Stefan-Boltzmann Law Calculator

Calculate thermal radiation power emitted by an object using the Stefan-Boltzmann law.

Sun: 5778 K, Human: ~310 K
Blackbody=1, Metal≈0.1, Human skin≈0.97
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How Temperature Controls Thermal Radiation

The Stefan–Boltzmann law describes the total thermal radiation emitted by a surface, and its most important feature is the fourth power of absolute temperature. Every object above absolute zero emits electromagnetic radiation. A perfect blackbody is the ideal emitter; real surfaces are represented by an emissivity ε between 0 and 1. The total emitted power also scales directly with emitting area.

The T4 dependence makes temperature extremely influential. If absolute temperature doubles while area and emissivity stay fixed, emitted power increases by 24 = 16. Temperature must be entered in kelvins because the law is tied to absolute thermal energy. A practical heat-transfer calculation often needs net radiation to surroundings rather than gross emitted power; for surroundings at Ts, the idealized net exchange is εσA(T4 − Ts4).

P = εσAT4     σ = 5.670374419 × 10−8 W·m−2·K−4
SymbolMeaningWhy it appears / units
PTotal emitted radiant powerWatts (J/s)
εEmissivityDimensionless, 0 to 1; measures emission relative to a blackbody
ARadiating surface aream2; twice the area gives twice the total emission
TAbsolute surface temperatureKelvins; appears to the fourth power

The result is total radiant power leaving the surface, not necessarily the object's net heat loss. Nearby walls, air, and other objects also radiate toward it. Convection and conduction may also be important, so Stefan–Boltzmann radiation is one part of a complete thermal energy balance.

Worked Examples

Example 1: Human body: T=310K, A=1.8m², ε=0.97
P = 0.97×5.67e-8×1.8×310⁴
Result: 914 W total radiation
This is why rooms feel warm with people
Example 2: Sun surface: T=5778K, A=6.08e18 m², ε=1
P = σAT⁴
Result: 3.85×10²⁶ W (luminosity)
Matches measured solar luminosity
Example 3: Small hot surface
ε = 0.80, A = 0.020 m2, T = 800 K → P = εσAT4
Result: P ≈ 372 W emitted
The small area still emits hundreds of watts because the fourth-power temperature factor is large at 800 K.
Example 4: Human body net radiation to a cooler room
ε = 0.97, A = 1.8 m2, T = 310 K, Ts = 293 K → Pnet = εσA(T4 − Ts4)
Result: Pnet ≈ 185 W
Gross emission is much larger than net radiative loss because the room radiates energy back toward the body.

Common Mistakes

⚠️
Using Celsius in T4

Convert to kelvins before applying the law. A Celsius value does not have the correct zero point for absolute thermal radiation.

⚠️
Forgetting the fourth power

Radiated power is not proportional to temperature itself. The T4 dependence is why hot objects radiate so strongly.

⚠️
Confusing emitted power with net heat loss

P = εσAT4 is gross emission. Net exchange with thermal surroundings requires subtracting their incoming radiation under the appropriate model.

⚠️
Assuming every surface has emissivity one

Real emissivity depends on material, finish, wavelength, and temperature. Polished metals can differ greatly from dark or oxidized surfaces.

Frequently Asked Questions

What is emissivity?
Emissivity (ε) ranges 0–1. A perfect blackbody has ε=1. Real surfaces emit less radiation. Polished metals: ε≈0.05. Human skin: ε≈0.97 in infrared.
How is this used in engineering?
Thermal management of satellites, electronics cooling, building insulation, and infrared thermometry all use the Stefan-Boltzmann law.
Why does the Stefan–Boltzmann law use kelvins?
The law depends on absolute thermodynamic temperature and predicts zero thermal emission only at absolute zero. Kelvin has that physical zero point. Using Celsius inside T4 would give meaningless ratios and severely incorrect radiant-power values.
What happens to radiated power if temperature doubles?
With emissivity and area fixed, the emitted power increases by 24 = 16. This fourth-power dependence is why relatively modest changes in very high temperatures can cause large changes in radiation heat transfer.
What is the difference between a blackbody and a real surface?
A blackbody is an ideal surface with emissivity ε = 1 and emits the maximum thermal radiation possible at a given temperature and area. A real surface generally has ε below 1, so its emitted power is reduced by that factor in the gray-body approximation.
Why can a warm person emit hundreds of watts but not cool at that rate by radiation?
The body emits infrared radiation, but surrounding walls and objects also emit infrared radiation back toward the body. Net radiative loss depends on the difference between fourth powers of body and surrounding temperatures, so it is much smaller than the body's gross emitted power in an ordinary room.

Formula Explorer connections

Interpretation: This formula tracks heat, temperature, work, entropy or transport in a thermodynamic system. Assumption: Use absolute temperature where required and consistent energy units. Constant properties, equilibrium, ideal gases or negligible losses may be assumed.

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