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Surface area (A) and time duration have a direct relationship with the amount of heat radiated (Q) (t). Therefore, it is practical to think about the amount of heat radiated per unit area per unit time (or power emitted per unit area). This is referred to as the body’s Emissive Power or Radiant Power, R, at a specific temperature, T. Dimensions of Emissive power are [L0M1T-3] and the SI unit is Jm-2s-1 or W/m2. The composition or polishing of the emitting surface is not a physical quantity. We compare objects made of various materials with the same geometry at the same temperature in order to talk about the material aspect. The maximum emissive power of a perfect blackbody occurs at a specific temperature. Because of this, comparing the emissive power of a given surface to that of the ideal blackbody at a given temperature is convenient.

## Emissive Power Formula

The following is the emissive power formula:

R = Q / At

Where,

• Q = Amount of heat radiated,
• A = Surface Area,
• t = Time duration.

### Coefficient of Emission or Emissivity

The difference between the emissive power R of a surface and the emissive power RB of a perfectly black surface, both at the same temperature, is known as the coefficient of emission or emissivity (e) of a given surface.

### Coefficient of Emission or Emissivity Formula

The Coefficient of Emission or Emissivity formula is as follows:

e = R / RB

Where,

• e = Emissivity or coefficient of emission,
• RB = Emissive power of a perfect black surface at a same temperature.

While e=0 for a perfect reflector, e=1 for a perfect blackbody.

The best blackbodies are not common objects. As a result, they radiate less than a blackbody at the same temperature. Along with temperature, the rate for these objects also depends on surface characteristics like colour and composition. The term “emissivity e” addresses each of these phenomena individually. For a regular body, 0<e<1 depends on the surface’s composition; for instance, copper has an emissivity of 0.3. Rough surfaces have a higher emissivity, whereas smooth and polished surfaces have a lower emissivity. Emissivity changes equally with temperature and radiation wavelength.

The amount of heat radiated by a body depends on,

• The absolute temperature of the body (T).
• The nature of the body – the material, nature of the surface – polished or not, etc.
• The surface area of the body (A).
• Time duration for which the body emits radiation (t).

## Kirchhoff’s Law of Heat Radiation

The thermal radiation emission and absorption by a body in thermal equilibrium are covered by Kirchhoff’s law of thermal radiation. It claims that for all wavelengths, the emissive power of a perfect blackbody at a given temperature is equal to the ratio of a body’s emissive power to coefficient of absorption for that body.

Kirchhoff’s law can also be stated as follows: for a body emitting and absorbing thermal radiation in thermal equilibrium, the emissivity is equal to its absorptivity. This is because we can define the emissive power of an ordinary body in comparison to a perfect blackbody through its emissivity.

Symbolically, a=e or more specifically a(λ) = e(λ).

The amount of heat radiated from a given region in a given amount of time is known as emissive power.

Quantity of radiant heat absorbed by body A = Quantity of heat emitted by body A

or

∴ aQ = R  …(Equation 1)

For the perfect blackbody B,

∴ Q = RB  …(Equation 2)

Dividing Equation 1 and Equation 2, we get

∴ a = R / RB

or

∴ RB = R / a

But R / RB = e  …(Emissive power)

∴ a = e

Hence, Kirchhoff’s law is theoretically proved.

## Solved Examples on Emissive Power

Question 1: A sphere with an area of 0.04 m2 has an emissive power of 0.7 Kcal/sm2. What is the temperature of the sphere’s surface after 30 seconds?

Given: A = 0.04 m2, R = 0.7 Kcal/sm2, t = 30 s

We have,

R = Q / At

∴ Q = R × A × t

∴ Q = 0.7 × 0.04 × 30

∴ Q = 0.84 Kcal

Question 2: A body with a surface area of 86 cm2 radiates 5003 J of energy in 3 minutes. Find the body’s radiant power?

Given: Q = 5003 J, t = 3 × 60 = 180 s, A = 86 cm2 = 86 × 10-4 m2

We have,

R = Q / At

∴ R = 5003 / 86 × 10-4 × 180

∴ R = 5003 / 15480 × 10-4

∴ R = 0.3231 × 10-4 J

Question 3: When the body’s surface area is 0.12 m2, the amount of heat emitted is 3984 J, and the time period is 49 s, determine the radiant power of the body.

Given: Q = 3984 J, t = 49 s, A = 0.12 m2

We have,

R = Q / At

∴ R = 3984 / 0.12 × 49

∴ R = 3984 / 5.88

∴ R = 677.5 J

Question 4: When the radiant power is 3.55 W/m2 and the emissive power of a perfectly black surface is 4.2, determine the coefficient of emission.

Given: R = 3.55 W/m2, RB = 4.2

We have,

e = R / RB

∴ e = 3.55 / 4.2

∴ e = 0.84

Question 5: When the radiant power is 2 Jm-2s-1 and the emissivity is 1, determine the emissive power of a perfect black surface.

Given: R = 2, e = 1

We have,

e = R / RB

∴ RB = R / e

∴ RB = 2 / 1

∴ RB = 2

## FAQs on Emissive Power

Question 1: What is Emissivity?

The coefficient of emission, or emissivity (e), of a given surface is the distinction between the emissive power R of a surface and the emissive power RB of a perfect black surface, both at the same temperature.

Question 2: Prove Kirchhoff’s law.

The amount of heat radiated from a given region in a given amount of time is known as emissive power.

Quantity of radiant heat absorbed by body A = Quantity of heat emitted by body A

or

∴ aQ = R  …(Equation 1)

For the perfect blackbody B,

∴ Q = RB  …(Equation 2)

Dividing Equation 1 and Equation 2, we get

∴ a = R / RB

or

∴ RB = R / a

But R / RB = e  …(Emissive power)

∴ a = e

Hence, Kirchhoff’s law is theoretically proved.

Question 3: Define Emissive power and write its SI unit.

The amount of radiant heat emitted by a body per unit area unit time at a given temperature is called as Emissive power. And its SI unit is Jm-2s-1 or W/m2.

Question 4: Write factors of the Amount of heat radiated by a body.