The document derives the Stefan-Boltzmann law from Planck's radiation law. It shows that Planck calculated the radiation power per unit area over all wavelengths of blackbody radiation. Integrating this and applying the limit results in the expression P=σT^4, where P is power radiated, σ is the Stefan-Boltzmann constant, and T is the absolute temperature. It then derives an expression for the temperature on the surface of the sun using energy balance considerations and the Stefan-Boltzmann law.

1. 1
AN ASSIGNMENT ON DERIVATION OF BASIC
EQUATION
NATIONAL INSTITUTE OE TECHNOLOGY
JAMSEDPUR
Submitted by :-
Rajbala Purnima Priya
rajbalapurnimapriya@gmail.com

2. 2
DERIVATION OF STEFAN BOLTZMANN LAW
STATEMENT: According to Stefan Boltzmann law, the amount of radiation
emitted from an area A per unit time of a black body at absolute temperature T is
directly proportional to the temperature to the power four.
P = SAT4
Where, P is power radiated by blackbody
S is Stefan’s constant is experimentally calculated as 5.67 × 10-8 W/m2 k4
U is total amount of radiation emitted by black body
A is area
T is temperature in absolute Kelvin
PROCEDURE:
Governing Equation: Law of conservation of energy.
We all know that the total Energy is conserved in the universe. Energy neither be
created nor be destroyed only it changes from one form to other form. By applying
energy conservation German Physicist Plank derive the Planck’s radiation law
for the spectral-energy distribution of radiation emitted by a blackbody.
Planck’s radiation law states that when an oscillator (atom in state of oscillation)
changes from a state of energy E1 to a state of lower energy E2, the discrete amount
of energy E1 − E2, or quantum of radiation, is equal to the product of the radiation
frequency, symbolized by the Greek letter ν and a constant h, now called which is
called Planck’s constant, that he determined from blackbody radiation data; i.e.
E1 − E2 = hν.
Here, h= 6.62607015 × 10−34 joule second

3. 3
He conducted the experiment and explained the ultraviolet catastrophe by
proposing that the energy of electromagnetic waves is quantized rather than
continuous. This can be illustrated by the following figure.
With that assumption, Planck calculated the following formula for the radiation
power per unit area over all the wavelengths of blackbody as :
FIGURE : Relationship betweenthe temperature of an object and the
spectrum of blackbody radiation

4. 4
𝑑𝑃
𝑑λ
*
1
𝐴
=
2∗𝜋∗ℎ∗𝐶2
λ5(e
h
λ∗kT −1)
Where,
 P is Power radiated.
 A is the surface area of a blackbody.
 λ is the wavelength of emitted radiation.
 h is Planck’s constant
 c is the velocity of light
 k is Boltzmann’s constant
 T is temperature.
We can obtain by integrating Plank’s radiation with respect to frequency and
applying the limit we have :
∫ 𝑑(𝑃/𝐴)
∞
0
=∫
2∗𝜋∗ℎ∗𝑐2
λ5(e
h
λ∗kT −1)
∞
0
d λ
This can be solved analytically and result obtained is as follows :
𝑃
𝐴
=
2𝜋(𝑘𝑇)4
ℎ3 𝑐2 *
𝜋5
15
This can be further written as

5. 5
𝑷
𝑨
= (
𝟐𝒌 𝟒∗𝝅 𝟓
𝟏𝟓𝒉 𝟑 𝒄 𝟐)T4
𝑃
𝐴
=ST4
hence , proved P= SAT4
Derivation of expression of temperature of sun on the periphery of
the sun using energy balance.
Parameters:
 σ – Stefan’s Boltzmann constant (W/m2K4)
 I – Extraterrestrial solar radiation or solar constant (W/m2)
 Rs – Radius of sun (m)
 As – Internal spherical surface area of sun (m2 )
 Ts – Temperature on the periphery of sun (K)
 Eb – Emissive power of sun assuming Sun as a black body (W)

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 Re – Radius of imaginary sphere equals to the average distance
between earth and sun.
 Ae – Surface area of imaginary sphere.(m)
Assumptions:
 Steady state radiation heat transfer takes place.
 Sun is assumed to be a black body.
By using Stefan’s Boltzmann law of radiation of emitting body , we
know that the total hemispherical emissive power of black body is
directly proportional to the fourth power of temperature of emitting
body.
i,e; Eb = σ * T4 (W/m2)
Eb = σ * As * T4 (W)
Eb = σ * 4π Rs
2
* T4
(As=4π*Rs
2)
By using the value of solar constant, which may be obtained
practically by using equipment available for solar radiation
calibration
Total energy measured on the horizontal surface per second is given
by

7. 7
= Gsc * Ae
= Gsc * 4πRe
2
Now by using energy balance, we can equate the energy leaving from
the sun intercepted by the imaginary sphere and the energy available
on the horizontal surface on the outer surface of earth atmosphere.
σ * 4πRs
2 * T4 = Gsc * 4πRe
2
T4 =
𝐺𝑠𝑐 ∗ 𝑅 𝑒
2
𝜎 ∗ 𝑅 𝑠
2
T = √
𝑮 𝒔𝒄∗ 𝑹 𝒆
𝟐
𝝈 ∗ 𝑹 𝒔
𝟐
𝟒