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Define spectral emissivity of a medium of thickness \(L\) in terms of the spectral absorption coefficient.

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Question: Define the spectral emissivity of a medium with thickness L in terms of the spectral absorption coefficient. Answer: The spectral emissivity (ε(λ)) of a medium with thickness L in terms of the spectral absorption coefficient (α(λ)) can be expressed as: $$ε(λ) = 1 - e^{- α(λ) L}$$

Step by step solution

01

Understanding the terms

Spectral emissivity (denoted by ε(λ)) is the ratio of the emissive power of a medium at a wavelength λ, to the emissive power of a blackbody at the same wavelength and temperature. It is a dimensionless quantity and varies with wavelength. The spectral absorption coefficient (denoted by α(λ)) is a measure of how much light is absorbed by a medium at each wavelength. It is the fraction of light absorbed per unit length of the medium and depends on the wavelength and the properties of the medium.
02

Relationship between spectral emissivity and spectral absorption coefficient

According to Kirchoff's law of thermal radiation, the spectral emissivity is equal to the spectral absorptivity (denoted by \(a(\lambda)\)) of a medium at the same wavelength and temperature: $$ε(λ) = a(λ)$$ In a medium of thickness L, the spectral absorptivity can be expressed in terms of the spectral absorption coefficient using the Bouguer-Lambert-Beer law: $$a(λ) = 1 - e^{- α(λ) L}$$
03

Expressing spectral emissivity in terms of spectral absorption coefficient

Now, we can express the spectral emissivity in terms of the spectral absorption coefficient by substituting the expression for spectral absorptivity from Step 2 into the equation from Step 1: $$ε(λ) = 1 - e^{- α(λ) L}$$ This is the definition of the spectral emissivity of a medium of thickness L in terms of the spectral absorption coefficient.

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Key Concepts

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

Understanding Spectral Absorption Coefficient
The spectral absorption coefficient, denoted by \( \alpha(\lambda) \), plays a critical role in the study of light and matter interactions. Imagine shining a beam of light at a specific wavelength \( \lambda \) through a substance. Some of this light gets absorbed by the medium. But just how much is absorbed? That's where the spectral absorption coefficient comes into play.

It quantifies the amount of light at a certain wavelength that is absorbed per unit length of the material. Think of it like a measure of the material's thirst for light: higher values mean the material is 'hungrier' for absorbing light at that wavelength.

A clear analogy would be sunglasses. Different lenses, depending on their material and coating, absorb varying amounts of sunlight. The spectral absorption coefficient is essentially the scientific version of this, specifying the 'darkness' or absorption property of the material for a given color of light.
Kirchoff's Law of Thermal Radiation
Kirchoff's law of thermal radiation is a game-changer in understanding thermal processes. It unveils a profound relationship: at thermal equilibrium, the amount of energy a body emits at a certain wavelength is equal to the amount it can absorb at that wavelength. In other words, good absorbers are good emitters, and vice versa.

This relationship is actually quite intuitive. Consider a perfectly black surface—it absorbs all incoming light and doesn't reflect any. According to Kirchoff's law, this black surface also emits maximum thermal radiation when heated. So, if you're ever by a campfire, you'll notice that darker colored rocks or logs tend to get hotter than lighter ones because they absorb and emit more heat.

Mathematically, Kirchoff's law connects the spectral emissivity \( \epsilon(\lambda) \) to the spectral absorptivity \( a(\lambda) \) through the simple equation \( \epsilon(\lambda) = a(\lambda) \). This fundamental principle allows us to predict and understand the behavior of real materials with respect to heat and light.
Exploring the Bouguer-Lambert-Beer Law
The Bouguer-Lambert-Beer law is a cornerstone in optical physics, remarkably handy for anyone studying how light is transmitted through substances. In essence, it describes how the intensity of light decreases as the light travels through an absorbing medium.

Picture a flashlight beam passing through fog. The light seems to fade away with distance, right? This law provides the math behind this fading effect, stating that the intensity of light diminishes exponentially with the distance traveled through the medium and depends on, you guessed it, the spectral absorption coefficient we discussed earlier.

To ground this scientifically, the law is expressed as \( I = I_0 e^{ - \alpha(\lambda) L} \), where \(I_0 \) is the initial light intensity, \( L \) is the thickness of the medium, and \( I \) is the intensity after passing through the medium. This law not only applies to light but extends to other types of waves, such as sound waves passing through the air or seismic waves traveling through the earth. It's a universal concept that you'll see popping up in various fields, from studying the deep sea to exploring the far reaches of space.

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Most popular questions from this chapter

Two thin radiation shields with emissivities of \(\varepsilon_{3}=0.10\) and \(\varepsilon_{4}=0.15\) on both sides are placed between two very large parallel plates, which are maintained at uniform temperatures \(T_{1}=600 \mathrm{~K}\) and \(T_{2}=300 \mathrm{~K}\) and have emissivities \(\varepsilon_{1}=0.6\) and \(\varepsilon_{2}=0.7\), respectively (Fig. P13-93). Determine the net rates of radiation heat transfer between the two plates with and without the shields per unit surface area of the plates, and the temperatures of the radiation shields in steady operation.

What is operative temperature? How is it related to the mean ambient and radiant temperatures? How does it differ from effective temperature?

This experiment is conducted to determine the emissivity of a certain material. A long cylindrical rod of diameter \(D_{1}=0.01 \mathrm{~m}\) is coated with this new material and is placed in an evacuated long cylindrical enclosure of diameter \(D_{2}=0.1 \mathrm{~m}\) and emissivity \(\varepsilon_{2}=0.95\), which is cooled externally and maintained at a temperature of \(200 \mathrm{~K}\) at all times. The rod is heated by passing electric current through it. When steady operating conditions are reached, it is observed that the rod is dissipating electric power at a rate of \(8 \mathrm{~W}\) per unit of its length and its surface temperature is \(500 \mathrm{~K}\). Based on these measurements, determine the emissivity of the coating on the rod.

A 2-m-internal-diameter double-walled spherical tank is used to store iced water at \(0^{\circ} \mathrm{C}\). Each wall is \(0.5 \mathrm{~cm}\) thick, and the \(1.5\)-cm-thick air space between the two walls of the tank is evacuated in order to minimize heat transfer. The surfaces surrounding the evacuated space are polished so that each surface has an emissivity of \(0.15\). The temperature of the outer wall of the tank is measured to be \(20^{\circ} \mathrm{C}\). Assuming the inner wall of the steel tank to be at \(0^{\circ} \mathrm{C}\), determine \((a)\) the rate of heat transfer to the iced water in the tank and \((b)\) the amount of ice at \(0^{\circ} \mathrm{C}\) that melts during a 24-h period.

13-59 This question deals with steady-state radiation heat transfer between a sphere \(\left(r_{1}=30 \mathrm{~cm}\right)\) and a circular disk \(\left(r_{2}=120 \mathrm{~cm}\right)\), which are separated by a center-to- center distance \(h=60 \mathrm{~cm}\). When the normal to the center of disk passes through the center of the sphere, the radiation view factor is given by $$ F_{12}=0.5\left\\{1-\left[1+\left(\frac{r_{2}}{h}\right)^{2}\right]^{-0.5}\right\\} $$ Surface temperatures of the sphere and the disk are \(600^{\circ} \mathrm{C}\) and \(200^{\circ} \mathrm{C}\), respectively; and their emissivities are \(0.9\) and \(0.5\), respectively. (a) Calculate the view factors \(F_{12}\) and \(F_{21}\). (b) Calculate the net rate of radiation heat exchange between the sphere and the disk. (c) For the given radii and temperatures of the sphere and the disk, the following four possible modifications could increase the net rate of radiation heat exchange: paint each of the two surfaces to alter their emissivities, adjust the distance between them, and provide an (refractory) enclosure. Calculate the net rate of radiation heat exchange between the two bodies if the best values are selected for each of the above modifications.

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