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Stefan-Boltzmann Calculator

Find the radiated power.

How it is calculated

The Stefan-Boltzmann law gives the heat a hot surface radiates: P = ε × σ × A × T4. Here P is the radiated power in watts (W), ε is the emissivity between 0 and 1, σ = 5.670374 × 10-8 W/(m2·K4) is the Stefan-Boltzmann constant, A is the surface area in square metres, and T is the absolute temperature in kelvin. Because of the fourth power, doubling the temperature increases the output by 16 times. If emissivity is left empty or is zero or less, the tool uses 1, an ideal black body. The result is printed in scientific notation with four significant digits. Area and temperature cannot be negative. This is the power emitted, not the net loss: a surface also absorbs radiation from its surroundings.

Worked example

A metal plate of 2 m2 at 500 K with an emissivity of 0.9. T4 = 5004 = 6.25 × 1010. Multiply by σ: 5.670374 × 10-8 × 6.25 × 1010 ≈ 3,544 W per square metre for a black body. Multiply by the area: 3,544 × 2 ≈ 7,088 W. Apply the emissivity: 7,088 × 0.9 ≈ 6,379 W, shown by the tool as 6.379 × 10³ W.

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Common mistakes

  • Entering the temperature in degrees Celsius or Fahrenheit; it must be in kelvin, and the fourth power magnifies the error.
  • Typing the area in cm2 instead of m2, which gives a result 10,000 times too small.
  • Entering emissivity as a percentage such as 90 instead of 0.9.
  • Reading the output as the net heat loss; a surface in a warm room also receives radiation, so the net flow depends on the surroundings' temperature as well.
  • Not noticing that a blank emissivity is treated as 1, which gives the maximum possible output.

Sources

About this calculator

This calculator finds the power radiated by an object using the Stefan-Boltzmann law, P = εσAT⁴, where ε is emissivity, σ is the Stefan-Boltzmann constant (5.67×10⁻⁸ W/m²K⁴), A is surface area, and T is absolute temperature in kelvin. Because power scales with the fourth power of temperature, small temperature changes produce large changes in radiated energy.

Physics and engineering students use it for thermodynamics and heat-transfer coursework, and it's the same law astronomers use to estimate a star's luminosity from its temperature and size, and engineers use to calculate heat loss from hot surfaces like furnaces or spacecraft.

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Comments (1)

  • Editorial team

    The result is emitted power, not net heat loss, because surroundings radiate back. For 1 m² at 300 K with emissivity 1 the tool gives about 459 W; at 293 K it gives about 418 W. Run both and subtract to estimate the net exchange, roughly 41 W. Convert Celsius first by adding 273.15. What surface are you looking at, and what emissivity are you assuming for it?

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