Stefan–Boltzmann Law Calculator
Calculate the thermal radiation emitted by any surface using the Stefan–Boltzmann law. Enter surface temperature, emissivity, and area to get radiated power, plus net heat transfer with the surroundings.
Find the Thermal Radiation Emitted by a Surface
Stefan–Boltzmann Law Formula
Where:
- \(P\) — total radiated power (W)
- \(\varepsilon\) — emissivity of the surface (dimensionless, 0 to 1)
- \(\sigma\) — Stefan–Boltzmann constant, \(5.670374419 \times 10^{-8}\ \mathrm{W/(m^2 \cdot K^4)}\)
- \(A\) — radiating surface area (m²)
- \(T\) — absolute temperature of the surface (K)
The formula gives the total electromagnetic power a surface emits by thermal radiation. Because \(T\) is raised to the fourth power, radiated power grows extremely fast with temperature: doubling the absolute temperature increases the radiation 16-fold.
If you also want the net exchange with cooler or hotter surroundings at temperature \(T_0\), the calculator uses:
\[ P_{net} = \varepsilon \, \sigma \, A \left( T^{4} - T_0^{4} \right) \]
A positive result means the surface loses heat; a negative result means it absorbs more radiation than it emits and heats up.
Stefan–Boltzmann Law – Calculation Example
Example: a metal plate with an emissivity of 0.9 and an area of 1 m² is heated to 500 K in a room whose walls are at 300 K.
\[ P = \varepsilon \, \sigma \, A \, T^4 = 0.9 \times 5.670374419 \times 10^{-8} \times 1 \times 500^4 \approx 3189.59\ \mathrm{W} \]
The plate emits about 3189.59 W (3.19 kW) of thermal radiation. To find the net exchange with the room:
\[ P_{net} = 0.9 \times 5.670374419 \times 10^{-8} \times 1 \times (500^4 - 300^4) \approx 3189.59 - 413.37 = 2776.22\ \mathrm{W} \]
Result: the plate radiates 3189.59 W in total, and its net radiative heat loss to the 300 K surroundings is 2776.22 W (about 2.78 kW).
How to Use the Calculator
- Enter the Surface temperature (T) and pick its unit — K, °C, or °F. The value is converted to kelvin automatically (e.g. an incandescent filament runs near 2800 K, boiling water near 373 K).
- Enter the Emissivity (ε) of the surface, between 0 and 1. Use 1 for an ideal blackbody, roughly 0.8–0.95 for most non-metallic surfaces, and 0.05–0.2 for polished metals.
- Enter the Surface area (A) and select m², cm², ft², or in².
- Optionally enter the Surroundings temperature (T₀) to also get the net radiative heat transfer. Leave this field blank if you only need the total radiated power.
- Read the results: total radiated power in watts, kilowatts, horsepower and BTU/h, plus the net heat transfer if a surroundings temperature was given.
What Is the Stefan–Boltzmann Law?
Every object above absolute zero emits electromagnetic radiation because of the thermal motion of its charged particles. At everyday temperatures almost all of this radiation is invisible infrared; hot objects like stove burners and incandescent filaments radiate enough visible light to glow. The Stefan–Boltzmann law, derived by Josef Stefan in 1879 and confirmed theoretically by Ludwig Boltzmann in 1884, quantifies this emission: the total radiated power is proportional to the fourth power of the absolute temperature.
The law strictly describes a blackbody — an idealized surface that absorbs and emits radiation perfectly at every wavelength. Real surfaces fall short of this ideal, which is accounted for by the emissivity \(\varepsilon\). A blackbody has \(\varepsilon = 1\); dark, rough, non-metallic surfaces typically fall between 0.8 and 0.95, while polished metals can be below 0.1, which is why a shiny aluminum surface radiates far less heat than a black-painted one at the same temperature.
Interpreting the result is straightforward. The total radiated power tells you how much energy the surface emits per second, regardless of what it receives back. The net radiative heat transfer is usually the more practical number: it is the difference between what the surface emits and what it absorbs from its surroundings. A hot object in a cool room has a positive net transfer (it cools by radiation); an object colder than its surroundings has a negative one (it warms up); and when the two temperatures are equal, emission and absorption cancel exactly.
The fourth-power dependence dominates practical thermal design. This is why radiant heaters, lava lamps, and re-entry spacecraft heat shields work the way they do, and why doubling the temperature of a kiln multiplies its wall losses by sixteen. To explore the wavelength where the radiation peaks at a given temperature, see the related Wien's Displacement Law calculator.
Assumptions, Limitations and Practical Tips
- The surface is isothermal and gray. The calculator assumes one uniform temperature and a constant emissivity across all wavelengths. Real materials have wavelength-dependent emissivity, which can introduce errors of a few percent.
- View factors are ignored in the net result. The net-transfer formula assumes the surroundings fully enclose the surface (a view factor of 1), so all radiation leaving the surface reaches the environment. For partially exposed surfaces the true net exchange is smaller.
- Only radiation is computed. Real objects also lose heat by convection and conduction; at moderate temperatures convection often dominates, so the radiated power alone rarely gives the total cooling rate.
- Temperature must be absolute. The law uses kelvin. The calculator converts °C and °F for you, but if you compute by hand, remember that 500 °C = 773.15 K — not 500 K — and the fourth power makes the difference enormous.
- Emissivity matters as much as temperature. When estimating heat from a real object, choosing \(\varepsilon\) properly (rough and oxidized surfaces radiate much more than polished ones) is often the largest source of uncertainty.
- Check the direction of the net result. A negative net power simply means the surroundings are hotter than the surface and it is absorbing energy — the surface will warm up.
For a related energy-conversion perspective, the Carnot efficiency calculator shows the maximum fraction of that heat energy a heat engine could convert to work.
Frequently Asked Questions
How do I calculate radiated power by hand?
Convert the temperature to kelvin, then multiply four factors: emissivity, the Stefan–Boltzmann constant \( 5.670374419 \times 10^{-8} \), the area in m², and \( T^4 \). For example, a 1 m² blackbody surface at 400 K radiates \( 1 \times 5.67 \times 10^{-8} \times 1 \times 400^4 \approx 1452 \) W. For net exchange, subtract \( T_0^4 \) from \( T^4 \) before multiplying by the other factors.
Why is my hand calculation so far off?
The most common mistake is plugging in Celsius or Fahrenheit instead of kelvin. Because \( T \) is raised to the fourth power, using 500 instead of 773.15 (500 °C in kelvin) understates the result by a factor of about 3.6. Always add 273.15 to °C, or convert °F with \( (T_F + 459.67) \times 5/9 \), and make sure the area is in m².
How much radiation does the human body emit?
An adult with roughly 1.8 m² of skin at about 305 K and emissivity near 0.97 radiates around 900–1000 W in total. However, the body also absorbs radiation from its ~293 K surroundings, so the net radiative loss is closer to 100–150 W — comparable to a bright light bulb. This is why you feel cold in a poorly heated room even when the air temperature seems acceptable.
Does this calculator account for convection and conduction?
No — it computes thermal radiation only. At room temperature, free convection often removes comparable or greater amounts of heat, so a real object usually cools faster than the radiated power alone suggests. In a vacuum (thermos flasks, spacecraft, the Sun-to-Earth path) radiation is the only mechanism, and the Stefan–Boltzmann result is then essentially the whole story.
What does a negative net heat transfer result mean?
It means the surroundings are hotter than the surface, so the surface absorbs more radiation than it emits and warms up. There is nothing wrong with the calculation — just reverse the interpretation. A positive value means the surface loses heat to cooler surroundings, and zero means the surface and its environment are in radiative equilibrium at the same temperature.
At what temperature does radiation become the dominant heat loss?
As a rule of thumb, radiation rivals or exceeds natural convection for surfaces around 100–200 °C and above, because radiated power grows as \( T^4 \) while convective losses grow roughly linearly with the temperature difference. That's why radiant heaters, kilns, and furnace design rely heavily on this law, while for a warm cup of coffee at 60 °C convection and evaporation matter more.
Can I use this calculator for the Sun or a light bulb filament?
Yes, as long as you know the surface temperature, area, and emissivity. The Sun's photosphere (~5778 K, \( \varepsilon \approx 1 \)) yields the familiar total output of about \( 3.8 \times 10^{26} \) W. For an incandescent filament near 2800 K, use the tungsten surface area and a temperature-dependent emissivity of roughly 0.3 — note that tungsten's emissivity varies with temperature, which limits accuracy.