Stefan Boltzmann Calculator
Calculate the power radiated by a body from its emissivity, surface area, and absolute temperature, using the Stefan-Boltzmann law.
Stefan Boltzmann Calculator
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Formula: P = εσAT⁴
Worked example — 63.2 MW/m²
Formula
P = εσAT⁴
Total radiated power from a blackbody. σ = 5.670374419×10⁻⁸ W/(m²·K⁴).
Worked Examples
Example 1: Sun surface
Problem:ε=1, A=1 m², T=5778 K
Solution:P=5.67e-8×5778⁴=6.32×10⁷ W/m²
Result:63.2 MW/m²
Frequently Asked Questions
What is the Stefan-Boltzmann law?
The Stefan-Boltzmann law states that the total power radiated by an object is P = εσAT⁴, where ε is emissivity (0–1), σ = 5.670×10⁻⁸ W/(m²·K⁴) is the Stefan-Boltzmann constant, A is surface area, and T is absolute temperature in Kelvin.
How do you calculate radiated power using the Stefan-Boltzmann law?
Multiply emissivity × σ × area × temperature⁴: P = εσAT⁴. For a 1 m² blackbody (ε = 1) at 300 K (room temperature): P = 1 × 5.67×10⁻⁸ × 1 × 300⁴ = 459 W. Doubling T increases power by 2⁴ = 16 times.
What units are used in Stefan-Boltzmann calculations?
Emissivity ε is dimensionless (0 to 1). Surface area A is in square metres (m²). Temperature T must be in Kelvin (K; K = °C + 273.15). Radiated power P is in Watts (W). The Stefan-Boltzmann constant σ = 5.670374419×10⁻⁸ W·m⁻²·K⁻⁴.
What is emissivity and how does it affect radiation?
Emissivity ε is the ratio of an object's radiation to that of a perfect blackbody at the same temperature. A blackbody has ε = 1. Polished metals have ε ≈ 0.02–0.05, human skin ε ≈ 0.98. Low-emissivity coatings on windows reduce heat radiation by up to 90%, improving insulation.
How do you calculate net radiative heat exchange with the surroundings?
An object both emits and absorbs, so the net rate is P_net = epsilon x sigma x A x (T^4 - T_surr^4). For skin with emissivity near 0.98, about 1.8 m-squared of area at 306 K in a 293 K room, the difference of the fourth powers is 1.40 x 10^9 K^4, giving roughly 140 W of net radiative loss. This is why a cold wall feels chilly even when the air is warm.
How sensitive is radiated power to an error in temperature?
Differentiating P proportional to T^4 gives a simple rule: a one percent error in absolute temperature produces a four percent error in radiated power. At 300 K that means a 3 K mistake shifts the answer by about 4 percent. It is also why forgetting to convert Celsius to kelvin is catastrophic here - entering 27 instead of 300.15 changes the result by a factor of well over ten thousand.
How does Wien's displacement law complement this calculation?
Stefan-Boltzmann gives the total power radiated across all wavelengths; Wien's law tells you where the spectrum peaks, at lambda_max = b/T with b = 2.898 x 10^-3 m K. The Sun's 5778 K surface peaks near 502 nm, in the blue-green, while a 300 K room-temperature object peaks near 9.7 micrometres in the far infrared. That is exactly why thermal cameras are built to see in the 8 to 14 micrometre band.
What is the difference between a blackbody, a grey body and a real surface?
A blackbody absorbs and emits perfectly, with epsilon = 1. A grey body has emissivity below 1 but constant across wavelengths, which is the assumption behind a single epsilon input. Real surfaces vary with wavelength, angle and finish, and Kirchhoff's law requires that at each wavelength emissivity equals absorptivity. Selective coatings exploit that, absorbing strongly in the visible while emitting weakly in the infrared.
How is radiated power used to estimate a planet's equilibrium temperature?
Balance absorbed sunlight against emitted thermal radiation. Absorption scales with the planet's disc area while emission covers the whole sphere, giving T = [S x (1 - a) / (4 x sigma)]^(1/4). Using a solar constant near 1361 W/m-squared and an albedo of 0.3 yields about 255 K for Earth. The actual surface is warmer because greenhouse gases absorb outgoing infrared and re-radiate part of it downward.
Why does radiation dominate heat loss at high temperature but not near room temperature?
Radiative loss climbs as T^4, while convective and conductive losses grow roughly in proportion to the temperature difference. Near ambient conditions the near-linear mechanisms usually carry most of the heat, so radiation is a modest contributor. As temperature rises the fourth-power term overtakes them quickly, which is why furnace walls, incandescent filaments and re-entry heat shields lose energy predominantly by radiation.
Reviewed for accuracy by Manoj Kumar, Mathematics Educator · Editorial policy
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