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Stefan-Boltzmann Power Calculator

Enter emissivity, surface area and temperature to find radiated power.

How it is calculated

The tool applies P = ε × σ × A × T⁴, with σ = 5.670374×10⁻⁸ W/(m²·K⁴). ε is the emissivity, a number from 0 to 1 (1 for an ideal black body), A is the radiating surface area in square metres and T is the surface temperature in kelvin. It first finds the power per unit area, flux = ε σ T⁴ in W/m², then multiplies by the area to get watts and divides by 1000 for kilowatts. Emissivity outside 0 to 1, or a temperature or area of zero or less, gives no result. This is the power the surface emits. It is not the net heat loss, because the surroundings radiate back; for net exchange you would use the difference T⁴ − T_surroundings⁴.

Worked example

Take a surface of 1.8 m² at 306 K (33 °C) with an assumed emissivity of 0.9. T⁴ = 306⁴ = 8,767,700,496. The flux is 5.670374×10⁻⁸ × 0.9 × 8,767,700,496 = 447.45 W/m². Multiplying by the area, 447.45 × 1.8 gives a radiated power of 805.40 W, or 0.8054 kW. This is how much that surface emits; if the room around it is at, say, 293 K it radiates much of that energy back, so the net loss is a fraction of 805 W.

Try this example →

Common mistakes

  • Entering the temperature in °C or °F. It must be in kelvin: add 273.15 to Celsius first. Entering 33 instead of 306 shifts the result by a factor of about 7,400.
  • Entering the emissivity as a percentage such as 90. The valid range is 0 to 1, so 90 gives no result; type 0.9.
  • Reading the answer as heat actually lost. The tool gives the emitted power only; the surroundings send radiation back, so the net loss is smaller.
  • Using the total surface area of an object when only part of it faces open space or radiates freely. Faces that touch another surface or are shielded should be left out of the area.
  • Expecting a doubling of power for a doubling of temperature. Power scales with T⁴, so doubling the kelvin temperature increases the power by 16 times.

About this calculator

The Stefan-Boltzmann law, P = εσAT⁴, gives the total thermal power radiated by a surface as electromagnetic radiation, where σ is the Stefan-Boltzmann constant (5.670374×10⁻⁸ W/m²K⁴), A is the surface area, T is the absolute temperature in kelvin, and ε is the emissivity — a dimensionless factor from 0 to 1 describing how efficiently a real surface radiates compared to an ideal black body (ε = 1).

Because power scales with the fourth power of temperature, radiated heat loss grows extremely fast as an object gets hotter — this is why the law is central to designing incandescent filaments, furnace and kiln linings, spacecraft thermal control surfaces, infrared thermography, and climate models that balance the Earth's absorbed sunlight against its radiated heat loss to space.

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Comments (1)

  • Editorial team

    Temperature enters to the fourth power, so doubling it multiplies emitted power by 16. The default 2 square metre body at 300 K emits 918.60 W, and at 600 K it emits 14,697.61 W. Convert Celsius first, since 27 degrees is about 300 K. Post the surface, its emissivity and where you got that value.

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