Question

The emissivity of a surface coated with aluminum oxide can be approximated to be \(0.15\) for radiation at wavelengths less than \(5 \mu \mathrm{m}\) and \(0.9\) for radiation at wavelengths greater than \(5 \mu \mathrm{m}\). Determine the average emissivity of this surface at (a) \(5800 \mathrm{~K}\) and (b) \(300 \mathrm{~K}\). What can you say about the absorptivity of this surface for radiation coming from sources at \(5800 \mathrm{~K}\) and \(300 \mathrm{~K}\) ?

Short answer

Question: Calculate the average emissivities of a surface coated with aluminum oxide at temperatures of 5800 K and 300 K, and discuss the absorptivity of the surface for radiation coming from sources at both temperatures. Answer: The average emissivities can be calculated by evaluating the integrals in Step 6 using numerical integration methods. For temperature 5800 K and 300 K, find the average emissivities by dividing the total power emitted by the surface by the power emitted by a black body at the same temperature. The absorptivity for radiation coming from sources at 5800 K and 300 K will have the same wavelength-dependent behavior (equal to 0.15 for wavelengths less than 5 μm and equal to 0.9 for wavelengths greater than 5 μm) due to Kirchhoff's Law of thermal radiation.

Step-by-step solution

Understand the Emissivity-Planck's Law relationship

Planck's Law gives the spectral intensity of blackbody radiation as a function of wavelength and temperature. We can relate the spectral emissivity (emissivity as a function of wavelength) to the spectral intensity using Planck's Law. The average emissivity is defined as the total emitted power divided by the total power emitted by a black body at the same temperature.

Write Planck's Law for blackbody radiation

Planck's Law for the spectral intensity of blackbody radiation as a function of wavelength and temperature is given by: \(B(\lambda, T) = \frac{2\pi hc^2}{\lambda^5}\frac{1}{e^{\frac{hc}{\lambda k_BT}} - 1}\) where \(B(\lambda, T)\) is the spectral intensity, \(h\) is the Planck constant, \(c\) is the speed of light, \(\lambda\) is the wavelength, \(k_B\) is the Boltzmann constant, and \(T\) is the temperature.

Write the power emitted by a black body

We know that the power emitted by a black body per unit area is given by Stefan-Boltzmann Law, which is: \(P_{BB}(T) = \sigma T^4\) where \(\sigma\) is the Stefan-Boltzmann constant and \(T\) is the temperature.

Calculate the spectral emissivity of the surface

The spectral emissivity at each wavelength, \(\epsilon(\lambda)\), can be assumed to be constant (either \(0.15\) or \(0.9\)) for the given intervals. Using this, we can determine the total power emitted by the surface: \(P_{surface}(T) = \int_{0}^{5 \mu m} 0.15 \times B(\lambda, T) d\lambda + \int_{5 \mu m}^{\infty} 0.9 \times B(\lambda, T) d\lambda\)

Calculate the average emissivity

Now we can determine the average emissivity at each temperature by dividing the total power emitted by the surface by the power emitted by a black body at the same temperature: \(\bar{\epsilon}(T) = \frac{P_{surface}(T)}{P_{BB}(T)}\)

Simplify the integrals

Although these integrals are difficult to compute analytically, we can use numerical integration methods (e.g., Simpson's rule, trapezoidal rule) to approximate the values of the integrals. For \(T = 5800\) K and \(T = 300\) K, we evaluate the integrals: \(\int_{0}^{5 \mu m} 0.15 \times B(\lambda, 5800 K) d\lambda\), \(\int_{5 \mu m}^{\infty} 0.9 \times B(\lambda, 5800 K) d\lambda\), \(\int_{0}^{5 \mu m} 0.15 \times B(\lambda, 300 K) d\lambda\), \(\int_{5 \mu m}^{\infty} 0.9 \times B(\lambda, 300 K) d\lambda\) Then, we find the average emissivities \(\bar{\epsilon}(5800 K)\) and \(\bar{\epsilon}(300 K)\) using the total powers emitted and the black body emissions.

Discuss absorptivity

By Kirchhoff's Law of thermal radiation, absorptivity at a given wavelength and temperature must be equal to emissivity at that same wavelength and temperature for a surface in thermal equilibrium. Therefore, we can say that the absorptivity of the surface for radiation coming from sources at \(5800 K\) and \(300 K\) will have the same wavelength-dependent behavior (i.e., equal to \(0.15\) for wavelengths less than \(5 \mu m\) and equal to \(0.9\) for wavelengths greater than \(5 \mu m\)).

Key concepts

Understanding Emissivity and Planck's Law

Emissivity is a material property that measures how efficiently a surface emits thermal radiation compared to a perfect emitter, known as a blackbody. The concept of a blackbody is crucial because it radiates energy at the maximum possible rate per unit area at each wavelength for any given temperature. Planck's Law allows us to describe this radiation with precision.

According to Planck's Law, the spectral intensity of the radiation emitted by a blackbody can be expressed mathematically. This relationship is given by the formula:
\[B(\lambda, T) = \frac{2\pi hc^2}{\lambda^5}\frac{1}{e^{\frac{hc}{\lambda k_BT}} - 1}\]
where \(B(\lambda, T)\) is the spectral intensity, \(h\) is Planck's constant, \(c\) is the speed of light, \(\lambda\) is the wavelength, \(k_B\) is Boltzmann's constant, and \(T\) is the absolute temperature of the blackbody. To grasp the concept of spectral intensity, imagine it as the brightness or strength of radiation at a specific wavelength and temperature.

The Role of Blackbody Radiation in Thermal Emission

Blackbody radiation serves as the standard to which all real-world objects are compared in terms of their thermal emission capabilities. In theoretical physics, a blackbody is an idealized object that perfectly absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. That is, it has an emissivity (\(\epsilon\)) of 1.

Since real objects do not behave as ideal blackbodies (having \(\epsilon\) less than 1), understanding the nature of blackbody radiation helps us explore and quantify the thermal emission of real materials. For instance, aluminum oxide coated surfaces don't completely absorb or emit radiation because their emissivity values range from 0.15 to 0.9 depending on the wavelength, making them selective emitters.

The Stefan-Boltzmann Law and Total Emitted Power

In the realm of thermodynamics, the Stefan-Boltzmann Law is a pivotal concept that relates the total energy radiated per unit surface area of a blackbody to the fourth power of its absolute temperature. This law is mathematically expressed as:
\[P_{BB}(T) = \sigma T^4\]
where \(P_{BB}(T)\) is the total power emitted per unit area, \(\sigma\) is the Stefan-Boltzmann constant, and \(T\) is the absolute temperature. The power calculation is integral in deriving the average emissivity of a surface by comparing its radiated power to the power that would be emitted if it were a blackbody at the same temperature.

Spectral Intensity and Material Emissivity

Spectral intensity plays a defining role in determining how much energy a material radiates at a specific wavelength. This per-wavelength perspective is crucial when materials emit and absorb radiation differently across the electromagnetic spectrum, such as the aluminum oxide coated surface mentioned in the exercise.

The spectral intensity of any material can be tied back to Planck's Law while incorporating the material's emissivity (\(\epsilon(\lambda)\)). The total power emitted by a surface for a given temperature takes into account the material's emissivity at each wavelength, thereby modifying the Planck's formula to include the multiplying factor of emissivity. The average emissivity overall wavelengths can then be calculated, which gives a measure of the material's total emissivity.

Kirchhoff's Law of Thermal Radiation

An essential principle when discussing radiation exchange between surfaces is Kirchhoff's Law of thermal radiation. It states that for a surface at thermal equilibrium, the emissivity (the ability to emit) at a given wavelength and temperature is equal to its absorptivity (the ability to absorb) at the same wavelength and temperature.

Applying this law to the surface with variable emissivity, depending on the wavelength, allows us to infer its absorptive behavior. It tells us that the absorptivity just like emissivity, will be 0.15 for wavelengths less than \(5 \mu m\) and 0.9 for those greater. This symmetrical relationship is vital in thermal analysis and helps understand how materials interact with incident radiation from various sources, at different temperatures.