Education Journalist | Study Abroad Lead
Radiant energy is one of the most important forms of energy, it is the electromagnetic energy that moves from one place to another in the form of electromagnetic waves.
- Human vision requires radiant energy to see anything.
- Only a small portion of it is visible to the human eye and most of the radiant energy such as radio waves and X- rays do not fall on the spectrum of visible light.
- We use radiant energy to communicate over long distances, to heat our homes and even to grow crops.
- Radiant energy is a type of kinetic energy, which means that it is related to motion.
- In the process of photosynthesis, plants collect radiant energy from the sun and convert it into chemical energy.
| Table of Content |
Radiant Energy
[Click Here for Sample Questions]
Radiant energy is the energy that is transferred from one place to another place by electromagnetic waves, such as X-rays, visible light, gamma rays and radio waves, which may be described in terms of continuous electromagnetic waves or discrete packets of energy called photons.
- Radiation is the process by which energy is transmitted in form of waves or particles through space or material medium.
- Radiant energy is also known as Electromagnetic Radiation or Electromagnetic radiant energy.
- Everybody, even at ordinary temperatures, emits energy in the form of radiant energy.
- Not all radiant energy is visible. Only photons with a very small range of energy can be seen by people, commonly known as visible light.
Read More:
Radiant Energy Formula
[Click Here for Sample Questions]
From Stephen’s law, the rate at which an object radiates energy is proportional to the fourth power of the absolute temperature of the object and it is given by
\(E = {Q \over t} = \sigma A \epsilon(T^4 - T^4_0)\)
Therefore, radiant energy per unit of time or radiant power is given by
\(Q = \sigma A \epsilon(T^4 - T^4_0)\)
- Where A is the surface area of the object
- σ is Stefan- Botzmann constant, σ = 5.67 x 10-8 Wm-2K-4
- ϵ is the emissivity of the body
- T is the absolute temperature of the body
- T0 is the absolute temperature of the surrounding
If surrounding temperature is T0 = 0 K, then
\(Q = \sigma A \epsilon T^4\)
If an object is not a source but it transmits radiant energy from the source, then energy transmitted or energy loss per unit of time from the object is given by
\(Q = \sigma A \tau T^4\)
Where, \(\tau\) is the transmissivity of the object
Emissivity
[Click Here for Sample Questions]
Emissivity (ϵ) is defined as the ratio of the emissive power of the surface to the emissive power of a perfectly black body.
- It is unitless and depends on the finishing of the surface.
- Its value lies between 0 and 1.
- For a black body, emissivity, ϵ = 1.
- For a fully reflected surface, ϵ = 0.
- Dark surfaces have emissivity close to 1.
- Shiny surfaces have emissivity close to 0.
- The emissivity of human skin is about 0.7.
Solved Examples[Click Here for Sample Questions] Ques. How much radiant energy is emitted by a perfect blackbody at 400K per unit area per second? Ans. We have, radiant energy, \(Q = \sigma A \epsilon T^4\) Therefore, radiant energy per unit area per second, E is given by \(E={Q \over A \times t} = \sigma \epsilon T^4\) For blackbody, emissivity, ϵ = 1 Given, absolute temperature, T = 400K Also, Stefan- Botzmann constant, σ = 5.67 x 10-8 Wm-2K-4 E = 5.67 x 10-8 x 1 x 4004 E = 1451.5 J Ques. A furnace inside temperature of 2250 K has a glass circular viewing of 6 cm diameter. If the transmissivity of glass is 0.08, then calculate heat loss from the glass window due to radiation. Ans. Given, Absolute temperature of the glass, T = 2250 K Transmissivity of glass, \(\tau\) = 0.08 Diameter of the glass window, d = 6 cm Therefore, the Area of the glass window, A= \(\pi\)r2 = \(\pi\)(d/2)2 = \(\pi\)(6/2)2 = 9 cm2 A = 28.28 x 10-4 m2 Heat energy loss Q from the glass window due to radiation can be calculated by using \(Q = \sigma A \tau T^4\) Q = 328.76 J/s Ques. A black body radiates heat energy at the rate of 2 x 105 Js-1m-2 at a temperature of 127oC. What is the temperature of the black body, at which the rate of heat radiation is 32 ×105 Js-1m-2? Ans. We know that, Radiant energy per unit area per second, E is given by \(E={Q \over A \times t} = \sigma \epsilon T^4\) Since emissivity ϵ and Stefan- Boltzmann constant σ is the same for both cases. Therefore, E ∝ T4 \({E_1 \over E_2}=({T_1 \over T_2})^4\) Given, at temperature, T1 = 127o C = 127 + 273 = 400 K, E1 = 2 x 105 Js-1m-2 We have to find at what temperature T2 , E2 = 32 x 105 Js-1m-2 T2 = 800 K = 800-273 = 527oC |
Things to Remember
- The energy that is transferred from one place to another place by electromagnetic waves is called Radiant energy.
- Radiant energy is a type of kinetic energy.
- The rate at which energy radiates from a body is directly proportional to the fourth power of absolute temperature.
- Radiant energy examples are X- rays, gamma rays and visible light.
- The formula of radiant energy is given by \(E = {Q \over t} = \sigma A \epsilon(T^4 - T^4_0)\)
- The ratio of the emissive power of the surface to the emissive power of a perfectly black body is called emissivity.
- For a black body, emissivity, ϵ = 1.
Sample Questions
Ques. The energy radiated by a black body is directly proportional to absolute temperature as – (2 marks)
1. T2
2. T3
3. T
4. T4
Ans. The correct option is (1)
Radiant energy per unit area per second, E is given by
\(E={Q \over A \times t} = \sigma \epsilon T^4\)
Where Q is radiant energy and σ is Stefan- Boltzmann constant
For blackbody, emissivity, ϵ = 1
Q ∝ T4
Ques. How much radiant energy is emitted by a perfect blackbody at 500 K per unit area per second? (3 marks)
Ans. We have, radiant energy, \(Q = \sigma A \epsilon T^4\)
Therefore, radiant energy per unit area per second, E is given by
\(E={Q \over A \times t} = \sigma \epsilon T^4\)
For blackbody, emissivity, ϵ = 1
Given, absolute temperature, T = 500K
Also, Stefan- Botzmann constant, σ = 5.67 x 10-8 Wm-2K-4
E = 5.67 x 10-8 x 625 x 108 = 3543.7 J
Ques. A black body radiates heat energy at the rate of 2 x 104 Js-1m-2 at a temperature of 227oC. What is the temperature of the black body, at which the rate of heat radiation is 162 ×104 Js-1m-2? (5 marks)
Ans. We know that Radiant energy per unit area per second, E is given by
\(E={Q \over A \times t} = \sigma \epsilon T^4\)
Since emissivity ϵ and Stefan- Boltzmann constant σ is the same for both cases.
Therefore, E ∝ T4
\({E_1 \over E_2}=({T_1 \over T_2})^4\)
Given, When temperature, T1 = 227o C = 227 + 273 = 500 K, then E1 = 2 x 104 Js-1m-2
We have to find at what temperature T2 , E1 = 162 x 104 Js-1m-2
∴ T2 = 1500 K =1500-273 = 1227oC
Ques. A furnace inside temperature of 3000 K has a glass circular viewing of 8 cm diameter. If the transmissivity of glass is 0.06, How much heat is lost from the glass window due to radiation? (5 marks)
Ans. Given, Absolute temperature of the glass, T = 3000 K
Transmissivity of glass, \(\tau\) = 0.06
Diameter of the glass window, d = 8 cm
Therefore, Area of glass window, A= \(\pi\)r2 = \(\pi\)(d/2)2 = \(\pi\)(8/2)2 = 16\(\pi\) cm2
A = 50.28 x 10-4 m2
Heat energy loss Q from the glass window due to radiation can be calculated by using
\(Q = \sigma A \tau T^4\)
Q = 1385.5 J/s
Ques. A metal ball of surface area 200 cm2 and temperature 527oC is surrounded by a vessel at 27oC. If the emissivity of the metal is 0.4, then calculate the rate of loss of heat from the ball. (5 marks)
Ans. Rate of energy radiated from an object is given by
\(E = {Q \over t} = \sigma A \epsilon(T^4 - T^4_0)\)
Given, A = 200 cm2 = 200 x 10-4 m2
ϵ = 0.4
T = 527 oC = 527+273 = 800 K
T0 = 27 oC = 27+273=300 K
Therefore, the rate of heat loss from the ball is,
E = 5.67 x 10-8 x 200 x 10-4 x 0.4 x (8004 – 3004)
E=182 J/s
Ques. The rate of radiation of a black body at 0 oC is E J/s. The rate of radiation of this black body at 273 oC will be (2 marks)
- 16E
- 8E
- 4E
- E
Ans. Correct option is (1)
The rate of radiation of a black is directly proportional to the fourth power of absolute temperature.
E ∝ T4
\({E_1 \over E_2}=({T_1 \over T_2})^4\)
Given, E1 = E, T1 = 0 oC = 273 K, T2 = 273 oC = 273+273 = 546 K
E2 = 16E
Ques. An object is at a temperature of 400 oC. At what temperature would it radiate energy twice as fast? (3 marks)
Ans. The rate at which object radiates energy is directly proportional to the fourth power of absolute temperature.
E ∝ T4
\({E_1 \over E_2}=({T_1 \over T_2})^4\)
If, E1 = E, then T1 = 400 oC = 673 K,
And if, E2 = 2E, then T2 is given by
1/1.189 = 673/T2
T2 = 1.189 x 673 =800 K
Ques. How much radiant energy emitted by a perfect blackbody at 600 K per unit area per second? (3 marks)
Ans. We have, radiant energy,\(Q = \sigma A \epsilon T^4\)
Therefore, radiant energy per unit area per second, E is given by
\(E={Q \over A \times t} = \sigma \epsilon T^4\)
For blackbody, emissivity, ϵ = 1
Given, absolute temperature, T = 600K
Also, Stefan- Botzmann constant, σ = 5.67 x 10-8 Wm-2K-4
E = 5.67 x 10-8 x 1296 x 108 = 7348 J
Ques. A black body radiates heat energy at the rate of 1 x 104 Js-1m-2 at a temperature of 227o C. What is the temperature of black body, at which the rate of heat radiation is 256 ×104 Js-1m-2? (5 marks)
Ans. We know that,radiant energy per unit area per second, E is given by
\(E={Q \over A \times t} = \sigma \epsilon T^4\)
Since, emissivity ϵ and Stefan- Botzmann constant σ is same for both cases.
Therefore, E ∝ T4
\({E_1 \over E_2}=({T_1 \over T_2})^4\)
Given, When temperature, T1 = 227o C = 227 + 273 = 500 K, then E1 = 1 x 104 Js-1m-2
We have to find at what temperature T2 , E1 = 256 x 104 Js-1m-2
∴ T2 = 2000 K = 2000 - 273 = 1727oC
Ques. How much radiant energy emitted by a perfect blackbody at 150 K per unit area per second? (3 marks)
Ans. We have, radiant energy, \(Q = \sigma A \epsilon T^4\)
Therefore, radiant energy per unit area per second, E is given by
\(E={Q \over A \times t} = \sigma \epsilon T^4\)
For blackbody, emissivity, ϵ = 1
Given, absolute temperature, T = 150K
Also, Stefan- Botzmann constant, σ = 5.67 x 10-8 Wm-2K-4
E= 5.67 x 10-8 x 5.06 x 108=28.7 J
Also Check:
Comments