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Chapter 6: Problem 12

What must be the temperature of a graybody with emissivity of 0.45 if it is to have the same total radiant exitance as a blackbody at \(5000 \mathrm{K} ?\)

Short Answer

Expert verified
The graybody temperature must be approximately 6021 K.

Step by step solution

01

Understanding the Problem

We need to find the temperature of a graybody that has the same total radiant exitance as a blackbody at 5000 K. The graybody has an emissivity (") of 0.45.
02

Using the Stefan-Boltzmann Law for Blackbody

First, we calculate the total radiant exitance of a blackbody at 5000 K using the Stefan-Boltzmann Law, \[ M = \sigma T^4 \] where \(\sigma\) is the Stefan-Boltzmann constant \(5.67 \times 10^{-8} \text{W/m}^2\text{/K}^4\). Therefore,\[ M_{blackbody} = 5.67 \times 10^{-8} \times (5000)^4 \text{W/m}^2.\]
03

Calculating the Blackbody Exitance

Compute the exitance for the blackbody:\[ M_{blackbody} = 5.67 \times 10^{-8} \times (5000)^4 = 7.085 \times 10^7 \text{W/m}^2.\] This is the total radiant exitance of the blackbody.
04

Applying the Stefan-Boltzmann Law to Graybody

A graybody's radiant exitance is given by\[ M_{graybody} = \epsilon \sigma T^4, \]where \(\epsilon = 0.45\) is the emissivity and \(T\) is the temperature we want to find. Set \(M_{graybody} = M_{blackbody}\):\[ 0.45 \times 5.67 \times 10^{-8} \times T^4 = 7.085 \times 10^7 \text{W/m}^2.\]
05

Solving for the Graybody Temperature

To find \(T\), rearrange and solve the equation:\[ T^4 = \frac{7.085 \times 10^7}{0.45 \times 5.67 \times 10^{-8}},\]\[ T^4 = \frac{7.085 \times 10^7}{2.5515 \times 10^{-8}},\]\[ T^4 = 2.777 \times 10^{15},\]\[ T = \sqrt[4]{2.777 \times 10^{15}} \approx 6020.8\text{K}.\]
06

Conclusion

The temperature of the graybody must be approximately 6021 K so that its total radiant exitance matches that of a blackbody at 5000 K.

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Key Concepts

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

Graybody
A graybody is a type of object that emits thermal radiation, similar to a blackbody, but with one key difference. Unlike a blackbody, which absorbs and emits radiation perfectly at all wavelengths, a graybody reflects a certain percentage of the incoming radiation. This partial reflectivity is described by a property called **emissivity**. The emissivity of a graybody is less than 1, meaning it is not a perfect emitter.
  • Graybodies are practical models for real-world objects that do not emit as much radiation as theoretical blackbodies.
  • Their radiation characteristics are influenced by the material properties and surface textures.
Graybodies are important in fields such as climate science, electronics cooling, and astrophysics, where realistic modeling of radiation is critical.
Emissivity
Emissivity is a measure of how effectively a surface emits thermal radiation compared to a blackbody. It ranges from 0 to 1, where 1 represents a perfect blackbody.
  • In simpler terms, it represents the efficiency of an object in radiating energy.
  • High emissivity means the surface emits radiation well, while low emissivity surfaces reflect more energy than they emit.
For example, in the provided exercise, we are dealing with a graybody that has an emissivity of 0.45. This means the graybody emits 45% of the radiation a blackbody would emit under the same conditions. Emissivity affects the rate of energy loss by radiation from objects, influencing how quickly they heat up or cool down.
Radiant Exitance
Radiant exitance, denoted by the symbol \(M\), is the power emitted by a surface per unit area. It is measured in watts per square meter (W/m虏). In thermodynamics and physics, radiant exitance is often used to quantify the thermal radiation from surfaces. The Stefan-Boltzmann Law gives us a way to calculate the radiant exitance of a blackbody:
  • For a blackbody, \( M = \sigma T^4 \) where \(\sigma\) is the Stefan-Boltzmann constant and \(T\) is the temperature in Kelvin.
  • For a graybody, the formula modifies to \( M = \epsilon \sigma T^4 \), where \(\epsilon\) is the emissivity.
In the exercise, we equate the radiant exitance of a blackbody to that of a graybody to find the temperature at which the graybody matches the blackbody in emitted power.
Temperature Calculation
Calculating the temperature of a graybody to match the radiant exitance of a blackbody involves rearranging the modified Stefan-Boltzmann equation. The problem in the exercise showed this process.
  • First, calculate the blackbody's exitance at a given temperature using \( M_{blackbody} = \sigma T^4 \).
  • For a graybody with known emissivity, solve \( M_{graybody} = \epsilon \sigma T^4 \).
In our task, once the radiant exitance of the blackbody was known, we set it equal to the graybody's exitance and solved for the temperature \(T\). After some algebra, this results in:\[ T = \sqrt[4]{\frac{M_{blackbody}}{\epsilon \sigma}} \]This gives the temperature at which the graybody radiates the same power as the blackbody, providing a vital tool for thermal calculations in engineering and scientific disciplines.

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Most popular questions from this chapter

A He-Ne laser has a beam waist (diameter) equal to about \(1 \mathrm{mm}\) a. What is its beam-spread angle in the far field? b. Estimate the diameter of this beam after it has propagated over a distance of \(1 \mathrm{km}\)

The allowed rotational energies \(E_{l}^{\text {rot }}\) of a diatomic molecule are given by $$E_{l}=\frac{l(l+1) \hbar^{2}}{2 I}$$ In this expression \(l\) is the rotational quantum number and can take the values \(l=0,1,2 \ldots ; I\) is the rotational inertia of the molecule about an axis through its center of mass; and \(\hbar=h / 2 \pi .\) The equilibrium separation of the two atoms in a diatomic hydrogen molecule \(H_{2}\) is about 0.074 nm. The mass of each hydrogen atom is about \(1.67 \times 10^{-27} \mathrm{kg}\) a. Show that the rotational inertia of the hydrogen molecule about an axis through its center of mass is about \(I=4.6 \times 10^{-48} \mathrm{kg} \cdot \mathrm{m}^{2}\) b. Find the difference in energy between the first excited rotational energy state and the ground rotational state. That is, find \(E_{1}^{\text {ot }}-E_{0}^{\text {fot }}\). Express the answer in both \(\mathrm{J}\) and \(\mathrm{eV}\). c. Find the relative likelihood \(P_{l=1} / P_{l=0}\) that a hydrogen molecule will be in its first excited rotational state in thermal equilibrium at room temperature, \(T=293 \mathrm{K}\) nore possible state degeneracies.

a. Will a photon of energy 5 eV likely be absorbed by a hydrogen atom originally in its ground state? b. What is the range of photon wavelengths that could ionize a hydrogen atom that is originally in its ground state? c. What is the range of photon wavelengths that could ionize a hydrogen atom that is originally in its \(n=2\) energy state?

The rate of decay of an assembly of atoms with population density \(N_{2}\) at excited energy level \(E_{2}\) when spontaneous emission is the only important process is $$\left(\frac{d N_{2}}{d t}\right)_{\mathrm{spont}}=-A_{21} N_{2}$$ Show that an initial population density \(N_{20}\) decreases to a value \(N_{20} / e\) in a time \(\tau\) equal to \(1 / A_{21} .\) That is, show that \(A_{21}\) is the inverse of the lifetime of the atomic level.

Derive the Stefan-Boltzmann law from the Planck blackbody spectral radiance formula. (Hint: Use a substitution of \(x=h c / \lambda k_{B} T\) to facilitate the integration.
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