Stefan-Boltzmann law

 

Explain the Stefan-Boltzmann law & its constant value

Introduction:-

Josef Stefan and Ludwig Boltzmann together proposed the law of black body radiation. 

The Stefan–Boltzmann law describes the power radiated from a black body in terms of its temperature.

What is Black Body?

A black body is a physical body that is capable of absorbing all electromagnetic radiation incidents on it, irrespective of the angle of incidence or its frequency. it can also known as a surface that is responsible for absorbing all the radiant energy that falls over it. 

Stefan Boltzmann Law Statement:-

the amount of radiation emitted per unit time from an area A of a black body at absolute temperature T is directly proportional to the fourth power of the temperature. 

j*=σT 4

where  σ is Stefan’s constant = 5.67 × 10-8 W/m2  k4

A body that is not a black body absorbs and hence emit less radiation, given by equation (1)

For such a body, u = e σ AT4 . . . . . . . (2)

where e = emissivity (which is equal to absorptive power) which lies between 0 to 1.

With the surroundings of temperature T0, net energy radiated by an area A per unit time.

Δu = u – uo = eσA [T4 – T04] . . . . . . (3)

Sample Problems on Stefan Boltzmann Law

Example1: A body of emissivity (e = 0.75), the surface area of 300 cm2 and temperature 227 ºC are kept in a room at temperature 27 ºC. Using the Stephens Boltzmann law, calculate the initial value of net power emitted by the body.

Using equation (3);

P = eσA (T4 – T04)

= (0.75) (5.67 × 10-8 W/m2 – k4) (300 × 10-4 m2) × [(500 K)4 – (300 K)4]

= 69.4 Watts.

Example 2: 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.

Solution:

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

e1/e2 = (T1/T2)4

⇒ e2 = (T2/T1)4 . e1 = (2)4 . 16

= 16.16 = 256 J m-2 s-1

Example 3: A black body has an emissivity of 0.1 and its area is 100m², at 200K. At what rate does it radiate energy?

Solution: Given,

Emissivity (ϵ) = 0.1

Area (A) = 100 m²

Temperature (T) = 200 K

Stefan Boltzmann Constant (σ) = 5.67 x 10⁻⁸ W/ m²K⁴

We know that:

Radiated Energy (P) = ϵ σ T⁴A = 0.1 x 5.67 x 10⁻⁸ W/ m²K⁴ x (200 K)⁴ x 100 m² = 907.2W

∴ Radiated Energy (P) = 907.2W.

Example: what is the temperature of the Sun?

You can use this Stefan Boltzmann law calculator to figure out what is the temperature of any celestial body - for example, of the Sun. To do it, you need to follow these steps:

Knowing the radius of Sun, R = 6.963 * 10⁸ m, calculate its surface. If you assume that Sun is a perfect sphere, then its area is equal to
A = 4πR² = 4π * (6.963 * 10⁸)² = 6.093 * 10¹⁸ m²

Then, find out what is the emissivity of the Sun. You can assume that it is a perfect black body with emissivity equal to 1.

You can measure how much solar power (on average) does the Earth receive. This value is called the solar constant and is equal to S = 1367 W/m². The power emitted by the Sun is dispersed uniformly around it, what means that if you multiply this solar constant by the area of a sphere of radius equal to the distance between Earth and Sun D, you will obtain

P = S * a = S * 4πD² = 1367 * 4π * (1.496 * 10¹¹)² = 3.845 * 10²⁶ W

Then, all you have to do is input these values into the Stefan Boltzmann equation to find the temperature of the Sun:
P = σ * ε * A * T⁴

T = [P / (σ * ε * A)] ^ (0.25)

T = [3.845 * 10²⁶ / (5.670367 * 10⁻⁸ * 1 * 6.093 * 10¹⁸)] ^ (0.25)

T = 5776 K

The temperature of the Sun is equal to 5776 K.