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Question

A body is placed inside a perfectly black enclosure and is allowed to reach thermal equilibrium. According to the Kirchhoff’s identity, which one of the following is equal to the emissivity of the body at a given wavelength?

The correct answer is
absorptivity of the body at the same wavelength

Kirchhoff's Identity Explained

Kirchhoff's Law of Thermal Radiation establishes a fundamental connection between how a body emits and absorbs thermal radiation.

The law specifically applies when a body is in thermal equilibrium with its surroundings. In this state, the condition of the body is stable, and there is no net flow of heat.

Applying Kirchhoff's Law

When a body is placed inside a perfectly black enclosure and reaches thermal equilibrium, its properties simplify:

  • Emissivity ($\epsilon$): This measures how effectively a surface emits thermal radiation compared to a perfect blackbody at the same temperature and wavelength.
  • Absorptivity ($\alpha$): This represents the fraction of incoming radiation at a specific wavelength that the body absorbs.

Under the conditions described (thermal equilibrium within a black enclosure), Kirchhoff's Law states that the emissivity of the body at any given wavelength is exactly equal to its absorptivity at that same wavelength.

Mathematically, this is expressed as:

$ \epsilon(\lambda, T) = \alpha(\lambda, T) $

Therefore, according to Kirchhoff's identity, the absorptivity of the body at a given wavelength equals its emissivity.

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Important Questions from Radiation

  1. Stefan Boltzmann's constant is expressed in the unit-

  2. A body whose absorptivity does not vary with temperature and wavelength of the incident ray is known as

  3. The process of heat transfer from a hot body to a cold body in a straight line, without affecting the intervening medium, is known as ______.

  4. Heat is transferred from an electric bulb by ______.

  5. Radiosity is defined as _______.
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