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Define the properties emissivity and absorptivity. When are these two properties equal to each other?

Short Answer

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Question: Explain the properties of emissivity and absorptivity and describe the condition in which they are equal to each other. Answer: Emissivity (ε) and absorptivity (α) are dimensionless properties that measure a material's effectiveness in emitting thermal radiation and absorbing incoming radiation, respectively, as compared to an ideal black body. They both range from 0 to 1, where 0 represents a perfect reflector while 1 represents a perfect emitter (black body). According to Kirchhoff's law of thermal radiation, emissivity and absorptivity are equal when a material is in thermal equilibrium and at a specific wavelength and temperature.

Step by step solution

01

Define Emissivity

Emissivity (denoted by ε) is a dimensionless property that measures a material's effectiveness in emitting thermal radiation as compared to the ideal black body. It ranges from 0 to 1, where 0 represents a perfect reflector (no emission) and 1 represents a perfect emitter (black body). The formula for emissivity is as follows: ε = (Power emitted by the material) / (Power emitted by an ideal black body)
02

Define Absorptivity

Absorptivity (denoted by α) is also a dimensionless property that quantifies a material's effectiveness in absorbing incoming radiation, compared to the ideal black body. It ranges from 0 to 1, where 0 indicates that the material does not absorb any radiation (perfect reflector), and 1 means that it absorbs all incident radiation (black body). The formula for absorptivity is as follows: α = (Power absorbed by the material) / (Power incident on the material)
03

Relate emissivity to absorptivity

By relating the two properties, we can find the condition when emissivity and absorptivity are equal. According to Kirchhoff's law of thermal radiation, for a given material in thermal equilibrium and at a specific wavelength and temperature, the emissivity is equal to the absorptivity: ε(λ,T) = α(λ,T) In other words, the emissivity and absorptivity are equal when the material is in thermal equilibrium, at a specific wavelength and temperature.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Emissivity
Emissivity, symbolized as \( \varepsilon \), is a key property in understanding how materials interact with thermal radiation. It tells us how efficiently a material emits energy as thermal radiation in comparison to an ideal black body. A black body is a theoretical object that emits the maximum possible amount of radiation for a given temperature. The emissivity of a material is a dimensionless number ranging from 0 to 1.
  • An emissivity of 1 means the material is a perfect emitter, just like a black body.
  • An emissivity of 0 means the material does not emit any radiation, acting like a perfect reflector.
In practical terms, knowing the emissivity of a material helps in calculating how much heat it can radiate. This property varies with temperature, surface condition, and wavelength of the radiation. A material's emissivity impacts its thermal balance in applications like heating systems, infrared thermography, and thermal insulation.
Absorptivity
Absorptivity, represented by \( \alpha \), measures how well a material absorbs thermal radiation. Like emissivity, it is a dimensionless quantity that varies between 0 and 1. This property provides insight into how materials interact with surrounding radiation and contribute to their thermal characteristics.
  • Absorptivity of 1 indicates that the material absorbs all incoming radiation, similar to a black body.
  • Absorptivity of 0 implies that the material reflects all radiation, similar to a perfect mirror.
Absorptivity is important in designs involving solar panels, thermal coatings, and building materials, influencing how effective they will be in capturing or reflecting heat. Factors like surface finish, temperature, and the wavelength of incident radiation can alter the absorptivity of the material.
Kirchhoff's Law
Kirchhoff's Law of Thermal Radiation elegantly links the concepts of emissivity and absorptivity. Introduced by Gustav Kirchhoff in 1859, the law states that, for a material in thermal equilibrium, the emissivity and absorptivity at a given wavelength and temperature are equal.
  • \( \varepsilon(\lambda,T) = \alpha(\lambda,T) \)
  • This equality implies that the more a material absorbs, the more it can emit at thermal equilibrium.
This relation is crucial in applications requiring radiative heat transfer analysis. It underpins the design of energy-efficient thermal systems, since realizing when emissivity equals absorptivity helps optimize materials for specific thermal tasks. Understanding this law assists in predicting a material's behavior under thermal stress or in energy transfer processes.
Thermal Equilibrium
Thermal equilibrium is a condition where a material or system reaches a state where temperature remains constant over time, as the net energy exchange is zero. At this point, the rate of energy leaving the material equals the rate of energy entering it.
  • Materials in thermal equilibrium do not experience a change in temperature.
  • This state is crucial for applying Kirchhoff's Law of Thermal Radiation, as it requires a constant temperature for emissivity to equal absorptivity.
Achieving thermal equilibrium is vital in processes ranging from industrial manufacturing to atmospheric science. The concept helps in balancing the energy budget of a system, ensuring energy efficient systems. It also plays a role in climate models and environmental monitoring, where understanding heat flows and exchanges can drive decisions for sustainable practices.

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Most popular questions from this chapter

A cordless telephone is designed to operate at a frequency of \(8.5 \times 10^{8} \mathrm{~Hz}\). Determine the wavelength of these telephone waves.

Consider a radio wave with a wavelength of \(10^{7} \mu \mathrm{m}\) and a \(\gamma\)-ray with a wavelength of \(10^{-7} \mu \mathrm{m}\). Determine the photon energies of the radio wave and the \(\gamma\)-ray, and the photon energy ratio of the \(\gamma\)-ray to the radio wave.

A small surface of area \(A=1 \mathrm{~cm}^{2}\) emits radiation as a blackbody at \(1800 \mathrm{~K}\). Determine the rate at which radiation energy is emitted through a band defined by \(0 \leq \phi \leq 2 \pi\) and \(45 \leq \theta \leq 60^{\circ}\), where \(\theta\) is the angle a radiation beam makes with the normal of the surface and \(\phi\) is the azimuth angle.

What is the solar constant? How is it used to determine the effective surface temperature of the sun? How would the value of the solar constant change if the distance between the earth and the sun doubled?

The absorber surface of a solar collector is made of aluminum coated with black chrome ( \(\alpha_{s}=0.87\) and \(\left.\varepsilon=0.09\right)\). Solar radiation is incident on the surface at a rate of \(600 \mathrm{~W} / \mathrm{m}^{2}\). The air and the effective sky temperatures are \(25^{\circ} \mathrm{C}\) and \(15^{\circ} \mathrm{C}\), respectively, and the convection heat transfer coefficient is \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). For an absorber surface temperature of \(70^{\circ} \mathrm{C}\), determine the net rate of solar energy delivered by the absorber plate to the water circulating behind it.

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