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Physics · Thermodynamics

Stefan-Boltzmann Law Calculator

Radiated power from a hot surface with P = εσAT⁴.

K
Absolute temperature in kelvin.
Radiating area in square metres.
0 to 1; 1 is a perfect black body.
Try a hot surface
Radiated power (P)
56.704kW

Power per unit area: 56.704 kW/m². Emissivity ε = 1.

Radiated power vs temperature — a steep T⁴ curve
Radiated power rising as the fourth power of temperature56.704 kW0 W0 K1000 K

A 1 m² black body at 1000 K radiates P = εσAT⁴ = 56,704 W (about 56.7 kW). Because power grows with the fourth power of temperature, doubling T multiplies radiated power by 2⁴ = 16 — so the same surface at 2000 K pours out roughly 907 kW.

Radiation from a hot surface

Every object above absolute zero radiates thermal energy. The Stefan-Boltzmann law gives the total power leaving a surface as P = εσAT⁴ — emissivity times the Stefan-Boltzmann constant σ times area times absolute temperature to the fourth power. The T⁴ term makes radiation extraordinarily sensitive to temperature, which is why a glowing filament or a star sheds so much more energy than a warm body.

P = ε × σ × A × T⁴  ·  T = (P ÷ εσA)^(1∕4)

P = radiated power (W), ε = emissivity (0–1), σ = 5.670374×10⁻⁸ W·m⁻²·K⁻⁴, A = area (m²), T = temperature (K)

Worked example

How much power does a 1 m² black body (ε = 1) radiate at 1000 K?

  1. 1
    Convert the temperature to kelvin. The law needs absolute temperature: here T = 1000 K already.
  2. 2
    Write the Stefan-Boltzmann law. P = ε × σ × A × T⁴ with σ = 5.670374×10⁻⁸ W·m⁻²·K⁻⁴.
  3. 3
    Raise T to the fourth power. 1000⁴ = 1×10¹² K⁴.
  4. 4
    Multiply everything together. P = 1 × 5.670374×10⁻⁸ × 1 × 1×10¹² = 56,704 W ≈ 56.7 kW.

Radiated power of a 1 m² black body

Perfect emitter (ε = 1, A = 1 m²). Power scales with T⁴, so it climbs fast.

Temperature (K)Radiated powerRough context
300459 WNear room temperature
5003.54 kWA hot oven surface
100056.7 kWDull red heat
2000907 kWBright orange-white
30004.59 MWIncandescent bulb region
577863.2 MWThe Sun’s surface

Reading the result

Temperature must be in kelvin. Because the law uses T⁴, feeding it Celsius gives nonsense — always convert first (K = °C + 273.15). A small error in T is magnified fourfold in the power.

The T⁴ dependence dominates. Radiated power is proportional to T⁴, so power rises far faster than temperature: raising T by 19 % nearly doubles the output, and doubling T multiplies it by 16.

Emissivity scales the whole result. A perfect black body has ε = 1; real surfaces sit below that (polished metal near 0.05, skin near 0.98). The radiated power is simply the black-body value multiplied by ε.

Why is temperature raised to the fourth power?
Integrating the Planck black-body spectrum over all wavelengths yields a total emitted power proportional to T⁴. This steep dependence means a modest temperature rise produces a large jump in radiated energy.
What is emissivity (ε)?
Emissivity is how efficiently a surface radiates compared with a perfect black body, on a 0-to-1 scale. A black body has ε = 1; polished metal can be near 0.05, human skin about 0.98. It multiplies the whole result.
What is the Stefan-Boltzmann constant σ?
σ = 5.670374×10⁻⁸ W·m⁻²·K⁻⁴. It sets the power per unit area a perfect black body radiates at a given absolute temperature, so P = εσAT⁴.
Do I have to use kelvin?
Yes. The law needs absolute temperature because T⁴ is meaningless with an offset scale. Convert Celsius with K = °C + 273.15 (and Fahrenheit to Celsius first) before entering T.
What is power per unit area?
Dividing radiated power by area gives the radiant flux εσT⁴ in W/m², independent of the object’s size. The Sun’s surface radiates about 63 MW per square metre.
Does this account for the surroundings?
No. P = εσAT⁴ is the power a surface emits. The net radiative exchange with surroundings at temperature T₀ is εσA(T⁴ − T₀⁴); this tool computes the outgoing term only.