Case 1
- Surface area
- 1m²
- Temperature
- 300K
459.30W
Open with these valuesStefan-Boltzmann Law Calculator
459.30W
Result: 459.30 WA blackbody radiates the Stefan-Boltzmann constant times its area times the fourth power of its absolute temperature. That fourth power is everything: double the temperature and the power goes up sixteenfold. Temperature must be in kelvin, so add 273.15 to a Celsius reading first.
Worked examples
Case 2
- Surface area
- 1m²
- Temperature
- 5772K
62,938,592.47W
Open with these valuesCase 3
- Surface area
- 2m²
- Temperature
- 500K
7,087.97W
Open with these valuesHow it's calculated
P = σ × A × T⁴
- StepEnter the radiating surface area in square metres.
- StepEnter the absolute temperature in kelvin — Celsius plus 273.15.
- StepRead the power an ideal blackbody of that size and heat radiates.
- ResultFor a real surface, multiply by its emissivity between 0 and 1.
Reference table
| Area, temperature | What that is | Radiated power (W) |
|---|---|---|
| 1, 255 | Earth's effective radiating temperature | 239.76 |
| 1, 300 | room temperature | 459.30 |
| 2, 500 | — | 7087.97 |
| 0.5, 1000 | — | 28351.87 |
| 1, 3000 | an incandescent filament | 4593003.28 |
| 1, 5772 | the Sun's surface | 62938592.47 |
Questions
How do I calculate radiated power with the Stefan-Boltzmann law?
Multiply the constant σ by the surface area and by the temperature raised to the fourth power: P = σ × A × T⁴. Use square metres and kelvin to get watts. One square metre at 300 K radiates about 459.3 W.
What is the Stefan-Boltzmann constant?
It is σ = 5.670374419 × 10⁻⁸ W/(m²·K⁴). Since the SI revision of 2019 it is an exact value rather than a measured one, because it follows from the defined Planck, Boltzmann and light-speed constants. It is the same for every material.
Why is the temperature raised to the fourth power?
Because radiated power grows with T⁴, a small rise in temperature produces a large rise in emitted power. Doubling the absolute temperature multiplies the power by sixteen, which is why a star's output is so sensitive to its surface temperature.
Does this apply to real surfaces?
The formula describes an ideal blackbody, whose emissivity is exactly 1. A real surface emits less, so multiply the result by its emissivity ε, a number between 0 and 1. Polished metal sits near 0.05, matt black paint near 0.95.
How do I convert Celsius to kelvin?
Add 273.15, so 20 °C becomes 293.15 K. The law only works with absolute temperature, because raising a Celsius value to the fourth power would be meaningless. Entering Celsius by mistake is the single most common error here.
Sources and last check
Information, not professional advice.