Stefan-Boltzmann Law
The Stefan-Boltzmann law gives the radiated power of a black body: it grows with the fourth power of the absolute temperature.
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Formula
P = \sigma \cdot A \cdot T^4Variables & units – Stefan-Boltzmann Law
| Symbol | Meaning | Unit |
|---|---|---|
| P | Radiated power | W |
| σ | Stefan-Boltzmann constant (5.670×10⁻⁸) | W/(m²·K⁴) |
| A | Radiating surface area | m² |
| T | Absolute temperature | K |
Derivation & background – Stefan-Boltzmann Law
Josef Stefan found the law empirically in 1879, Ludwig Boltzmann derived it thermodynamically in 1884; today it follows from the Planck radiation law by integrating over all wavelengths. Real surfaces radiate less than the ideal black body, corrected by the emissivity ε ≤ 1: P = ε·σ·A·T⁴. A body in surroundings at temperature T_U radiates net P = ε·σ·A·(T⁴ − T_U⁴). The T⁴ dependence makes radiation the dominant heat transfer at high temperatures.
Exam blueprint
Validity range
Holds exactly for an ideal black body. Real surfaces need the emissivity ε ≤ 1 (P = ε·σ·A·T⁴); the net exchange with the surroundings uses the difference T⁴ − T_U⁴. T always in kelvin.
Derivation steps
Integrating the Planck radiation spectrum over all wavelengths yields the T⁴ dependence.
- 1The spectral radiance of the black body is summed (integrated) over all wavelengths.
- 2The integral gives P/A ∝ T⁴; the constant σ bundles fundamental constants (σ = 2π⁵k⁴/(15h³c²)).
Rearrangements
Temperature
This is how stellar surface temperatures are determined.
Area
Yields, for example, stellar radii from luminosity and temperature.
Task variant
What power does the Sun radiate? (R = 6.96×10⁸ m, T = 5778 K)
A = 4πR² ≈ 6.09×10¹⁸ m², T⁴ ≈ 1.115×10¹⁵ K⁴. P = 5.67×10⁻⁸ × 6.09×10¹⁸ × 1.115×10¹⁵ ≈ 3.8×10²⁶ W.
How much does a person radiate net? (A = 1.7 m², skin 306 K, surroundings 293 K, ε ≈ 1)
P = σ·A·(T⁴ − T_U⁴) = 5.67×10⁻⁸ × 1.7 × (8.77 − 7.37)×10⁹ ≈ 135 W.
Common mistakes
Substituting temperatures in degrees Celsius.
Because of the fourth power kelvin is mandatory: 25 °C = 298 K.
Forgetting the back radiation of the surroundings.
Net emission is only σ·A·(T⁴ − T_U⁴).
Treating real surfaces as perfect black bodies.
Include the emissivity ε; polished metal has ε of only 0.02 to 0.1.
Forgetting the fourth root when solving for T.
T = (P/(σA))^(1/4), do not divide by 4.
Exam context
- Tasks on stellar luminosity and radius, filaments and the Earth radiation balance; the factor-16 effect of doubling the temperature is popular.
These mistakes cost points in real exams. The set drills them until they stick.
Formula cluster
Thermal radiation
The third heat transfer mechanism besides conduction and convection.
Worked example
Filament: T = 2500 K, A = 10 mm² = 10⁻⁵ m²: P = 5.67×10⁻⁸ × 10⁻⁵ × 2500⁴ ≈ 22 W. Due to T⁴ the power already doubles at about 19 % more temperature.
Applications
Lamps and filaments, stellar luminosity and surface temperature, thermal imaging cameras, climate physics (Earth radiation balance), thermography
Quanta exam set
Curated exam set for "Stefan-Boltzmann Law":
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Which formula describes Stefan-Boltzmann Law?
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How do you rearrange P = σ·A·T⁴ for Temperature?
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Which common mistake happens with Stefan-Boltzmann Law?
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Scientific sources
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Frequently asked questions about Stefan-Boltzmann Law
How do you calculate the radiated power of a hot body?+
Insert area and absolute temperature into P = σ·A·T⁴, with the Stefan-Boltzmann constant σ = 5.670×10⁻⁸ W/(m²·K⁴). The temperature must be in kelvin. Filament example: A = 10 mm² = 10⁻⁵ m² at T = 2500 K gives P = 5.67×10⁻⁸ × 10⁻⁵ × 2500⁴ ≈ 22 W, since 2500⁴ = 3.9×10¹³. For real, non-ideal surfaces additionally multiply by the emissivity ε (0 to 1). Compute the fourth power step by step or with the power key and check the powers of ten at the end; they are the most common error source.
Why is the fourth power of temperature so decisive?+
Because small temperature changes cause huge power changes. If the absolute temperature doubles, the radiation grows by a factor of 2⁴ = 16. Just 19 % more temperature already doubles the power, since 1.19⁴ ≈ 2. That is why thermal radiation dominates all other transport paths at high temperatures: a hotplate at 500 °C (773 K) radiates per area about 48 times more than the same plate at room temperature (293 K), calculation: (773/293)⁴ ≈ 48. Conversely, for astronomy this means: a star with twice the surface temperature shines 16 times brighter at the same size, which is why hot blue stars are far more luminous than red ones of similar radius.
Why must the temperature be inserted in kelvin?+
The law describes radiation as a function of absolute temperature, which starts at absolute zero, where a body radiates nothing. The Celsius scale instead places its zero arbitrarily at the freezing point of water; a body at 0 °C still radiates strongly (about 315 W/m²). Because of the fourth power, Celsius errors are catastrophic: for 25 °C, computing with T = 25 instead of 298 K yields a value too small by the factor (298/25)⁴ ≈ 20,000, and for negative Celsius values the fourth power would even mask the sign problem. So always convert: T in K = temperature in °C + 273.15.
How do you calculate the luminosity and temperature of stars?+
Stars radiate to a good approximation like black bodies, so L = σ·4πR²·T⁴ with the stellar radius R. For the Sun (R = 6.96×10⁸ m, T = 5778 K) this gives L = 5.67×10⁻⁸ × 6.09×10¹⁸ × 1.115×10¹⁵ ≈ 3.8×10²⁶ W. Astronomers use the relation in both directions: from measured luminosity and spectral temperature follows the radius R = √(L/(4πσT⁴)), which is how giant stars and white dwarfs were recognised as such. The temperature itself comes from the Wien displacement law using the colour of the star (λ_max·T = 2.898×10⁻³ m·K). Together the two laws make radius, temperature and power determinable from light alone.
Why does a body cool more slowly than P = σAT⁴ suggests?+
Because the surroundings radiate back. A body at temperature T emits σ·A·T⁴ but simultaneously receives σ·A·T_U⁴ from surrounding surfaces at temperature T_U. Net, it only loses P = ε·σ·A·(T⁴ − T_U⁴). Human example: skin at 306 K in a room at 293 K gives a net of about 135 W over 1.7 m²; without the back radiation it would be an unrealistic 845 W. For small temperature differences the net loss grows approximately linearly with ΔT (Newton law of cooling). This is why we feel cold near cold windows despite warm room air: the cold pane radiates less back, so our net loss towards it increases.
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How do you calculate with Stefan-Boltzmann Law?
Here is how to work through a typical Stefan-Boltzmann Law (P = σ·A·T⁴) task step by step:
- 1
Task
What power does the Sun radiate? (R = 6.96×10⁸ m, T = 5778 K)
Solution path
A = 4πR² ≈ 6.09×10¹⁸ m², T⁴ ≈ 1.115×10¹⁵ K⁴. P = 5.67×10⁻⁸ × 6.09×10¹⁸ × 1.115×10¹⁵ ≈ 3.8×10²⁶ W.
- 2
Task
How much does a person radiate net? (A = 1.7 m², skin 306 K, surroundings 293 K, ε ≈ 1)
Solution path
P = σ·A·(T⁴ − T_U⁴) = 5.67×10⁻⁸ × 1.7 × (8.77 − 7.37)×10⁹ ≈ 135 W.