Physics · Thermodynamics

Stefan–Boltzmann law: emission and net thermal radiation

Q: How do you calculate the radiated power of a hot body?

For a black surface, use $P_{\mathrm{emit}}=\sigma AT^4$ with actual emitting area in square metres and absolute temperature in kelvin. An area of $10 mm^{2}$ equals $1.0\times{}10^{-5} m^{2}$; at 2500 K it emits about 22.15 W. A real gray surface additionally requires an appropriate emissivity ε. This is total thermal emission, rather than only visible light or one detector channel. Net heat exchange also requires absorbed environmental radiation. First establish whether the problem asks for emission, radiation exchange or electrical heating power. These coincide only under additional balance assumptions; conduction and convection can also carry energy.

Q: How do you calculate the luminosity and temperature of stars?

For a spherical star, $L=4\pi R^2\sigma T_{\mathrm{eff}}^4$ defines effective temperature from total luminosity and radius. With $R=6.96\cdot10^8\,\mathrm m$ and $T_{\mathrm{eff}}=5778\,\mathrm K$, luminosity is about $3.847\times{}10^{26} W$. Use the entire spherical surface, not $\pi{}R^{2}$. An observed radiant flux density yields luminosity only with distance and suitable geometry and absorption corrections. A spectrum can provide a color temperature, but this need not equal Teff exactly. Wien’s law also requires the correct spectral-density convention: a peak per wavelength interval cannot be obtained by merely converting the coordinate of a peak per frequency interval.

Q: Why does a body cool more slowly than P = σAT⁴ suggests?

$P=\sigma AT^4$ counts emission, while the body simultaneously absorbs radiation. For a fully viewed large isothermal enclosure, the diffuse gray model gives $P_{\mathrm{net,out}}=\varepsilon\sigma A(T^4-T_{\mathrm u}^4)$. At $T=T_{\mathrm u}$, net exchange vanishes even though both individual powers remain positive. An ideal black $1.7 m^{2}$ surface at 306 K facing surroundings at 293 K emits about 845.17 W but loses about 134.73 W net. This is a simplified surface model, not a complete measurement of human heat balance. Air temperature does not always replace Tu: cold windows and differing view factors change incoming radiation. Cooling rate also requires heat capacity and other energy transfers.

Total emitted radiant power from a black surface grows as $T^{4}$ . Emitted power and net heat exchange are different quantities.

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Formula

Calculate

P = σ \cdot A \cdot T^{4}

P in W; σ ≈ 5.670374419×10⁻⁸ W/(m²·K⁴); A in m²; T in K. ε is dimensionless.

Variables & units

Variables & units
Symbol	Meaning	Unit
$P$	Total power emitted by the black surface	W
$σ$	Stefan–Boltzmann constant derived from exact SI defining constants h, c and kB	W/(m²·K⁴)
$A$	Actual emitting surface area, not merely projected area	m²
$T$	Absolute surface temperature	K

LaTeX

P = \sigma \cdot A \cdot T^4

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Diagram: P over T as a very steeply rising curve, dashed guide lines show that P rises from P₁ at T₁ to 2·P₁ at 1.19·T₁. — The radiated power P grows with T⁴: just 19 % more temperature doubles P.

Stefan-Boltzmann law calculator

Find the radiated power or surface temperature of a uniformly warm grey surface, including the net loss to cooler surroundings.

Calculation model

Find

Total emitting area A

Total radiating surface area, e.g. 0.02 m² for a small hotplate.

Emissivity ε

Emissivity between 0 and 1, e.g. 0.9 for a dark surface; 1 means a black body.

Surface temperature T

Surface temperature, ideally in kelvin, e.g. 500 K, which is about 227 °C.

Filled with example values. Type your own and the result updates straight away.

Result: Power P

P = 63.79 W

With A = 0.02 m², ε = 0.9 and T = 500 K, P = 63.79 W.

Working · rearrange for P, substitute, evaluate

1. Rearrange for P
$P_{\mathrm{emit}}=\varepsilon\sigma AT^4$
2. Substitute values (in SI units)
$0.9\cdot 5.67037\times10^{-8}\cdot 0.02\cdot500^4=63.7917$
3. Result
P = 63.79 W

σ ≈ 5.670374419·10⁻⁸ W/(m²·K⁴); the decimal representation is rounded. Use kelvin.

Quick tips

Estimate: 1 m² of black surface at 300 K radiates about 460 W. Doubling the kelvin temperature gives 16 times as much, because P grows with T⁴.
For the net loss raise both kelvin temperatures to the fourth power first, then subtract: T⁴ − Tᵤ⁴, not (T − Tᵤ)⁴. Working in °C gives a wrong result.

Assumptions and limits

Isothermal diffuse gray surface with specified constant emissivity. Entire emitting area, all wavelengths and outgoing hemisphere; kelvin. Emission only, not net loss or detector power.

Derivation & background

Stefan formulated the empirical law in 1879; Boltzmann derived it thermodynamically in 1884. The fourth-power law follows by integrating the Planck spectrum over wavelength and using angular integration $M_\lambda=\pi B_\lambda$ . Gray surfaces require an appropriate emissivity ε. Net exchange additionally needs incident radiation and geometry. The simple difference $\varepsilon\sigma A(T^4-T_{\mathrm u}^4)$ applies to a small diffuse gray surface fully viewing a large isothermal radiative enclosure. Air temperature alone does not determine counter-radiation; conduction and convection are separate energy-balance terms.

Exam blueprint

Validity range

$P=\sigma AT^4$ gives total power emitted over all wavelengths and the outgoing hemisphere by a black surface at absolute temperature T. A gray surface uses εσAT⁴ with appropriate hemispherical emissivity. The net form $\varepsilon\sigma A(T^4-T_{\mathrm u}^4)$ requires a small diffuse gray surface fully viewing a large isothermal radiative enclosure through a transparent gap. Tu is radiation temperature, not automatically air temperature. Convection and conduction are additional terms.

Derivation steps

Integrate the Planck spectrum over the outgoing hemisphere and all wavelengths, then isolate temperature dependence with a dimensionless variable.

1Define $B_\lambda$ as blackbody spectral radiance per projected area, solid angle and wavelength interval. $T$ is absolute; $\lambda$ is vacuum wavelength.
2For direction-independent blackbody radiance, $\int_{\mathrm{hemisphere}}\cos\theta\,\mathrm d\Omega=\pi$ gives spectral exitance $M_\lambda=\pi B_\lambda$ .
3Integrate $M_\lambda$ over all wavelengths. $M_{\mathrm b}=\int_0^\infty M_\lambda\,\mathrm d\lambda$ is power per area, not yet total power.
4Set $x=\frac{hc}{\lambda k_{\mathrm B}T}$ . Substitution including the differential gives $M_{\mathrm b}=\frac{2\pi k_{\mathrm B}^4T^4}{h^3c^2}\int_0^\infty\frac{x^3}{e^x-1}\,\mathrm dx$ .
5The integral equals $\frac{\pi^4}{15}$ . Define $\sigma=\frac{2\pi^5k_{\mathrm B}^4}{15h^3c^2}\approx5.670374419\cdot10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}$ .
6Multiply by emitting area: $P_{\mathrm{emit}}=\sigma AT^4$ . A gray model adds $\varepsilon$ . A fully viewed large black isothermal enclosure gives $P_{\mathrm{net,out}}=\varepsilon\sigma A(T^4-T_{\mathrm u}^4)$ .

Rearrangements

Temperature

T = \sqrt[4]{\frac{P}{\sigma \cdot A}}

Blackbody emission with known area and total emitted power. In a stellar model this defines effective temperature; detector power additionally needs geometry.

Area

A = \frac{P}{\sigma \cdot T^4}

For blackbody emission. A spherical stellar model additionally has $A=4\pi R^2$ ; luminosity is total emitted power. Include $\varepsilon$ for gray emission.

Task variant

Model the Sun as a spherical blackbody with radius $6.96\cdot{}10^{8} m$ and effective temperature 5778 K. Find total emitted power, not irradiance measured at Earth’s surface.

Approach: the whole sphere emits, $A=4\pi R^2$ .
Thus $P_{\mathrm{emit}}=4\pi R^2\sigma T^4\approx3.85\cdot10^{26}\,\mathrm W$ .

Check: doubling radius at unchanged temperature would quadruple power. This is effective total blackbody power, not a local solar spectrum.

As a simplified radiation model of a person, take an uncovered isothermal black surface of $1.7 m^{2}$ at 306 K, fully viewing a large black enclosure at 293 K. Find net radiative loss. Clothing, airflow, perspiration and other heat-transfer paths are excluded.

Approach: emission minus absorbed enclosure radiation.
$P_{\mathrm{net,out}}=\sigma A(T^4-T_{\mathrm u}^4)\approx134.728\,\mathrm W$ .
Positive power leaves the modelled surface.

Check: equal surface and enclosure temperatures would give zero net power despite radiation in both directions. This is not a measured total heat loss for a real clothed person.

Common mistakes

Substituting temperatures in degrees Celsius.

Use absolute temperatures: 25 °C equals 298.15 K. Do not raise Celsius values to a power or use them in temperature ratios.

Forgetting the back radiation of the surroundings.

Subtract absorbed irradiation. Only for this gray-surface model fully viewing a large black isothermal enclosure is $P_{\mathrm{net,out}}=\varepsilon\sigma A(T^4-T_{\mathrm u}^4)$ .

Treating real surfaces as perfect black bodies.

Use emissivity appropriate to the surface, temperature and spectral weighting. Oxidation and roughness affect it; one universal value for all metal surfaces is inappropriate.

Forgetting the fourth root when solving for T.

For a blackbody with known emitted power, $T=\sqrt[4]{\frac{P_{\mathrm{emit}}}{\sigma A}}$ . Gray emission adds $\varepsilon$ to the denominator; net loss additionally requires enclosure conditions.

Exam context

Tasks on stellar luminosity and radius, filaments and the Earth radiation balance; the factor-16 effect of doubling the temperature is popular.

Formula cluster

Thermal radiation

The third heat transfer mechanism besides conduction and convection.

Q = m·c·ΔT

Heat Capacity: Q = m·c·ΔT

η = 1 − T_k/T_w

Carnot Efficiency

Worked example

A surface modeled as black, with $A=10\,\mathrm{mm^2}=1.0\cdot10^{-5}\,\mathrm{m^2}$ and $T=2500\,\mathrm K$ , emits about 22.15 W. A real filament requires its emissivity. At fixed A and ε, multiplying absolute temperature by $2^{1/4}\approx1.1892$ doubles emitted power; net power with incoming radiation follows a different relation.

Applications

Model thermal radiation balances, incandescent bodies, infrared measurements and stellar luminosity. Climate and surface models additionally need counter-radiation, spectral properties, geometry and other transport mechanisms.

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
Model the Sun as a spherical blackbody with radius $6.96\cdot{}10^{8} m$ and effective temperature 5778 K. Find total emitted power, not irradiance measured at Earth’s surface.
Solution path
1. Approach: the whole sphere emits, $A=4\pi R^2$ .
2. Thus $P_{\mathrm{emit}}=4\pi R^2\sigma T^4\approx3.85\cdot10^{26}\,\mathrm W$ .
Check: doubling radius at unchanged temperature would quadruple power. This is effective total blackbody power, not a local solar spectrum.
2
Task
As a simplified radiation model of a person, take an uncovered isothermal black surface of $1.7 m^{2}$ at 306 K, fully viewing a large black enclosure at 293 K. Find net radiative loss. Clothing, airflow, perspiration and other heat-transfer paths are excluded.
Solution path
1. Approach: emission minus absorbed enclosure radiation.
2. $P_{\mathrm{net,out}}=\sigma A(T^4-T_{\mathrm u}^4)\approx134.728\,\mathrm W$ .
3. Positive power leaves the modelled surface.
Check: equal surface and enclosure temperatures would give zero net power despite radiation in both directions. This is not a measured total heat loss for a real clothed person.

Understand and apply the equation

$P=\sigma AT^4$ counts all thermally emitted power from an ideal blackbody. It is not automatically heat lost to surroundings or power reaching one detector. Integrate the spectrum with the correct angular factor, calculate a gray surface inside a radiative enclosure, then infer temperature from net heating.

What radiant power does the Stefan–Boltzmann law give?

A is actual emitting surface area, not the projected area in a photograph. Pemit integrates all wavelengths and the outgoing hemisphere. An ideal blackbody fully absorbs incident thermal radiation and emits the black spectrum at its temperature. It emits even at 0°C because its absolute temperature is 273.15 K.[2]

P_{\mathrm{emit}}=A M_b=A\sigma T^4

Mb is total blackbody radiant exitance per surface area in W/m².[1]

The body simultaneously receives radiation. Emitted power can therefore be large while net exchange is small. At equal body and surrounding radiation temperatures, absorption and emission cancel; each separate power remains nonzero. Energy over a time interval requires integrating power over time.

Obtain the fourth-power law from the Planck spectrum

Stefan formulated the empirical relation in 1879; Boltzmann provided a thermodynamic derivation in 1884. That historical argument did not require the later Planck spectrum. The integration below shows how the quantum spectrum yields the same total law and additionally determines its constant. Planck’s radiation law describes the spectral distribution before integration.[6]

Spectral radiance Bλ counts power per projected area, solid angle and wavelength interval. Blackbody radiance is direction independent. Exitance nevertheless requires the projection cosθ: its hemisphere integral is π. Integrating Bλ over wavelength alone gives radiance per solid angle rather than exitance.[3]

\begin{aligned}M_{\mathrm b}&=\pi\int_0^\infty B_\lambda(T)\,\mathrm d\lambda\\&=\frac{2\pi k_{\mathrm B}^4T^4}{h^3c^2}\int_0^\infty\frac{x^3}{e^x-1}\,\mathrm dx\\x&=\frac{hc}{\lambda k_{\mathrm B}T}\end{aligned}

The substitution reverses limits; its negative differential restores a positive integral.[3]

Evaluate the integral

The dimensionless integral is $\frac{\pi^4}{15}$ . Expand $\frac1{e^x-1}=\sum_{n=1}^{\infty}e^{-nx}$ and integrate each term: $\int_0^\infty x^3e^{-nx}\,\mathrm dx=\frac6{n^4}$ , hence $6\sum_{n=1}^{\infty}\frac1{n^4}=\frac{\pi^4}{15}$ . The fourth temperature power remains outside a constant factor. This derives scaling, but not a real coating’s emissivity.

\sigma=\frac{2\pi^5 k_B^4}{15h^3c^2}\approx5.670374419\times10^{-8}\,\mathrm{W\,m^{-2}K^{-4}}

The SI defining values of h, c, kB are exact; the decimal representation of σ here is rounded.[4]

What emissivity assumes

P_{\mathrm{emit}}=\varepsilon\sigma AT^4,\qquad0\le\varepsilon\le1

The gray model uses an appropriate hemispherical emissivity specified as constant.

Real surfaces may emit differently with wavelength, direction and temperature. Constant gray emissivity idealises these properties. Visible color alone does not determine thermal-infrared emission. A thermal camera also detects reflected surrounding radiation and propagation losses. Thus ε cannot be assigned just from a visible color name.[3]

A 10 mm² area at 2500 K gives about 22.15 W for $\varepsilon=1$ . This is a black model surface, not a claimed real filament emissivity. At fixed area and unchanged ε, emitted power doubles for $\frac{T_{\mathrm{new}}}{T_{\mathrm{old}}}=2^{1/4}\approx1.189207$ . These are absolute temperatures.

How do you calculate net radiative power to the surroundings?

P_{\mathrm{net,out}}=\varepsilon\sigma A(T^4-T_u^4)

Positive means net energy leaves the surface; negative means net absorption.[1][5]

This difference assumes a small opaque diffuse gray surface fully exposed to a large isothermal radiative enclosure. The intervening medium is transparent for the radiation considered. Incident black surrounding radiation is then σTu⁴, with matching absorption ε. Tu is the radiation temperature of visible surroundings, not automatically air temperature.[1][5]

Finite surfaces, walls at different temperatures, shields or specular surfaces need view factors and possibly multiple reflections. Simply using $\varepsilon\sigma A(T^4-T_{\mathrm u}^4)$ would omit geometry. Convection and conduction also need separate terms. High temperature alone therefore does not prove radiation dominates every other heat-transfer mechanism.[5]

Emission stays positive while net flow changes sign

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Area, emissivity and enclosure are those of the first worked example. The separation between curves is the constant absorbed counter-radiation. Negative net power means absorption.[1][2]

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Example: emitted versus net rejected power

A model surface in a radiative enclosure

Given: $A=0.0250\,\mathrm{m^2}$ , $T=420\,\mathrm K$ , $\varepsilon=0.800$ . Large isothermal radiative enclosure at 295 K; transparent gap, diffuse gray model.
Find: Calculate emission, absorbed counter-radiation and net power separately.

Approach: Separate emission from net transfer to the large isothermal surroundings. $\begin{aligned}&P_{\mathrm{emit}}=\varepsilon\sigma AT^4\\{}&P_{\mathrm{net}}=\varepsilon\sigma A(T^4-T_{\mathrm{amb}}^4)\text{.}\end{aligned}$ Calculate emission from surface temperature.
$P_{\mathrm{emit}}=0.800\sigma\cdot0.0250\cdot420^4=35.28896\,\mathrm W$
Subtract absorbed surrounding radiation.
$P_{\mathrm{abs}}=0.800\sigma\cdot0.0250\cdot295^4=8.58875\,\mathrm W$
The difference is net rejection.
$P_{\mathrm{net,out}}=\varepsilon\sigma A(420^4-295^4)\approx26.70022\,\mathrm W$

Result: Net rejection is about 26.70 W rather than 35.29 W emission. These numbers exclude additional convection or conduction.

Check: Direct check: $\frac{P_{\mathrm{net}}}{P_{\mathrm{emit}}}=1-\left(\frac{295}{420}\right)^4\approx0.75662$ . Counter-radiation is positively absorbed; equal temperatures would give zero difference.

Example: infer a steady radiative temperature

An idealised radiative heater

Given: $A=0.0400\,\mathrm{m^2}$ , $\varepsilon=0.600$ , $T_{\mathrm u}=300\,\mathrm K$ . Net electrical heating 200 W; at steady state radiation is the only heat rejection.
Find: Find steady surface temperature.

Approach: In the steady radiation-only model, input balances net output. $P_{\mathrm{in}}=\varepsilon\sigma A(T^4-T_{\mathrm{amb}}^4)$ . Set storage to zero; input equals net rejection.
$200=0.600\sigma\cdot0.0400(T^4-300^4)$
Include counter-radiation before taking the fourth root.
$T=\left(\frac{200}{0.600\sigma\cdot0.0400}+300^4\right)^{\frac{1}{4}}=627.51890\,\mathrm K$

Result: T≈627.52 K, or 354.37°C, in the specified steady model.

Check: Pabs≈11.02321 W and Pemit≈211.02321 W return 200 W net. Neglecting counter-radiation gives a smaller temperature; additional heat-transfer paths also change equilibrium.

Example: two temperatures on one surface

Why mean temperature underestimates emission

Given: An ideal diffuse gray surface has total area 0.0200 m². Equal subareas are at 300 and 500 K, both with $\varepsilon=0.800$ . Each fully views a large black 300 K enclosure. Other heat paths and mutual irradiation are excluded.
Find: Find emission and net power, then compare with a calculation using mean temperature 400 K.

Approach: evaluate the local fourth power on each subarea before adding powers. Emissivity is the same supplied value, not inferred from temperature.
$\begin{aligned}P_{\mathrm{emit}}&=\varepsilon\sigma\frac A2(T_1^4+T_2^4)\\P_{\mathrm{abs}}&=\varepsilon\sigma A T_u^4\\P_{\mathrm{net,out}}&=P_{\mathrm{emit}}-P_{\mathrm{abs}}\end{aligned}$
Evaluate the equal-area contributions and counter-radiation separately.
$\begin{aligned}P_{\mathrm{emit}}&\approx32.0263\,\mathrm W\\P_{\mathrm{abs}}&\approx7.34881\,\mathrm W\\P_{\mathrm{net,out}}&\approx24.6775\,\mathrm W\end{aligned}$
The mean-temperature calculation uses a different fourth power and therefore gives lower emission.
$\begin{aligned}\overline T&=400\,\mathrm K\\P_{\mathrm{mean}}&=\varepsilon\sigma A\overline T^4\approx23.2259\,\mathrm W\\\frac{P_{\mathrm{emit}}}{P_{\mathrm{mean}}}&=\frac{300^4+500^4}{2\cdot400^4}=1.37890625\end{aligned}$

Result: Model emission is about 37.9% greater than the mean-temperature estimate. Net output is 24.68 W; the 300 K subarea exchanges no net energy with the enclosure.

Check: Independent check: only the 500 K half contributes net outward power. εσ(A/2)(500⁴−300⁴) again gives 24.6775 W. This is spatial accounting, not a temperature distribution derived from a heat-conduction equation.

Explore further: near equilibrium and stellar models

\begin{aligned}T&=T_{\mathrm u}+\Delta T,\quad |\Delta T|\ll T_{\mathrm u}\\P_{\mathrm{net,out}}&\approx4\varepsilon\sigma AT_{\mathrm u}^3\Delta T\end{aligned}

Linearisation concerns net power near equilibrium, not total emission.

Thermal response additionally depends on heat capacity: a simple lumped model has $C(T)\frac{\mathrm dT}{\mathrm dt}=P_{\mathrm{in}}-P_{\mathrm{net,out}}$ . Even constant ε gives no constant cooling rate beyond small temperature offsets because of T⁴. Cooling measurements also need conduction, convection and a meaningful uniform body temperature.

A spherical black stellar model has luminosity $L=4\pi R^2\sigma T_{\mathrm{eff}}^4$ . $R=6.96\cdot10^8\,\mathrm m$ and $T=5778\,\mathrm K$ give about 3.847×10²⁶ W. Observed flux implies luminosity only with distance and propagation assumptions. Teff is effective temperature defined by total flux; a color-dependent spectral temperature need not coincide.[2]

What the law does not determine

σAT⁴ does not give spectral distribution, detector power or luminous output in lumens. Those require Bλ, projected area, accepted solid angle, bandwidth and detector response. A narrow infrared channel collects only part of bolometric emission. Nonequilibrium laser or luminescent emission is not described by this thermal blackbody model.[3]

With nonuniform temperature or emissivity, integrate local contributions over the surface. Taking a temperature average first and then raising it to the fourth power is generally wrong. Extra calculation digits are model checks; real material values, areas and temperatures have their own uncertainties.

Determine total emitting area and use kelvin. For heat loss, subtract appropriate counter-radiation; a detector additionally needs its actual geometry and spectral response. Add local powers when temperature varies across the surface so that the fourth-power dependence is preserved.

Five solved radiation exercises

Transfer · Foundation

A blackbody keeps its area while temperature rises from 300 to 600 K. How does emission change?

Show hint

Compare absolute temperatures.

Learning goal: Understand fourth-power emission scaling.

Show solution and calculation

Approach: area and the Stefan–Boltzmann constant cancel in the power ratio.
$\frac{P_{\mathrm{new}}}{P_{\mathrm{old}}}=\left(\frac{600}{300}\right)^4=16$ .
This does not fix net loss if surroundings also change.

Check: doubling power twice would give only four; here temperature is raised to the fourth power.

\frac{P_2}{P_1}=16

Calculate · Application

A 0.0100 m² black surface at 280 K fully sees a large 300 K enclosure. Find net outward power.

Show hint

The sign specifies direction.

Learning goal: Assign sign to net absorption in warmer surroundings.

Show solution and calculation

Approach: $P_{\mathrm{net,out}}=\sigma A(T^4-T_{\mathrm u}^4)$ .
$P_{\mathrm{net,out}}=\sigma\cdot0.0100\,\mathrm{m^2}\bigl((280\,\mathrm K)^4-(300\,\mathrm K)^4\bigr)\approx-1.10767\,\mathrm W$ : the surface absorbs net heat.

Check: the cooler surface must gain net radiative heat; the negative sign confirms this.

P_{\mathrm{net,out}}\approx-1.10767\,\mathrm W

Calculate · Application

At unchanged A and ε, double emitted power from an initial 400 K. Find new temperature.

Show hint

Take the fourth root of the power ratio.

Learning goal: Find temperature for doubled emission.

Show solution and calculation

Approach: $T_2=T_1\sqrt[4]{\frac{P_2}{P_1}}$ .
$T_{\mathrm{new}}=400\cdot2^{1/4}\,\mathrm K\approx475.683\,\mathrm K$ .

Check: $\left(\frac{475.683}{400}\right)^4\approx2$ . Increase is about 18.92%, not 100%.

T_2=400\,2^{\frac{1}{4}}\,\mathrm K\approx475.683\,\mathrm K

Interpret · Application

Two equal black surfaces are at 400 K. One fully views a large 300 K enclosure, the other a 400 K enclosure. Compare emitted and net outward power per area; exclude other heat-transfer modes.

Show hint

Emission depends on the emitter; net power subtracts incoming radiation.

Learning goal: Distinguish equal emission from equal net loss.

Show solution and calculation

Approach: Both emit equal power per area.
In the 400 K enclosure incoming radiation exactly balances emission, giving zero net power.
In the 300 K enclosure net outward flux is positive, approximately 992.315 W/m².
Zero net heat flux does not mean zero radiation in either direction.

\begin{aligned}\frac{P_{\mathrm{emit}}}{A}&=\sigma(400\,\mathrm K)^4\\{}\frac{P_{\mathrm{net},1}}A&=\sigma(400^4-300^4)\,\mathrm{K^4}\approx992.315\,\mathrm{W\,m^{-2}}\\{}P_{\mathrm{net},2}&=0\end{aligned}

[1]

Error analysis · Application

At fixed area and emissivity temperature rises from 20 to 40 °C. Someone claims emission rises sixteenfold because temperature doubled. Find the correct ratio using kelvin; σ cancels.

Show hint

Absolute temperatures are 293.15 and 313.15 K.

Learning goal: Reject Celsius ratios in thermal radiation.

Show solution and calculation

Approach: Take the ratio of absolute temperatures and raise it to the fourth power.
The factor is approximately 1.30212, an increase around 30.2%.
Celsius zero is an offset and cannot be used in this ratio law.
This concerns emission; net loss additionally requires enclosure temperature.

\frac{P_2}{P_1}=\left(\frac{313.15\,\mathrm K}{293.15\,\mathrm K}\right)^4\approx1.30212

[1]

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

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This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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October 2, 2026Refined question headings and added scientifically relevant paragraph links; retained existing derivations, examples, diagrams, exercises and sources.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Refined the learning sequence, symbolic exercise approaches and model limits; added a third original example and quantitative figure, with suitable further explanations optional.

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Quanta. Stefan-Boltzmann Law. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/stefan-boltzmann-strahlung

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Frequently asked questions about Stefan-Boltzmann Law

How do you calculate the radiated power of a hot body?+

For a black surface, use $P_{\mathrm{emit}}=\sigma AT^4$ with actual emitting area in square metres and absolute temperature in kelvin. An area of $10 mm^{2}$ equals $1.0\times{}10^{-5} m^{2}$ ; at 2500 K it emits about 22.15 W. A real gray surface additionally requires an appropriate emissivity ε. This is total thermal emission, rather than only visible light or one detector channel. Net heat exchange also requires absorbed environmental radiation. First establish whether the problem asks for emission, radiation exchange or electrical heating power. These coincide only under additional balance assumptions; conduction and convection can also carry energy.

Why is the fourth power of temperature so decisive?+

At unchanged area and emissivity, emitted power grows as the fourth power of absolute temperature. Doubling T therefore multiplies emission by sixteen, while an increase of about 18.92% doubles it. Comparing 773 K with 293 K gives an emission ratio of about 48.44. This does not directly state net heat loss or whether radiation dominates convection. Incoming radiation, surface properties and heat-transfer conditions are still needed. Physically, the $T^{4}$ dependence follows from integrating the entire Planck spectrum. Increasing temperature raises spectral emission and changes its wavelength distribution, so a narrow detector band need not follow the same fourth-power ratio.

Why must the temperature be inserted in kelvin?+

The equation uses absolute temperature. Celsius has a shifted zero: $\frac T{\mathrm K}=\frac t{{}^\circ\mathrm C}+273.15$ . A black surface at 0°C therefore has $T=273.15\,\mathrm K$ and emits about 315.66 W per square metre. Substituting zero into $T^{4}$ would incorrectly remove this physically present energy flow. Temperature ratios such as $\frac{T_2}{T_1}$ must also use kelvin values. Although temperature differences have the same numerical value in kelvin and Celsius degrees, this does not permit raising Celsius readings to the fourth power. Net power subtracts fourth powers of two absolute temperatures; $T^4-T_{\mathrm u}^4$ is not generally equal to $(T-T_{\mathrm u})^4$ .

How do you calculate the luminosity and temperature of stars?+

For a spherical star, $L=4\pi R^2\sigma T_{\mathrm{eff}}^4$ defines effective temperature from total luminosity and radius. With $R=6.96\cdot10^8\,\mathrm m$ and $T_{\mathrm{eff}}=5778\,\mathrm K$ , luminosity is about $3.847\times{}10^{26} W$ . Use the entire spherical surface, not $\pi{}R^{2}$ . An observed radiant flux density yields luminosity only with distance and suitable geometry and absorption corrections. A spectrum can provide a color temperature, but this need not equal Teff exactly. Wien’s law also requires the correct spectral-density convention: a peak per wavelength interval cannot be obtained by merely converting the coordinate of a peak per frequency interval.

Why does a body cool more slowly than P = σAT⁴ suggests?+

$P=\sigma AT^4$ counts emission, while the body simultaneously absorbs radiation. For a fully viewed large isothermal enclosure, the diffuse gray model gives $P_{\mathrm{net,out}}=\varepsilon\sigma A(T^4-T_{\mathrm u}^4)$ . At $T=T_{\mathrm u}$ , net exchange vanishes even though both individual powers remain positive. An ideal black $1.7 m^{2}$ surface at 306 K facing surroundings at 293 K emits about 845.17 W but loses about 134.73 W net. This is a simplified surface model, not a complete measurement of human heat balance. Air temperature does not always replace Tu: cold windows and differing view factors change incoming radiation. Cooling rate also requires heat capacity and other energy transfers.

Quanta exam set

Curated exam set for "Stefan-Boltzmann Law":

Question (front)

Which formula describes Stefan-Boltzmann Law?

Answer in your set

Question (front)

How do you rearrange Stefan-Boltzmann Law for Temperature?

Answer in your set

Question (front)

Which common mistake happens with Stefan-Boltzmann Law?

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Stefan–Boltzmann law: emission and net thermal radiation

Total emitted radiant power from a black surface grows as

T^{4}

. Emitted power and net heat exchange are different quantities.

AdvancedUpper secondaryBachelor foundationsSubject specialization

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Symbol

Meaning

Unit

P

Total power emitted by the black surface

σ

Stefan–Boltzmann constant derived from exact SI defining constants h, c and kB

W/(m²·K⁴)

A

Actual emitting surface area, not merely projected area

m²

T

Absolute surface temperature

Derivation & background

Exam blueprint

Validity range

Derivation steps

Integrate the Planck spectrum over the outgoing hemisphere and all wavelengths, then isolate temperature dependence with a dimensionless variable.

1Define $B_\lambda$ as blackbody spectral radiance per projected area, solid angle and wavelength interval. $T$ is absolute; $\lambda$ is vacuum wavelength.
2For direction-independent blackbody radiance, $\int_{\mathrm{hemisphere}}\cos\theta\,\mathrm d\Omega=\pi$ gives spectral exitance $M_\lambda=\pi B_\lambda$ .
3Integrate $M_\lambda$ over all wavelengths. $M_{\mathrm b}=\int_0^\infty M_\lambda\,\mathrm d\lambda$ is power per area, not yet total power.
4Set $x=\frac{hc}{\lambda k_{\mathrm B}T}$ . Substitution including the differential gives $M_{\mathrm b}=\frac{2\pi k_{\mathrm B}^4T^4}{h^3c^2}\int_0^\infty\frac{x^3}{e^x-1}\,\mathrm dx$ .
5The integral equals $\frac{\pi^4}{15}$ . Define $\sigma=\frac{2\pi^5k_{\mathrm B}^4}{15h^3c^2}\approx5.670374419\cdot10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}$ .
6Multiply by emitting area: $P_{\mathrm{emit}}=\sigma AT^4$ . A gray model adds $\varepsilon$ . A fully viewed large black isothermal enclosure gives $P_{\mathrm{net,out}}=\varepsilon\sigma A(T^4-T_{\mathrm u}^4)$ .

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:

Task

Model the Sun as a spherical blackbody with radius $6.96\cdot{}10^{8} m$ and effective temperature 5778 K. Find total emitted power, not irradiance measured at Earth’s surface.

Solution path

Approach: the whole sphere emits, $A=4\pi R^2$ .
Thus $P_{\mathrm{emit}}=4\pi R^2\sigma T^4\approx3.85\cdot10^{26}\,\mathrm W$ .

Check: doubling radius at unchanged temperature would quadruple power. This is effective total blackbody power, not a local solar spectrum.

Task

Solution path

Approach: emission minus absorbed enclosure radiation.
$P_{\mathrm{net,out}}=\sigma A(T^4-T_{\mathrm u}^4)\approx134.728\,\mathrm W$ .
Positive power leaves the modelled surface.

Check: equal surface and enclosure temperatures would give zero net power despite radiation in both directions. This is not a measured total heat loss for a real clothed person.

Understand and apply the equation

P=\sigma AT^4

counts all thermally emitted power from an ideal blackbody. It is not automatically heat lost to surroundings or power reaching one detector. Integrate the spectrum with the correct angular factor, calculate a gray surface inside a radiative enclosure, then infer temperature from net heating.

What radiant power does the Stefan–Boltzmann law give?

P_{\mathrm{emit}}=A M_b=A\sigma T^4

Mb is total blackbody radiant exitance per surface area in W/m².[1]

Obtain the fourth-power law from the Planck spectrum

\begin{aligned}M_{\mathrm b}&=\pi\int_0^\infty B_\lambda(T)\,\mathrm d\lambda\\&=\frac{2\pi k_{\mathrm B}^4T^4}{h^3c^2}\int_0^\infty\frac{x^3}{e^x-1}\,\mathrm dx\\x&=\frac{hc}{\lambda k_{\mathrm B}T}\end{aligned}

The substitution reverses limits; its negative differential restores a positive integral.[3]

Evaluate the integral

\sigma=\frac{2\pi^5 k_B^4}{15h^3c^2}\approx5.670374419\times10^{-8}\,\mathrm{W\,m^{-2}K^{-4}}

The SI defining values of h, c, kB are exact; the decimal representation of σ here is rounded.[4]

What emissivity assumes

P_{\mathrm{emit}}=\varepsilon\sigma AT^4,\qquad0\le\varepsilon\le1

The gray model uses an appropriate hemispherical emissivity specified as constant.

How do you calculate net radiative power to the surroundings?

P_{\mathrm{net,out}}=\varepsilon\sigma A(T^4-T_u^4)

Positive means net energy leaves the surface; negative means net absorption.[1][5]

Emission stays positive while net flow changes sign

QUANTA STUDYquanta-study.de

Area, emissivity and enclosure are those of the first worked example. The separation between curves is the constant absorbed counter-radiation. Negative net power means absorption.[1][2]

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Example: emitted versus net rejected power

A model surface in a radiative enclosure

Given: $A=0.0250\,\mathrm{m^2}$ , $T=420\,\mathrm K$ , $\varepsilon=0.800$ . Large isothermal radiative enclosure at 295 K; transparent gap, diffuse gray model.
Find: Calculate emission, absorbed counter-radiation and net power separately.

Approach: Separate emission from net transfer to the large isothermal surroundings. $\begin{aligned}&P_{\mathrm{emit}}=\varepsilon\sigma AT^4\\{}&P_{\mathrm{net}}=\varepsilon\sigma A(T^4-T_{\mathrm{amb}}^4)\text{.}\end{aligned}$ Calculate emission from surface temperature.
$P_{\mathrm{emit}}=0.800\sigma\cdot0.0250\cdot420^4=35.28896\,\mathrm W$
Subtract absorbed surrounding radiation.
$P_{\mathrm{abs}}=0.800\sigma\cdot0.0250\cdot295^4=8.58875\,\mathrm W$
The difference is net rejection.
$P_{\mathrm{net,out}}=\varepsilon\sigma A(420^4-295^4)\approx26.70022\,\mathrm W$

Result: Net rejection is about 26.70 W rather than 35.29 W emission. These numbers exclude additional convection or conduction.

Example: infer a steady radiative temperature

An idealised radiative heater

Given: $A=0.0400\,\mathrm{m^2}$ , $\varepsilon=0.600$ , $T_{\mathrm u}=300\,\mathrm K$ . Net electrical heating 200 W; at steady state radiation is the only heat rejection.
Find: Find steady surface temperature.

Approach: In the steady radiation-only model, input balances net output. $P_{\mathrm{in}}=\varepsilon\sigma A(T^4-T_{\mathrm{amb}}^4)$ . Set storage to zero; input equals net rejection.
$200=0.600\sigma\cdot0.0400(T^4-300^4)$
Include counter-radiation before taking the fourth root.
$T=\left(\frac{200}{0.600\sigma\cdot0.0400}+300^4\right)^{\frac{1}{4}}=627.51890\,\mathrm K$

Result: T≈627.52 K, or 354.37°C, in the specified steady model.

Check: Pabs≈11.02321 W and Pemit≈211.02321 W return 200 W net. Neglecting counter-radiation gives a smaller temperature; additional heat-transfer paths also change equilibrium.

Example: two temperatures on one surface

Why mean temperature underestimates emission

Given: An ideal diffuse gray surface has total area 0.0200 m². Equal subareas are at 300 and 500 K, both with $\varepsilon=0.800$ . Each fully views a large black 300 K enclosure. Other heat paths and mutual irradiation are excluded.
Find: Find emission and net power, then compare with a calculation using mean temperature 400 K.

Approach: evaluate the local fourth power on each subarea before adding powers. Emissivity is the same supplied value, not inferred from temperature.
$\begin{aligned}P_{\mathrm{emit}}&=\varepsilon\sigma\frac A2(T_1^4+T_2^4)\\P_{\mathrm{abs}}&=\varepsilon\sigma A T_u^4\\P_{\mathrm{net,out}}&=P_{\mathrm{emit}}-P_{\mathrm{abs}}\end{aligned}$
Evaluate the equal-area contributions and counter-radiation separately.
$\begin{aligned}P_{\mathrm{emit}}&\approx32.0263\,\mathrm W\\P_{\mathrm{abs}}&\approx7.34881\,\mathrm W\\P_{\mathrm{net,out}}&\approx24.6775\,\mathrm W\end{aligned}$
The mean-temperature calculation uses a different fourth power and therefore gives lower emission.
$\begin{aligned}\overline T&=400\,\mathrm K\\P_{\mathrm{mean}}&=\varepsilon\sigma A\overline T^4\approx23.2259\,\mathrm W\\\frac{P_{\mathrm{emit}}}{P_{\mathrm{mean}}}&=\frac{300^4+500^4}{2\cdot400^4}=1.37890625\end{aligned}$

Result: Model emission is about 37.9% greater than the mean-temperature estimate. Net output is 24.68 W; the 300 K subarea exchanges no net energy with the enclosure.

Explore further: near equilibrium and stellar models

\begin{aligned}T&=T_{\mathrm u}+\Delta T,\quad |\Delta T|\ll T_{\mathrm u}\\P_{\mathrm{net,out}}&\approx4\varepsilon\sigma AT_{\mathrm u}^3\Delta T\end{aligned}

Linearisation concerns net power near equilibrium, not total emission.

What the law does not determine

Five solved radiation exercises

Transfer · Foundation

A blackbody keeps its area while temperature rises from 300 to 600 K. How does emission change?

Show hint

Compare absolute temperatures.

Learning goal: Understand fourth-power emission scaling.

Show solution and calculation

Approach: area and the Stefan–Boltzmann constant cancel in the power ratio.
$\frac{P_{\mathrm{new}}}{P_{\mathrm{old}}}=\left(\frac{600}{300}\right)^4=16$ .
This does not fix net loss if surroundings also change.

Check: doubling power twice would give only four; here temperature is raised to the fourth power.

\frac{P_2}{P_1}=16

Calculate · Application

A 0.0100 m² black surface at 280 K fully sees a large 300 K enclosure. Find net outward power.

Show hint

The sign specifies direction.

Learning goal: Assign sign to net absorption in warmer surroundings.

Show solution and calculation

Approach: $P_{\mathrm{net,out}}=\sigma A(T^4-T_{\mathrm u}^4)$ .
$P_{\mathrm{net,out}}=\sigma\cdot0.0100\,\mathrm{m^2}\bigl((280\,\mathrm K)^4-(300\,\mathrm K)^4\bigr)\approx-1.10767\,\mathrm W$ : the surface absorbs net heat.

Check: the cooler surface must gain net radiative heat; the negative sign confirms this.

P_{\mathrm{net,out}}\approx-1.10767\,\mathrm W

Calculate · Application

At unchanged A and ε, double emitted power from an initial 400 K. Find new temperature.

Show hint

Take the fourth root of the power ratio.

Learning goal: Find temperature for doubled emission.

Show solution and calculation

Approach: $T_2=T_1\sqrt[4]{\frac{P_2}{P_1}}$ .
$T_{\mathrm{new}}=400\cdot2^{1/4}\,\mathrm K\approx475.683\,\mathrm K$ .

Check: $\left(\frac{475.683}{400}\right)^4\approx2$ . Increase is about 18.92%, not 100%.

T_2=400\,2^{\frac{1}{4}}\,\mathrm K\approx475.683\,\mathrm K

Interpret · Application

Two equal black surfaces are at 400 K. One fully views a large 300 K enclosure, the other a 400 K enclosure. Compare emitted and net outward power per area; exclude other heat-transfer modes.

Show hint

Emission depends on the emitter; net power subtracts incoming radiation.

Learning goal: Distinguish equal emission from equal net loss.

Show solution and calculation

Approach: Both emit equal power per area.
In the 400 K enclosure incoming radiation exactly balances emission, giving zero net power.
In the 300 K enclosure net outward flux is positive, approximately 992.315 W/m².
Zero net heat flux does not mean zero radiation in either direction.

\begin{aligned}\frac{P_{\mathrm{emit}}}{A}&=\sigma(400\,\mathrm K)^4\\{}\frac{P_{\mathrm{net},1}}A&=\sigma(400^4-300^4)\,\mathrm{K^4}\approx992.315\,\mathrm{W\,m^{-2}}\\{}P_{\mathrm{net},2}&=0\end{aligned}

[1]

Error analysis · Application

At fixed area and emissivity temperature rises from 20 to 40 °C. Someone claims emission rises sixteenfold because temperature doubled. Find the correct ratio using kelvin; σ cancels.

Show hint

Absolute temperatures are 293.15 and 313.15 K.

Learning goal: Reject Celsius ratios in thermal radiation.

Show solution and calculation

Approach: Take the ratio of absolute temperatures and raise it to the fourth power.
The factor is approximately 1.30212, an increase around 30.2%.
Celsius zero is an offset and cannot be used in this ratio law.
This concerns emission; net loss additionally requires enclosure temperature.

\frac{P_2}{P_1}=\left(\frac{313.15\,\mathrm K}{293.15\,\mathrm K}\right)^4\approx1.30212

[1]

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 2, 2026Refined question headings and added scientifically relevant paragraph links; retained existing derivations, examples, diagrams, exercises and sources.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Refined the learning sequence, symbolic exercise approaches and model limits; added a third original example and quantitative figure, with suitable further explanations optional.

Cite this reference page

Quanta. Stefan-Boltzmann Law. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/stefan-boltzmann-strahlung

When citing a specific scientific statement, also cite the specialist source linked to that statement.

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Frequently asked questions about Stefan-Boltzmann Law

How do you calculate the radiated power of a hot body?+

Why is the fourth power of temperature so decisive?+

Why must the temperature be inserted in kelvin?+

How do you calculate the luminosity and temperature of stars?+

Why does a body cool more slowly than P = σAT⁴ suggests?+

Stefan–Boltzmann law: emission and net thermal radiation

Variables & units

Stefan-Boltzmann law calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Thermal radiation

Worked example

Applications

How do you calculate with Stefan-Boltzmann Law?

What radiant power does the Stefan–Boltzmann law give?

Obtain the fourth-power law from the Planck spectrum

What emissivity assumes

How do you calculate net radiative power to the surroundings?

Emission stays positive while net flow changes sign

Example: emitted versus net rejected power

A model surface in a radiative enclosure

Example: infer a steady radiative temperature

An idealised radiative heater

Example: two temperatures on one surface

Why mean temperature underestimates emission

Explore further: near equilibrium and stellar models

What the law does not determine

Five solved radiation exercises

A blackbody keeps its area while temperature rises from 300 to 600 K. How does emission change?

A 0.0100 m2 black surface at 280 K fully sees a large 300 K enclosure. Find net outward power.

At unchanged A and ε, double emitted power from an initial 400 K. Find new temperature.

Two equal black surfaces are at 400 K. One fully views a large 300 K enclosure, the other a 400 K enclosure. Compare emitted and net outward power per area; exclude other heat-transfer modes.

At fixed area and emissivity temperature rises from 20 to 40 °C. Someone claims emission rises sixteenfold because temperature doubled. Find the correct ratio using kelvin; σ cancels.

Sources for this explanation

Frequently asked questions about Stefan-Boltzmann Law

Subject references and further reading

Related formulas

More formulas: Heat and thermodynamics

Retain Stefan-Boltzmann Law for exams

Stefan–Boltzmann law: emission and net thermal radiation

Variables & units

Stefan-Boltzmann law calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Thermal radiation

Worked example

Applications

How do you calculate with Stefan-Boltzmann Law?

What radiant power does the Stefan–Boltzmann law give?

Obtain the fourth-power law from the Planck spectrum

What emissivity assumes

How do you calculate net radiative power to the surroundings?

Emission stays positive while net flow changes sign

Example: emitted versus net rejected power

A model surface in a radiative enclosure

Example: infer a steady radiative temperature

An idealised radiative heater

Example: two temperatures on one surface

Why mean temperature underestimates emission

Explore further: near equilibrium and stellar models

What the law does not determine

Five solved radiation exercises

A blackbody keeps its area while temperature rises from 300 to 600 K. How does emission change?

A 0.0100 m2 black surface at 280 K fully sees a large 300 K enclosure. Find net outward power.

At unchanged A and ε, double emitted power from an initial 400 K. Find new temperature.

Two equal black surfaces are at 400 K. One fully views a large 300 K enclosure, the other a 400 K enclosure. Compare emitted and net outward power per area; exclude other heat-transfer modes.

At fixed area and emissivity temperature rises from 20 to 40 °C. Someone claims emission rises sixteenfold because temperature doubled. Find the correct ratio using kelvin; σ cancels.

Sources for this explanation

Frequently asked questions about Stefan-Boltzmann Law

Subject references and further reading

Related formulas

A 0.0100 m² black surface at 280 K fully sees a large 300 K enclosure. Find net outward power.

A 0.0100 m² black surface at 280 K fully sees a large 300 K enclosure. Find net outward power.