Stefan–Boltzmann
P = εσAT⁴.
About this calculator
The Stefan-Boltzmann law says every object radiates thermal energy at a rate proportional to its surface area and the fourth power of its absolute temperature: P = εσAT⁴, where σ is the Stefan-Boltzmann constant (5.670374419×10⁻⁸ W/(m²·K⁴)) and ε is emissivity, a 0-to-1 factor describing how efficiently a real surface radiates compared to a perfect blackbody. This calculator plugs your temperature (in Kelvin), area (in square meters), and emissivity straight into that formula to return total radiated power in watts — no approximation, just the direct physics. Because temperature is raised to the fourth power, small changes matter enormously: doubling absolute temperature increases radiated power sixteenfold, which is why thermal radiation dominates so completely at stellar temperatures despite being a minor factor in everyday room-temperature heat transfer.
The defaults model a resting human body (310 K, about 1.8 m² of skin, emissivity 0.97, since skin radiates almost like a perfect blackbody regardless of visible skin color, which only affects reflectance in visible light, not infrared emissivity), but the same formula scales from a light bulb filament to a star's photosphere. One important limitation: this calculates gross power radiated outward only — it doesn't net out power absorbed from the surrounding environment, so it will overstate real-world heat loss for anything sitting in a warm room or in sunlight, where two-way radiative exchange applies. Temperature must be entered in Kelvin, not Celsius or Fahrenheit — a common input mistake, since the T⁴ dependence makes using the wrong scale produce wildly wrong results rather than a small error.
Inputs
Results
Power (W)
914.33
How to Use This Calculator
- Enter the blackbody surface temperature T (K).
- Set the emitting surface area (m2).
- Enter emissivity (1.0 for perfect blackbody, 0.95 for most non-metallic surfaces, 0.05 for polished aluminum).
- Review total radiated power in watts.
- Use for thermal engineering, building energy analysis, or stellar luminosity estimates.
How the result changes with T (K)
| T (K) | Power (W) |
|---|---|
| 155 | 57.146 |
| 233 | 291.796 |
| 465 | 4,628.793 |
| 775 | 35,715.998 |
What each input means
- T (K)
- Absolute temperature in Kelvin. Human body is about 310 K; Sun surface is about 5778 K.
- Area (m²)
- Surface area of the radiating body in square meters. Average human body surface area is about 1.7-1.9 m2.
- ε
- Emissivity of the surface (0-1). Human skin is about 0.97; polished metal can be below 0.1.
How this is calculated
Formula
P = ε × σ × A × T⁴Engine last updated . Checked against 4 independently-derived tests — how we verify calculators. Built by Paul Gunder, a software engineer, not a licensed financial, medical, or legal professional.
Frequently Asked Questions
Why does the calculator use 0 as a valid temperature or area input instead of falling back to a default?
The engine deliberately uses `?? default` rather than `|| default` for tempK, areaM2, and emissivity, because 0 is a physically meaningful value for each: absolute zero, zero radiating area, or a perfect reflector that emits nothing. With the older `||` pattern, entering 0 for any of these would have silently snapped back to the non-zero default instead of correctly returning P = 0 watts.
Why does doubling the temperature increase radiated power by 16x instead of 2x?
Power depends on temperature raised to the fourth power (T⁴) in the Stefan-Boltzmann law, so doubling absolute temperature multiplies power by 2⁴ = 16. This is why thermal radiation is negligible at everyday room temperatures but completely dominates energy loss at stellar temperatures — the T⁴ term amplifies small temperature differences enormously.
Why does the calculator ask for temperature in Kelvin instead of Celsius or Fahrenheit?
Because P = εσAT⁴ requires absolute temperature — Celsius or Fahrenheit values include an offset from absolute zero that breaks the fourth-power relationship entirely. Entering a Celsius value where Kelvin is expected won't just shift the answer slightly; it will produce a wildly wrong result since the formula assumes T is measured from true zero.
Does the result account for heat the object absorbs from its surroundings?
No — this calculates only the gross power radiated outward from the surface, with no subtraction for radiation absorbed back from a warm room, sunlight, or other nearby sources. For an object sitting in a warm environment, the real net heat loss is smaller than this figure, since two-way radiative exchange isn't modeled here.
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