Stefan Boltzmann Law
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.
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 ().
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 is surrounded by a body at a temperature
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).