How this instrument works
Josef Stefan announced the fourth-power rule in Vienna in 1879, having teased it out of John Tyndall's measurements on a platinum wire glowing from dull red up to dazzling white. Five years later his student Ludwig Boltzmann rebuilt the same result from thermodynamics alone, running a Carnot cycle on a cylinder of radiation and leaning on Maxwell's finding that light exerts pressure equal to one third of its energy density. Stefan then turned his own law around and placed our Sun's surface near 5700 K — first estimate of any stellar temperature that was not simply guesswork.
Emissivity ε is the honesty factor. Gustav Kirchhoff had shown by 1860 that at thermal equilibrium a surface emits exactly as readily as it absorbs, so a perfect absorber — a blackbody — must also be a perfect emitter, ε = 1. Nothing real quite reaches that, though matt black paint, human skin and weathered brick all sit between 0.93 and 0.98, and a small hole drilled into a heated cavity comes closer still. That is why calibration blackbodies are built as cavities rather than as plates. Polished metals behave in the opposite way: bright aluminium foil emits at roughly 0.04, and that single fact is why an infrared thermometer aimed at shiny steel reports nonsense while the same instrument reads a painted wall to within a degree.
Written as P = εσAT⁴, this is gross emission into surroundings held at absolute zero, which is rarely anyone's situation. A person at 306 K is not shedding 800 W into a 293 K room, because that room radiates back and only their difference εσA(T⁴ − T_surr⁴) actually departs — around 100 W, which is why a full lecture theatre warms up. Three further assumptions deserve watching. Any surface here is treated as grey, its emissivity identical at every wavelength, an idea that fails badly for selective coatings and for glass. Nothing is presumed to block its view, so enclosed or mutually facing geometries need view factors. And an emitter must be large compared with wavelengths it produces: across sub-micron gaps, evanescent coupling ferries heat far past anything this equation permits.
- Put the emitting surface into Radiating area — square centimetres, square metres or square feet. Use the face that actually sees open space, not the whole object.
- Enter Absolute temperature (K) in kelvin, never Celsius — add 273.15 first. Type 27 where you meant 300 K and the answer lands fifteen thousand times low, since the error is raised to the fourth power.
- Set Emissivity between 0 and 1: use 1 for an ideal blackbody, 0.95 for matt black paint or skin, 0.8 for oxidised steel, 0.04 for polished aluminium.
- For a sphere such as a planet or star, enter 4πR² as the Radiating area; for a flat panel that radiates from one side only, enter just that single face.
- Read Radiated power in watts, kilowatts or megawatts. It is gross output — subtract the surroundings' own εσAT⁴ if what you want is the net loss.
Worked example — a square metre of furnace mouth at 1000 K
A muffle furnace is opened for loading, and one square metre of its glowing interior faces the room. Radiating area = 1, Absolute temperature (K) = 1000 — that is 727 °C — and Emissivity = 1, since a deep cavity behaves as an almost perfect blackbody. Because 1000⁴ is exactly 10¹², the arithmetic collapses to a decimal shift: P = 1 × 5.670374419 × 10⁻⁸ × 1 × 10¹², giving a Radiated power of 56703.74419 W, near enough 56.7 kW.
Fifty-seven kilowatts off one square metre is nineteen electric kettles pointed at you, and it explains why furnace doors are opened briefly and from behind a face shield. Almost none of that is light you can see: Wien's displacement law puts the spectral peak at 2.898 × 10⁻³ ⁄ 1000 = 2.9 µm, deep in the mid-infrared, so a 1000 K cavity looks a modest cherry-red while pouring out invisible heat. Cool the same aperture to 500 K and the output does not halve — it drops sixteenfold, to 3.54 kW. That is the entire character of a fourth power.
Questions
Why must the temperature be entered in kelvin?
Because the law rests on absolute temperature, where zero means no thermal motion whatever. Celsius and Fahrenheit place their zeros at arbitrary points, so ratios between their readings carry no physical meaning — and this formula takes a ratio, then raises it to the fourth power. Enter 27 instead of 300 K and the answer comes back roughly fifteen thousand times too small. Add 273.15 to any Celsius figure before it goes into Absolute temperature (K). The one quantity where Celsius survives intact is a temperature difference, which this sheet never asks you for.
What emissivity should I use for my surface?
Match it to the finish rather than the bulk material, since a micron of oxide or paint governs the whole result. Rough guides near room temperature: human skin 0.98, water 0.96, matt black paint 0.95, brick and concrete 0.90 to 0.93, wood 0.90, oxidised steel 0.75 to 0.85, as-rolled stainless 0.16, polished aluminium 0.04. Emissivity also drifts with temperature and with viewing angle, so treat handbook figures as ±0.05 unless somebody measured your actual sample. Where doubt remains, paint a test patch matt black and aim at that.
Is this the heat my object actually loses?
No — it is gross emission, computed as though surroundings sat at absolute zero. Everything nearby radiates back at you, so your net figure is εσA(T⁴ − T_surr⁴). Run both temperatures through this sheet and subtract: one 320 K panel in a 295 K room emits 595 W per square metre at ε = 1 but nets only about 165 W. Out in deep space, surroundings sit at 2.7 K, whose fourth power is negligible, and gross emission becomes honest again — which is why spacecraft radiators are textbook cases for this bare formula.
Does the law hold for a gas or a flame?
Only loosely. Stefan–Boltzmann describes a surface, whereas gases radiate from a volume and do so in narrow bands set by their molecules rather than across a smooth continuum. Carbon dioxide and water vapour emit strongly at a few wavelengths and stay nearly transparent between them, so a clean blue flame radiates far less than its temperature would suggest. Sooty flames differ again: incandescent carbon particles behave much like tiny grey bodies, which is why a yellow candle feels warmer across a room than a hotter blue gas ring.
How does this relate to Wien's displacement law?
They answer different questions about one spectrum. Stefan–Boltzmann returns total power under Planck's curve; Wien locates where that curve peaks, at λ_max = 2.898 × 10⁻³ ⁄ T metres. At 300 K that peak lands near 9.7 µm, which is precisely why thermal cameras are built for an 8 to 14 µm window. At 5772 K, our Sun's effective temperature, it falls at 502 nm, green and near mid-range of human vision. Each follows from Planck's law, one by integrating and one by differentiating; neither can be derived from its partner.
Why is the Stefan–Boltzmann constant an exact number now?
Because σ is no longer measured on its own — it falls out of Planck's law as 2π⁵k⁴ ⁄ (15h³c²). When the SI was revised in May 2019, the Boltzmann constant k, the Planck constant h and the speed of light c were every one of them fixed by definition, so any combination of the three inherited that exactness. CODATA prints 5.670374419 × 10⁻⁸ W·m⁻²·K⁻⁴, and those digits are simply where a non-terminating exact expression has been truncated to fit a table.