About the Stefan-Boltzmann Law Calculator
This Stefan-Boltzmann law calculator works out how much power a hot surface radiates as thermal (black-body) radiation, using P = εσAT⁴. Enter the temperature, the emissivity of the surface and its area — or the radius of a sphere such as a star or planet — and it returns the emitted power, the power absorbed from the surroundings, the net radiative heat loss, the radiated flux per square metre and the wavelength at which the emission peaks. A second mode works backwards from a known power to the surface temperature.
It is used by physics and astronomy students (the Sun’s luminosity follows directly from its radius and temperature), engineers estimating radiant heat loss from pipes, furnaces and heaters, and anyone curious how much heat the human body radiates. A chart shows how steeply power rises with temperature: doubling the absolute temperature multiplies the radiated power sixteen times.
Emissivity runs from 0 to 1: 1 is a perfect black body, human skin and matte paint are about 0.95–0.98, oxidised steel about 0.8 and polished aluminium only about 0.05. The net-exchange formula assumes the surroundings are large compared with the object and behave like a black body at their temperature, and that the object is grey (same emissivity at all wavelengths).
How to use the stefan-boltzmann law calculator
- 1Choose whether to find the radiated power or the temperature.
- 2Enter the emissivity of the surface.
- 3Enter the surface area, or choose a sphere and enter its radius.
- 4Enter the surface temperature (or the power) and pick °C, K or °F.
- 5Keep “account for surroundings” on for net heat loss, or switch it off for total emission.
Formula and method
The Stefan–Boltzmann law states that the total power radiated per unit area by a black body is σT⁴, where T is the absolute temperature in kelvin and σ = 5.670374419 × 10⁻⁸ W/(m²·K⁴) is the Stefan–Boltzmann constant (exact in the 2019 SI). A real surface radiates a fraction ε of that, its emissivity. Multiplying by the area gives the total emitted power; for a sphere such as a star the area is 4πr².
An object also absorbs radiation from its surroundings, so the net loss is εσA(T⁴ − Tₛ⁴), which assumes large, black-body surroundings and a grey surface. Solving that for T gives the temperature mode. Wien’s displacement law, λ_peak = b/T with b = 2.897771955 × 10⁻³ m·K, gives the wavelength at which the emission is strongest. Temperatures are converted to kelvin before any calculation.
- P
- Radiated power (W)
- ε
- Emissivity, 0–1
- σ
- Stefan–Boltzmann constant, 5.670374419 × 10⁻⁸ W/(m²·K⁴)
- A
- Radiating surface area (m²)
- T
- Absolute surface temperature (K)
- Tₛ
- Absolute temperature of the surroundings (K)
- b
- Wien displacement constant, 2.897771955 × 10⁻³ m·K
Worked examples
Heat radiated by a person at 33 °C skin temperature in a 20 °C room
Emitted: 0.98 × 5.6704 × 10⁻⁸ × 1.8 × 306.15⁴ = 878.7 W. The 293.15 K room radiates 738.7 W back, so the net loss is about 140 W — more than the roughly 100 W a resting adult produces, which is why an unclothed person feels cool in a 20 °C room. In practice clothing and the parts of the body that face each other (arms and torso) reduce the effective radiating area. The emission peaks near 9.5 µm in the infrared.
Luminosity of the Sun
Using the IAU nominal solar radius of 695,700 km and effective temperature of 5,772 K, A = 4π(6.957 × 10⁸ m)² = 6.08 × 10¹⁸ m², and P = 5.6704 × 10⁻⁸ × A × 5772⁴ ≈ 3.828 × 10²⁶ W — the Sun’s nominal luminosity.
Temperature of a 1 kW black-body emitter of 100 cm²
T = (1000 / (5.6704 × 10⁻⁸ × 0.01))^¼ = (1.7635 × 10¹²)^¼ ≈ 1,152 K, about 879 °C — a bright red-orange glow.
Oxidised steel plate at 500 °C in a 25 °C workshop
P_net = 0.8 × 5.6704 × 10⁻⁸ × 0.5 × (773.15⁴ − 298.15⁴) ≈ 7.9 kW of radiant heat loss, which is why furnace walls are insulated and shielded.
Frequently asked questions
What is the Stefan–Boltzmann law?+
It says the power radiated by a black body is proportional to the fourth power of its absolute temperature: P = σAT⁴. For real surfaces an emissivity factor ε (0–1) is added, giving P = εσAT⁴.
What is the value of the Stefan–Boltzmann constant?+
σ = 5.670374419 × 10⁻⁸ W·m⁻²·K⁻⁴. Since the 2019 redefinition of the SI it is exact, because it is calculated from the exact Planck constant, Boltzmann constant and speed of light.
Why must temperature be in kelvin?+
The law uses absolute temperature raised to the fourth power. Using Celsius would give zero radiation at 0 °C, which is wrong — a 0 °C object (273.15 K) still radiates about 315 W per square metre.
What is emissivity?+
Emissivity is how efficiently a surface radiates compared with a perfect black body at the same temperature. Matte black paint and skin are 0.9–0.98; bare oxidised metals 0.6–0.9; polished silver or aluminium can be below 0.05.
How do I find a star’s luminosity from its temperature?+
Treat the star as a black-body sphere: L = 4πR²σT⁴. With the Sun’s radius of 695,700 km and effective temperature 5,772 K this gives about 3.83 × 10²⁶ W.
What is the difference between emitted and net radiated power?+
Emitted power is everything the surface radiates. Net power subtracts the radiation it absorbs from its surroundings, εσA(T⁴ − Tₛ⁴). Net power is what matters for real heat loss or gain.