What this calculator does
Every object radiates heat, and the amount depends on the fourth power of its absolute temperature. That fourth power is the striking part: doubling the temperature multiplies the radiated power by sixteen.
It explains why a fire feels so much hotter than a radiator, why stars are so luminous, and why radiative losses dominate in any high-temperature process while being nearly negligible at room temperature.
The formula
Multiply emissivity by the Stefan-Boltzmann constant, by the surface area, and by the absolute temperature raised to the fourth power. Temperature must be in kelvin, since the relationship depends on absolute temperature.
| Term | Meaning |
|---|---|
| Emissivity (ε) | How effectively a surface radiates compared with an ideal blackbody. 1 is a perfect emitter; polished metal can be below 0.1. |
| Stefan-Boltzmann constant (σ) | 5.670374419 × 10⁻⁸ W/(m²·K⁴). |
| Blackbody | An idealised surface that absorbs and emits perfectly at every wavelength. |
The inputs explained
| Field | What to enter |
|---|---|
| Emissivity (1 = ideal blackbody) | Emissivity between 0 and 1. Most non-metallic surfaces are around 0.9; bright polished metals are far lower. |
| Surface area (m²) | The radiating surface area in square metres. |
| Temperature (K) | The absolute temperature in kelvin. Room temperature is about 293 K. |
When to use it
Estimating radiative heat loss
For anything hot, radiation can dominate the total heat loss, and this gives its size.
Understanding stellar output
A star's luminosity follows directly from its surface temperature and area, which is how stellar properties are deduced.
Comparing surface finishes
Emissivity varies enormously between materials, so a polished surface radiates far less than a painted one at the same temperature.
Worked examples
Every figure in the tables below is produced by this page’s own calculator at build time, so the numbers and the tool always agree. Select any row to load that scenario.
How steeply does radiated power rise with temperature?
One square metre of perfect emitter at a range of temperatures.
| Temperature | Radiated power | Power at double the temperature |
|---|---|---|
| 300 K | 459.3003279 W | 7,348.805247 W (16x from T⁴) |
| 500 K | 3,543.984012 W | 56,703.74419 W (16x from T⁴) |
| 1,000 K | 56,703.74419 W | 907,259.907 W (16x from T⁴) |
| 5,778 K | 63,200,699.73 W | 1,011,211,196 W (16x from T⁴) |
Questions
Why the fourth power?
It comes out of integrating the blackbody spectrum across all wavelengths. As temperature rises both the intensity at each wavelength and the range of wavelengths emitted increase, and combining the two gives the fourth-power dependence.
Does a cool object radiate at all?
Yes, everything above absolute zero radiates. Net heat transfer depends on the difference between what is emitted and what is absorbed from the surroundings, which is why a room-temperature object in a room-temperature room shows no net loss.
Why does emissivity matter so much?
Because it multiplies the whole result. A polished aluminium surface with emissivity around 0.05 radiates roughly a twentieth of what a painted surface at the same temperature does, which is exactly why thermal insulation often uses reflective foil.
Must temperature be in kelvin?
Yes, absolutely. The fourth power of a celsius value is meaningless, and at room temperature using celsius rather than kelvin would understate the result by many orders of magnitude.
For the wavelength at which that radiation peaks, see the Wien's displacement law calculator. For total stellar output, see the luminosity calculator.