Stefan-Boltzmann Calculator, Thermal Radiation (P = εσAT⁴)

The Stefan-Boltzmann calculator solves the thermal radiation equation P = εσAT⁴ for the power radiated by a surface at absolute temperature T, with optional surroundings temperature for net flux. It includes a back-solve mode that finds the surface temperature that would produce a given radiated power, and a library of preset emissivities for common surfaces.

What is the Stefan-Boltzmann law?

The Stefan-Boltzmann law states that the total thermal radiation emitted per unit area by a blackbody is proportional to the fourth power of its absolute temperature. For a real surface with emissivity ε, the law reads:

P = ε · σ · A · T⁴

where P is radiated power in watts, A is the surface area in m², T is the absolute temperature in kelvin, ε is the dimensionless emissivity (0 to 1), and σ is the Stefan-Boltzmann constant:

σ = 5.670374419 × 10⁻⁸ W / (m² · K⁴)   (CODATA 2018)

The relationship was first observed empirically by the Slovene-Austrian physicist Josef Stefan in 1879 and given a thermodynamic derivation five years later by Ludwig Boltzmann, using Maxwell's equations and the second law of thermodynamics.

What does emissivity mean?

Emissivity ε is a dimensionless number between 0 and 1 that describes how efficiently a surface radiates compared to a perfect blackbody at the same temperature.

• ε = 1, perfect blackbody (theoretical maximum; only approximated by some practical surfaces like soot or lampblack)
• ε = 0.95, human skin, fresh snow, asphalt
• ε = 0.05, polished aluminium, polished silver
• ε ≈ 0, perfect reflector

Emissivity is wavelength-dependent. Most engineering calculations use a single grey-body emissivity averaged over all wavelengths; for highly selective surfaces (solar absorbers, infrared mirrors), a wavelength-resolved calculation is required.

How to convert from °C to K

The Stefan-Boltzmann law requires absolute temperature in kelvin. To convert:

T(K) = T(°C) + 273.15

Examples:

• 0 °C = 273.15 K
• 25 °C = 298.15 K
• 100 °C = 373.15 K (boiling water at 1 atm)
• 1000 °C = 1273.15 K
• 2500 K = 2226.85 °C (typical tungsten bulb filament)

If you accidentally plug in degrees Celsius instead of kelvin, the answer will be off by a factor of millions, a very common pitfall.

Worked examples

Example 1: A human in a cold room

A person with skin temperature 33 °C (306.15 K), emissivity 0.98, and total skin area 1.7 m² stands in a room at 20 °C (293.15 K). How much net heat does the body lose by radiation?

P_net = ε · σ · A · (T⁴ − T_env⁴)
     = 0.98 × 5.67e-8 × 1.7 × (306.15⁴ − 293.15⁴)
     = 0.98 × 5.67e-8 × 1.7 × (8.79e9 − 7.39e9)
     = 0.98 × 5.67e-8 × 1.7 × 1.40e9
     ≈ 132 W

About 132 W of net radiative heat loss, most of the body's basal metabolic rate at rest (about 100 W) goes to making up this loss plus the convective and evaporative losses.

Example 2: A 100 W incandescent bulb

A tungsten filament operates at 2500 K with ε ≈ 0.30. What surface area does the filament need to radiate 100 W?

P = ε · σ · A · T⁴
A = P / (ε · σ · T⁴)
  = 100 / (0.30 × 5.67e-8 × 2500⁴)
  = 100 / (0.30 × 5.67e-8 × 3.91e13)
  = 100 / 665
  ≈ 0.15 cm²

A surprisingly tiny area, about 0.15 cm² of glowing tungsten is all that's needed to radiate 100 W. Real filaments are coiled into a longer wire to fit inside a standard bulb.

Example 3: Earth as a blackbody

The Earth absorbs about 240 W/m² of solar radiation on average (after reflection by clouds and ice). What temperature does a perfect blackbody at 240 W/m² sit at?

P/A = σ · T⁴
T = (240 / σ)^(1/4) = (240 / 5.67e-8)^(1/4) = (4.23e9)^(1/4) ≈ 254.8 K

About -18.4 °C, close to the actual mean emission temperature of Earth's surface, accounting for clouds and greenhouse effect.

Example 4: Radiative cooling roof

A roof at 30 °C (303.15 K) with ε = 0.90 radiates to a clear night sky at 0 °C (273.15 K). What is the net cooling power per square metre?

P_net/A = ε · σ · (T⁴ − T_sky⁴)
        = 0.90 × 5.67e-8 × (303.15⁴ − 273.15⁴)
        = 0.90 × 5.67e-8 × (8.44e9 − 5.56e9)
        ≈ 147 W/m²

About 147 W/m² of cooling, enough to drop a roof's surface temperature well below ambient air temperature on a clear night. This is the principle behind "radiative cooling paint" being developed to reduce air-conditioning demand.

Common applications

Incandescent lighting

A 60 W incandescent bulb radiates about 60 W from a tiny tungsten filament at 2500 K. The radiated power spectrum peaks in the near-infrared, with only 2-5% of the radiation falling in the visible band, the rest is waste heat. Modern LEDs are more efficient because they emit narrow-spectrum light rather than a broad Planck distribution.

Low-emissivity windows

A double-pane window with a Low-E coating has an emissivity of about 0.05 in the thermal infrared (compared to ε ≈ 0.84 for uncoated glass). The coating reflects interior infrared radiation back into the room rather than letting it pass through and escape, cutting heating losses by 30-50% in cold climates.

Solar thermal collectors

A good solar absorber should have:

• High ε in the visible/near-IR (to absorb sunlight)
• Low ε in the mid-IR (to keep absorbed heat from re-radiating)

Selective surfaces achieve this with multilayer coatings of metal and dielectric materials.

Fire detection and thermal imaging

A thermal imaging camera measures emitted infrared radiation and converts it to a temperature map using the Stefan-Boltzmann law. The camera's emissivity setting compensates for surfaces that are not perfect blackbodies, human skin (ε ≈ 0.98) reads correctly, but polished metal (ε ≈ 0.05) reads as much colder than its true temperature.

Cryogenic insulation

In a cryogenic vacuum system, the only heat transfer mechanism is radiation. Even a small ε at room temperature can dump tens of watts into a 4 K surface. Multilayer insulation (MLI), alternating layers of highly reflective metalized plastic, reduces the effective emissivity to below 0.001.

Greenhouse effect

Glass is nearly transparent to visible light but strongly absorbing in the infrared. Sunlight enters a greenhouse, is absorbed by the interior, and re-radiates as longer-wavelength infrared, which the glass absorbs and re-radiates back inward. The Stefan-Boltzmann law applied to the interior surfaces gives the equilibrium temperature, which can be 20-30 K warmer than ambient.

Solar energy

The sun's surface temperature is about 5778 K. At ε = 1 (treat it as a blackbody), the radiated flux is:

P/A = σT⁴ = 5.67e-8 × 5778⁴ ≈ 6.3 × 10⁷ W/m² ≈ 63 MW/m²

At Earth's distance, the flux drops to about 1360 W/m² (the solar constant), reduced by the inverse-square law.

Why the T⁴ dependence?

The fourth-power dependence comes from integrating the Planck blackbody spectrum over all wavelengths and summing over all directions. Mathematically:

P/A = ∫₀^∞ ∫₄π I(λ, T) dΩ dλ

where I(λ, T) is the spectral radiance given by Planck's law:

I(λ, T) = (2hc²/λ⁵) / (exp(hc/λkT) − 1)

Carrying out the integral and picking up the factor of (1/4) from the cosine-of-angle averaging over a hemisphere gives:

P/A = (2π⁵ k⁴ / 15 c² h³) T⁴ ≡ σ T⁴

So σ is not an independent constant, it can be derived from Boltzmann's constant k, the speed of light c, and Planck's constant h.

The historical origins

Josef Stefan (1835-1893) was a Slovene-Austrian physicist working in Vienna. In 1879 he observed the radiation from a heated platinum wire and noticed that the radiated power per unit area scaled as the fourth power of the absolute temperature, about 5.67 × 10⁻⁸ W/(m²·K⁴) became known as Stefan's constant.

His former student Ludwig Boltzmann (1844-1906) provided the theoretical derivation in 1884, using Maxwell's electromagnetic theory and a thermodynamic argument involving the pressure of light. Boltzmann's derivation was an early triumph of the then-new theory of thermodynamics.

The constant was re-measured with increasing precision over the 20th century, and the current CODATA value of 5.670374419 × 10⁻⁸ W/(m²·K⁴) is exact to 9 significant figures.

Common mistakes

Using °C instead of K

The single most common error. Plugging in 25 °C instead of 298.15 K gives an answer that's wrong by a factor of millions. Always convert to kelvin first.

Using surface area instead of effective radiating area

For non-flat surfaces, the effective radiating area can differ from the geometric area. A cylindrical wire has the same geometric area as a flat strip of the same dimensions, but the radiation is emitted from the entire cylinder, with cosine-of-angle intensity distribution. For most engineering purposes, geometric area is a good approximation; for precise work, account for view factors.

Forgetting emissivity

If you assume ε = 1 but your real surface has ε = 0.1, you will over-predict the radiated power by 10×. Polished metal surfaces in particular are easily mistaken.

Net flux vs gross flux

When you compute P = εσAT⁴ you get the total power the surface emits. But a surface in a warm environment also absorbs radiation. The net flux P_net = εσA(T⁴ − T_env⁴) accounts for both.

Treating one temperature as average

In a complex geometry, different surfaces can have very different temperatures. The Stefan-Boltzmann law applies to a single surface temperature; summing over multiple surfaces requires integration or numerical methods.

Why metals have low emissivity

In a metal, free electrons dominate the optical response. A photon entering a metal interacts primarily with the electron gas rather than the atomic lattice; the photon is typically reflected before it can thermalise. The metal therefore has low absorption (and low emissivity, by Kirchhoff's law).

Polished silver is the most extreme case at ε ≈ 0.02. Aluminium is similar at ε ≈ 0.05. As the metal roughens, oxidises, or is coated, the emissivity rises dramatically, a roughened aluminium surface can have ε ≈ 0.3. This is why a thermal-camera image of a vehicle shows the painted panels glowing brightly while the chrome trim stays dark, even at the same temperature.

Kirchhoff's law

For any opaque body in thermal equilibrium, emissivity equals absorptivity at every wavelength:

ε(λ) = α(λ)

A surface that absorbs strongly at wavelength λ also emits strongly at that wavelength. This is why a blackbody is also a perfect absorber: ε = 1 everywhere means α = 1 everywhere.

For a grey body (ε constant across wavelengths), the same relation holds in integrated form. Kirchhoff's law is the bridge between the Stefan-Boltzmann law (which describes emission) and the problem of absorption (which describes how a surface responds to incoming radiation).

Frequently Asked Questions

What is the exact value of the Stefan-Boltzmann constant?

The CODATA 2018 recommended value is exactly 5.670374419 × 10⁻⁸ W / (m² · K⁴). It can be derived from more fundamental constants as σ = (2π⁵ k⁴) / (15 c² h³), where k = 1.380649 × 10⁻²³ J/K (Boltzmann's constant), c = 299,792,458 m/s (speed of light), and h = 6.62607015 × 10⁻³⁴ J·s (Planck's constant).

Is the Stefan-Boltzmann law only for blackbodies?

The strict form P/A = σT⁴ is for a blackbody. For a real surface, multiply by the emissivity: P/A = εσT⁴. This is the "grey-body" approximation, accurate for most engineering surfaces, less so for highly selective ones.

What temperature is the sun, and does it follow Stefan-Boltzmann?

The sun's surface (photosphere) is about 5778 K. At that temperature with ε = 1, the Stefan-Boltzmann law predicts about 6.3 × 10⁷ W/m², close to the actual solar flux. This confirms the sun's photosphere behaves approximately as a blackbody (which is why the solar spectrum matches a 5778 K blackbody so well).

How does this relate to climate change?

The Stefan-Boltzmann law applied to Earth's energy balance sets the equilibrium temperature. A doubling of CO₂ concentration reduces the outgoing infrared flux by about 4 W/m². To restore equilibrium, Earth's temperature must rise by about 1 °C (a rough estimate based on the derivative of σT⁴ at 255 K). The exact number depends on feedback loops, but Stefan-Boltzmann is the starting point.

Can a colder object radiate more than a hotter object?

No, at the same emissivity, hotter always wins. But emissivity varies, so a polished metal at 1000 K (ε = 0.05) radiates less than a sooty surface at 300 K (ε = 0.95). However, for the same surface (fixed ε), hotter wins decisively.

What is the wavelength of peak emission?

That's Wien's displacement law, not Stefan-Boltzmann: λ_max = b/T, where b ≈ 2.898 × 10⁻³ m·K. At 300 K, λ_max ≈ 9.66 µm (far infrared). At 5778 K, λ_max ≈ 501 nm (visible green), which is why the sun is white.

Why is incandescent lighting so inefficient?

A tungsten filament at 2500 K radiates ~60 W/cm², but the peak wavelength is about 1.16 µm, in the near-infrared, not the visible spectrum. The visible portion (about 400-700 nm) is only about 5% of the radiated power. LEDs are more efficient because they emit narrow-spectrum light directly, with most output in the visible band.

Does the formula work in vacuum?

Yes, thermal radiation is purely electromagnetic and propagates through vacuum perfectly. In fact, in vacuum (space), radiation is the only way a satellite loses waste heat. Convection and conduction both require a medium.

What's the difference between emissivity and reflectance?

By Kirchhoff's law, for an opaque body: ε + ρ = 1 (where ρ is reflectance). A surface with ρ = 0.95 reflects 95% of incident radiation and emits only 5% as much as a blackbody would. A surface with ρ = 0.05 absorbs 95% and emits 95% as much as a blackbody.

Can the formula predict the heat death of the universe?

In a sense. The Stefan-Boltzmann law applied to every radiating body means all temperature differences must eventually smooth out. In the very far future, all matter approaches a uniform temperature near absolute zero, radiating an infinitesimal amount. This is "heat death", the thermodynamic equilibrium state predicted by the second law.

References

• Encyclopaedia Britannica, Stefan-Boltzmann Law. https://www.britannica.com/science/Stefan-Boltzmann-law
• CODATA recommended value of the Stefan-Boltzmann constant, NIST. https://physics.nist.gov/cgi-bin/cuu/Value?stefan