary. It emerges from the deeper Planck radiation law, which describes how energy is distributed across different wavelengths. When you integrate Planck's law over all wavelengths, the total power comes out proportional to $T^4$ . In other words, the fourth power is a consequence of the shape of the black-body spectrum — not a guess.

Note

The law applies strictly to black bodies — ideal surfaces that absorb all incident radiation and emit the maximum possible energy at each temperature. Real objects emit less; their emission is given by $E = \varepsilon \sigma T^{4}$ , where $\varepsilon$ (emissivity) is between 0 and 1.

A Quick Check with Numbers

Take a black body at room temperature, 300 K. Its radiated power per square metre is:

E = (5.67 \times 10^{-8}) \times (300)^4 \approx 459 \; \text{W/m}^2

That is substantial — about five 100-watt bulbs per square metre. But we don't feel it because our surroundings radiate back at us at nearly the same rate. The net energy exchange depends on the temperature difference.

Now heat the same surface to 600 K. The power becomes:

E = (5.67 \times 10^{-8}) \times (600)^4 \approx 7348 \; \text{W/m}^2

That is 16 times larger — exactly $2^4$ .

Why It Matters

The Stefan–Boltzmann law is the backbone of thermal radiation physics. It explains: …

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The total energy radiated per second per unit area by a black body is proportional to the fourth power of its absolute temperature: E = sigma T^4.

Stefan-Boltzmann Law states: The amount of thermal radiation energy emitted per unit time per unit surface area by a perfect black body is directly proportional to the fourth power of its absolute temperature.

Mathematically:

E = sigma T^4

where:

E = radiant energy emitted per second per unit area (W/m^2)

sigma = Stefan-Boltzmann constant = 5.67 x 10^-8 W m^-2 K^-4

T = absolute temperature of the body (in kelvin)

…

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Q.State Stefan - Boltzmann Law and write its expression.

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