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Stefan Boltzmann Law

Stefan Boltzmann Law

Edited By Vishal kumar | Updated on Sep 25, 2024 03:30 PM IST

Ever wondered why a campfire always seems hotter when you are closer to it, or why stars shine so brilliantly at night? All these phenomena have an explanation rooted in the basic principles of thermal physics through the Stefan-Boltzmann Law. This is named for physicists Josef Stefan and Ludwig Boltzmann and describes the relation of temperature to the total energy radiated by a black body. Knowing this law allows us to decipher the radiative properties of an object, from household heaters through to the furthest stars.

Stefan Boltzmann Law
Stefan Boltzmann Law

In this article, we will cover the concept of Stefan Boltzmann's Law with mathematical formulation. This concept is the part of Properties of Solids and Liquids which is a crucial chapter in Class 11 physics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), National Eligibility Entrance Test (NEET), and other entrance exams such as SRMJEE, BITSAT, WBJEE, VITEEE and more. In the last ten years (2013-2023) two questions were asked from this concept in the NEET exam. But no direct question was asked in JEE from this concept.

Stefan Boltzmann Law

According to Stefan Boltzmann's law, the radiant energy emitted by a perfectly black body per unit area per sec is directly proportional to the fourth power of its absolute temperature, or the emissive power of the black body is directly proportional to the fourth power of its absolute temperature (\theta).

i.e Eαθ4⇒E=σθ4

where

σ= Stefan's constant and its value is
σ=5.67×10−8 W/m2K4

For Ordinary body

1. Emissive power:

It is given by e=ϵE

So according to Stefan Boltzmann law

e=ϵE=ϵσθ4

where ϵ represents the emissivity of the material.

2. Radiant energy:

If Q is the total energy radiated by the ordinary body then

e=QA×t=ϵσθ4⇒Q=Aϵσθ4t

3. Radiant power (P): I

t is defined as the energy radiated per unit area.

i.e i.e P=Qt=Aϵσθ4

4. If an ordinary body at a temperature \theta is surrounded by a body at a temperature \theta _0

Then according to Stefan Boltzmann law,

e=ϵσ(θ4−θ04)

We can understand better through video.

Solved Example Based On Stefan Boltzmann Law

Example 1: The rate of radiation of the black body is E J/sec. The rate of radiation (in E) of this black body at 273oC will be :

1) 8

2) 16

3) 4

4) 1

Solution:

Eαθ4⇒E=σθ4E2E1=(T2T1)4

Where T is the Temperature in kelvin
⇒(273+273273+0)4=16

Hence, the answer is option (2).

Example 2: An object is at a temperature of 400oC. At what temperature would it radiate energy twice as fast?

1) 200oC

2) 200K

3) 800oC

4) 800K

Solution:

From Stefan's Law,

E∝θ4⇒E=σθ4 - wherein σ= Stefan's constant σ=5.67×10−8 W/m2 K4

(E2E1)=(T2T1)4

Where T is the Temperature in kelvin

(21)=(T400+273)4 br T=214⋅673=800K

Hence, the answer is option (4).

Example 3: The area of a hose of a heat furnace is 10−4 m2. It radiates 1.58×105cal of heat per hour. If the emissivity of the furnace is 0.80 , then its temperature (in K ) is :

1)1500

2) 2000

3) 2500

4) 3000

Solution:

For Ordinary Body -

Emissive power is given by e=ϵE=Aϵσθ4

- wherein

ϵ= represent emissivity of the material

E=σϵAT41.58×105×4.260×60=5.6×10−8×10−4×0.8×T4T≈2500 K

Hence, the answer is option (3).

Example 4: Assuming the sun to be a spherical body of radius R at the temperature of TK, evaluate the total radiant power, incident on earth, at a distance r from the sun (when r0 is the radius of the earth and σ is Stefan's constant):

1) R2σT4r2 2) 4πr02R2σT4r2 3) πr02R2σT4r2 4) r02R2σT44πr2

Solution:

Total power radiated by the sun =σ(4πR2)⋅T4
The intensity of this radiation at a distance of
r=σ⋅(4πR2)⋅T44πr2=σT4(R2r2)

Amount of energy received on the earth =σT4(Rr)2⋅πr02

Hence, the answer is option (3).

Example 5: The dimensions of σb4 ( σ= Stefan's constant and b= Wein's constant) are :

1) [M0L0T0]
2) [ML4T−3]
3) [ML−2T2]
4) [ML6T−3]

Solution:

Dimension of Work, Potential Energy, Kinetic Energy, Torque - ML2T−2

As λmT=b or b4=λm4T4 and energy area × time =σT4

or σ= energy area × timeT ∴σb4=( energy area × time )λm4

or [σb4]=[ML2T−2][L2][T][L4]=[ML4T−3]

Hence, the answer is option (2).

Summary

The Stefan-Boltzmann Law specifies that the total energy radiated per unit surface area of a blackbody in one unit of time is directly proportional to the fourth power of its absolute temperature. The law had explained to us the thermal radiation emitted by objects and is very relevant to astrophysics, climatology, and engineering fields. It tells why hotter objects glow more of their energy and how this energy scales with temperature.

FAQ's
What is the Stefan-Boltzmann Law?
Stefan-Boltzmann law, often known as Stefan's law, asserts that the total radiant heat power emitted by a surface is proportional to the fourth power of its absolute temperature.
How does the Stefan-Boltzmann Law account for the energy sent out by stars?
Clearly, however, the law he put forward shows that the star's energy emission increases with temperature, which allows an easy explanation for the huge radiation from hot stars.
What is radiant emittance?
The radiant emittance is the total amount of energy radiated per unit surface area of a black body per unit of time.
How is Stefan-Boltzmann's Law used in climate science?
The Stefan-Boltzmann law helps to appreciate the Earth's radiation balance and temperature changes in connection with the energy the planet radiates, which is very important in modelling the climate.
Why do hotter objects emit more energy according to the Stefan Boltzmann Law?
Since the energy radiated is proportional to the fourth power in temperature, even small rises in temperature result in large increases in radiated energy.


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