# What is Stefan-boltzmann: Definition and 31 Discussions

The Stefan–Boltzmann law describes the power radiated from a black body in terms of its temperature. Specifically, the Stefan–Boltzmann law states that the total energy radiated per unit surface area of a black body across all wavelengths per unit time

j

{\displaystyle j^{\star }}
(also known as the black-body radiant emittance) is directly proportional to the fourth power of the black body's thermodynamic temperature T:

j

=
σ

T

4

.

{\displaystyle j^{\star }=\sigma T^{4}.}
The constant of proportionality σ, called the Stefan–Boltzmann constant, is derived from other known physical constants. Since 2019, the value of the constant is

σ
=

2

π

5

k

4

15

c

2

h

3

=
5.670374

×

10

8

W

m

2

K

4

,

{\displaystyle \sigma ={\frac {2\pi ^{5}k^{4}}{15c^{2}h^{3}}}=5.670374\ldots \times 10^{-8}\,\mathrm {W\,m^{-2}\,K^{-4}} ,}
where k is the Boltzmann constant, h is Planck's constant, and c is the speed of light in a vacuum. The radiance from a specified angle of view (watts per square metre per steradian) is given by

L
=

j

π

=

σ
π

T

4

.

{\displaystyle L={\frac {j^{\star }}{\pi }}={\frac {\sigma }{\pi }}T^{4}.}
A body that does not absorb all incident radiation (sometimes known as a grey body) emits less total energy than a black body and is characterized by an emissivity,

ε
<
1

{\displaystyle \varepsilon <1}
:

j

=
ε
σ

T

4

.

{\displaystyle j^{\star }=\varepsilon \sigma T^{4}.}

j

{\displaystyle j^{\star }}
has dimensions of energy flux (energy per unit time per unit area), and the SI units of measure are joules per second per square metre, or equivalently, watts per square metre. The SI unit for absolute temperature T is the kelvin.

ε

{\displaystyle \varepsilon }
is the emissivity of the grey body; if it is a perfect blackbody,

ε
=
1

{\displaystyle \varepsilon =1}
. In the still more general (and realistic) case, the emissivity depends on the wavelength,

ε
=
ε
(
λ
)

{\displaystyle \varepsilon =\varepsilon (\lambda )}
.
To find the total power radiated from an object, multiply by its surface area,

A

{\displaystyle A}
:

P
=
A

j

=
A
ε
σ

T

4

.

{\displaystyle P=Aj^{\star }=A\varepsilon \sigma T^{4}.}
Wavelength- and subwavelength-scale particles, metamaterials, and other nanostructures are not subject to ray-optical limits and may be designed to exceed the Stefan–Boltzmann law.

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1. ### The Second Law of Thermodynamics and the Stefan-Boltzmann Law

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2. ### B Understanding the Stefan-Boltzmann Law (when the surroundings are hotter)

1.If so what would the law mean if ##T_{surroundings}>T##? 2. Stefan-Boltzmann Law is formulated as ##H = A\sigma T^4## where ##H## is the energy emitted per unit time, ##A## is the area of the object, ##T## is the absolute temperature of the object and (3.) I am unclear about whether...
3. ### A Stefan-Boltzmann Equation question (qualitative)

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6. ### I Stefan-Boltzmann Law: IvdvcosΘdw Explained

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9. ### Stefan-Boltzmann law, luminosity, brightness and magnitude?

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11. ### Questions About Stefan-Boltzmann Law

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14. ### Derive Stefan-Boltzmann law from Wien’s law.

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16. ### Use Planck’s radiation law to derive the Stefan-Boltzmann

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17. ### Heat transfer stefan-Boltzmann

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18. ### Stephan-Boltzmann Law Help for Science Fair Project - Needed Now!

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19. ### The Stefan-Boltzmann Law Demonstrator Device

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20. ### Stefan-boltzmann solved for temperature?

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21. ### Stefan-Boltzmann law question (astronomy)

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22. ### Derivation of Stefan-Boltzmann law from Thermodynamics

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23. ### Derive Stefan-Boltzmann Law from Planck Distribution for blackbody radiation

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24. ### Integral in Stefan-Boltzmann law

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25. ### Radiation - solving for Q - Stefan-Boltzmann law

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26. ### Why Can't Stefan-Boltzmann Law Be Explained in Classical Mechanics?

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29. ### Linearizing Stefan-Boltzmann equation

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31. ### Understand Stefan-Boltzmann Law: Where Does the T^4 Come From?

I've been reading about the Stefan-Boltzmann law, but there is one thing that I don't understand. Why is it T^4? I can't think of anywhere that the 4 is coming from and can't find anything about this with Google searches.