Stefan Boltzmann Constant: Value, Formula, Black Body, Applications

Collegedunia Team logo

Collegedunia Team Content Curator

Content Curator

Stefan Boltzmann Constant is the constant σ of the Stefan-Boltzmann law. This law was initially experimented by Stefan and later theoretically proved by Boltzmann. Its SI unit value is 5.67 × 10–8 W m–2 K–4.

Keyterms: Electromagnetic energy, Radiation, Temperature, Emissivity, Thermal equilibrium, Parameters, Energy


Stefan Boltzmann Law

[Click Here for Sample Questions]

Energy can be transferred by radiation over enormous distances, without a medium (i.e., in vacuum). At absolute temperature T, the total electromagnetic energy radiated by a body is related to its size, ability to radiate (called emissivity), and temperature. 

The energy emitted per unit time (H) for a perfect radiator is given by H = AT4, where A is the area and T is the absolute temperature of the body.

Stefan Boltzmann Law

Stefan Boltzmann Law

According to Stefan Boltzmann's law:

The total amount of radiation energy produced by a black body per unit time from a specific area at absolute temperature is directly proportional to the black body's fourth power.


Black Body and Stefan Boltzmann Constant

[Click Here for Previous Year Questions]

A black body is a hypothetical physical object having enough mass all the energy directed at it, containing it fully and not reflecting any of it back, regardless of the wavelength or angle of incidence of the incident radiations. 

Since it absorbs all light that touches its surface, it is described as a "black body." A black body emits electromagnetic black-body radiation when it is at a constant temperature and consequently in thermal equilibrium. 

Black Body

Black Body

The most significant feature of the blackbody radiation curves is that they are universal. They depend on the temperature and are unaffected by the blackbody's size, shape, or composition. Planck's law states that the spectrum of emitted radiation is defined solely by the temperature of the black body and no other parameters. 

At constant temperature, a black body has two desired properties:

  1. It is an ideal emitter: a black body emits as much or more thermal radiative energy than any other body at the same temperature.
  2. It is a diffuse emitter: the energy emitted iso-tropically, regardless of direction, when measured per unit area perpendicular to the direction of the back body.

Also Read:


Planck's Law and Stefan Boltzmann Constant

[Click Here for Sample Questions]

Planck's law asserts that when the temperature of a radiation-emitting body rises, the total radiated energy of the body increases, and the emitted radiation has shorter wavelengths, as established by German physicist Max Planck in 1900.

  • B is the shadowy radiance of the body
  • v is the frequency
  • T is the absolute temperature
  • k B (= 1.380649×10−23 JK−1) is the Boltzmann constant
  • h (= 6.62607015×10−34 JHz−1) is the Planck constant

Mathematical Representation of Stefan Boltzmann Law

[Click Here for Previous Year Questions]

From the statement of Stefan Boltzmann law, j* = σT4

where

  • j* is known as the black-body radiant emittance
  • σ (= 5.67 × 10-8 W/m2 k4) is called the Stefan Boltzmann constant and is a constant of proportionality
  • T is the temperature of the black body

For a body that is not black, the radiant emittance is given by:

u = e σ AT4

Taking into consideration the surrounding temperature T 0, the net energy released per unit area is given by:

Δu = u – u0 = eσA [T4 – T04]

Where e is called emissivity, which is defined as the ability of a surface to radiate energy as thermal radiation. It can also be defined as the ratio of the thermal radiation from a surface to the radiation from an ideal black body at the same temperature, given by Stefan Boltzmann law.

Emissivity lies between 0 to 1, and for ideal black bodies, an emissivity of 1 is desired.

Also Check: Thermal Expansion


Value of Stefan Boltzmann Constant 

[Click Here for Sample Questions]

σ = 5.670367(13) × 10-8 W ⋅ m-2 K-4

From the above, the SI unit of Stefan's constant can be written as W ⋅ m-2 K-4.

Here,

  • w stands for Watt.
  • m for Metre.
  • K for Kelvin.

Value of Stefan Boltzmann Constant 

Value of Stefan Boltzmann Constant

Dimensional formula for Stefan Boltzmann law constant will be [M] 1 [T]-3 [?]-4


Elementary Definitions

[Click Here for Previous Year Questions]

  • Total Emissive Power (e)

At a certain temperature, a body's total emissive power is defined as the total amount of radiant energy radiated every second per unit area of the body's surface. It is represented by the symbol e.

Its SI unit is J m-2 s-1.

Total emissive power of a black body is represented by the symbol E.

  • Emissivity ε

Emissivity of a body is the ratio of the total emissive power of the body to the total emissive power of a black body. It is represented by the symbol ε. It means that.

ε =e/E or e=εE. For a block body, ε=1.

  • Total Absorptive Power (a)

The total absorptive power of a body is defined as the ratio of the total radiant energy absorbed by the body in a certain interval of time to the total energy falling upon it in the same interval of time. It is represented by the symbol a.

Total absorptive power of a black body is represented by the symbol A. By definition A = 1.

Elementary Definitions

Elementary Definitions


Applications of Stefan Boltzmann Constant

[Click Here for Sample Questions]

The Stefan Boltzmann constant has many uses in physics. Some of them are:

  • It is used to calculate the quantity of heat emitted by the dark body.
  • It can convert temperature (K) to intensity (Wm-2) units, which is essentially power per unit area.

So, the Stefan Boltzmann constant, Dimensional formula, Stefan Boltzmann constant value, formula, terms and units, and applications are an important part of physics.

Also Check:


Things to Remember

  • Stefan’s law states that the total energy radiated per unit surface area per unit time by a black body at all the wavelengths is proportional to the fourth power of its absolute temperature.
  • A blackbody is the perfect absorber and emitter of light. It absorbs any light that falls on it. A perfect blackbody is also a perfect radiator. 
  • The better an object is at absorbing light, the better it is at emitting it. So a perfect absorber should be the most efficient radiator possible; but at the same time, if an object is a perfect absorber it will not reflect any radiation, and so it will look black. 
  • The absolute temperature of an object is the temperature on a scale, where 0 is taken as absolute zero. This is also known as thermodynamic temperature; absolute temperature scales are Kelvin, degree units Celsius and Rankine degree unit Fahrenheit.
  • Absolute zero is the temperature at which a system is in the state of lowest possible energy i.e., minimum energy. When molecules approach this temperature, their movements drop towards zero. It is the lowest temperature that a gas thermometer can measure. You know electronic devices will not work at this temperature. Finally, the Kinetic Energy of the molecules becomes negligible or ze.

Previous Year Questions


Sample Questions

Ques. What are the Stefan constant units? (1 Mark)

Ans. The Unit of the Stefan-Boltzmann constant is 5.67 x 10-8 watt per meter squared per kelvin to the fourth (Wm2K-4).

Ques. Why is Stefan's law significant? (2 Marks)

Ans. The Stefan-Boltzmann Law describes how much energy the Sun emits in relation to its temperature (or allows scientists to figure out how hot the sun is based on how much power strikes the Earth in a square metre). The law also forecasts how much heat is radiated into space by the Earth.

Ques. What are the limitations of Stefan's law? (2 Marks)

Ans. Because of Stefan's Law, Newton's Law of Cooling is derived (or deduced). Newton's Law of Cooling's Limitations: When a body's excess warmth above its surroundings is small, this law applies (about 40°C) When the body cools, the ambient temperature is considered remaining constant.

Ques. A hot black body emits the energy at the rate of 16 J m-2 s-1 and its most intense radiation corresponds to 20,000 Å. When the temperature of this body is further increased and its most intense radiation corresponds to 10,000 Å, then find the value of energy radiated in Jm-2 s-1. (2 Marks)

Ans. Wein’s displacement law is, λm.T = b

i.e. T∝ [1/ λm]

Here, λm becomes half, the Temperature doubles.

Now from Stefan Boltzmann Law, e = sT4

e,/e2 = (T1/T2)4

⇒ e2 = (T2/T1)4 . e1 = (2)4.16

= 16.16 = 256 J m-2 s-1

Ques. What is the Stefan Boltzmann Constant Value? (3 Marks)

Ans. Stefan Boltzmann Constant Value

Types of Units Stefan Boltzmann’s Constant Value Units
CGS units σ ≈ 5.6704×105 erg. cm2.s1.K4
Thermochemistry σ ≈ 11.7×108 Cal. cm2. day1. K4
US Customary units σ ≈ 1.714×109 BTU.hr1.ft2.°R4.

Ques. What is Stefan-Boltzmann constant? Write its dimensional formula. (5 Marks)

Ans. The Stefan-Boltzmann law states that the intensity of the wavelength of the energy radiated from a body is increased with the increase in the difference between the temperature of the body and the surroundings.

The rate of heat energy loss or the power radiated by any body is proportional to the emissivity of the material of the body, the surface area of the body which can emit radiation and the fourth power of the temperature difference between the body and the surroundings. 

Mathematically, this is shown as:

\(P \propto eAT^4\)

The proportionality is made into an equation by a constant known by the Stefan-Boltzmann constant. It has a value given as:

\(\sigma = 5.67 \times 10^{-8} W m^{-2}K^{4}\)

From the Stefan-Boltzmann law, we can deduce the constant as:

\(P = \sigma e AT^4 \)

Therefore, 

\(\sigma =\frac{P}{EAT^4}\)

The ratio of the R.H.S. will always be a constant for any material. The actual value of the Stefan-Boltzmann constant is given by the equation:

\(\sigma = \frac{2\pi^5kB^4}{15h^2c^2}\)

\(\therefore \sigma = 5.67 \times 10^-8 Wm ^{-2}K^{-4}\)

Where, kB is the Boltzmann constant,

h is the Planck's constant,

c is the speed of light.

The dimensional formula for each physical quantity can be given as:

[kB] = JK-1

⇒ [kB] = [ML2T-2K1]

∴  [kb]4 = [M4L8T-8K4]

[h] = Js

⇒ [h] = [ML2T-1]

∴  [h]3 = [M3L6T-3]

[c] = ms-2

⇒ [c] = [LT-2]

∴  [c]2 = [L2T-4]

The dimensional formula for the constant is:


Do Check Out:

CBSE CLASS XII Related Questions

  • 1.
    Two statements are given, one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C), and (D) as given below.
    Assertion (A): In double slit experiment, if one slit is closed, diffraction pattern due to the other slit will appear on the screen.
    Reason (R): For interference, at least two waves are required.

      • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
      • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
      • Assertion (A) is true, but Reason (R) is false.
      • Both Assertion (A) and Reason (R) are false.

    • 2.
      In the figure, curved lines represent equipotential surfaces. A charge \( Q \) is moved along different paths A, B, C, and D. The work done on the charge will be maximum along the path:
       curved lines represent equipotential surfaces

        • A
        • B
        • C
        • D

      • 3.
        Two infinitely long conductors kept along XX' and YY' axes are carrying current \( I_1 \) and \( I_2 \) along -X axis and -Y axis respectively. Find the magnitude and direction of the net magnetic field produced at point P(X, Y).


          • 4.

            Two slits 0.1 mm apart are arranged 1.20 m from a screen. Light of wavelength 600 nm from a distant source is incident on the slits. How far apart will adjacent bright interference fringes be on the screen? 


              • 5.
                The distance of an object from the first focal point of a biconvex lens is \( X_1 \) and distance of the image from second focal point is \( X_2 \). The focal length of the lens is:

                  • \( X_1 X_2 \)
                  • \( \sqrt{X_1 + X_2} \)
                  • \( \sqrt{X_1 X_2} \)
                  • \( \frac{X_2}{X_1} \)

                • 6.
                  A current flows through a cylindrical conductor of radius \( R \). The current density at a point in the conductor is \( j = \alpha r \) (along its axis), where \( \alpha \) is a constant and \( r \) is the distance from the axis of the conductor. The current flowing through the portion of the conductor from \( r = 0 \) to \( r = \frac{R}{2} \) is proportional to:

                    • \( R \)
                    • \( R^2 \)
                    • \( R^3 \)
                    • \( R^4 \)
                  CBSE CLASS XII Previous Year Papers

                  Comments


                  No Comments To Show