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Chapter 12: Problem 26

The sun can be treated as a blackbody at \(5780 \mathrm{~K}\). Using EES (or other) software, calculate and plot the spectral blackbody emissive power \(E_{b \lambda}\) of the sun versus wavelength in the range of \(0.01 \mu \mathrm{m}\) to \(1000 \mu \mathrm{m}\). Discuss the results.

Short Answer

Expert verified
Answer: Planck's Law is used to find the spectral blackbody emissive power, given by: \[E_{b \lambda}(\lambda, T) = \frac{2\pi hc^2}{\lambda^5} \frac{1}{e^{\frac{hc}{\lambda k T}} - 1}\] where \(h\) is Planck's constant, \(c\) is the speed of light, \(k\) is Boltzmann's constant, \(\lambda\) is the wavelength, and \(T\) is the temperature of the blackbody. The peak wavelength is found using Wien's Displacement Law, given by: \[\lambda_{max} = \frac{b}{T}\] where \(b\) is Wien's displacement constant.

Step by step solution

01

Understand Planck's Law and Wien's Displacement Law

Planck's Law for spectral blackbody emissive power is given by: \[E_{b \lambda}(\lambda, T) = \frac{2\pi hc^2}{\lambda^5} \frac{1}{e^{\frac{hc}{\lambda k T}} - 1}\] where \(h\) is Planck's constant, \(c\) is the speed of light, \(k\) is Boltzmann's constant, \(\lambda\) is the wavelength, and \(T\) is the temperature of the blackbody. Wien's Displacement Law states that the wavelength corresponding to the maximum emissive power depends on the temperature and is given by: \[\lambda_{max} = \frac{b}{T}\] where \(b\) is Wien's displacement constant. This law is useful for detecting the peak of the spectral blackbody emissive power when analyzing the plots.
02

Calculate the blackbody emissive power for each wavelength

Using EES or other software, calculate the spectral blackbody emissive power \(E_{b \lambda}\) for each wavelength in the range of \(0.01 \mu \mathrm{m}\) to \(1000 \mu \mathrm{m}\). The temperature of the sun is given as \(5780 \mathrm{~K}\). Use Planck's Law to find \(E_{b \lambda}\) for each wavelength.
03

Create a plot of \(E_{b \lambda}\) versus wavelength

Plot the calculated values of \(E_{b \lambda}\) on the y-axis versus the wavelength on the x-axis. The plot should cover the entire range of wavelengths from \(0.01 \mu \mathrm{m}\) to \(1000 \mu \mathrm{m}\).
04

Analyze the peak wavelength using Wien's Displacement Law

With the plot generated, we can obtain the peak wavelength by comparing it with the values calculated using Wien's Displacement Law. The peak wavelength, \(\lambda_{max}\), can be found using the formula: \[\lambda_{max} = \frac{b}{T}\] where \(T = 5780 \mathrm{~K}\). Make sure the peak wavelength calculated matches the peak point on the plot.
05

Discuss the results

The plot shows the distribution of sun's spectral blackbody emissive power across different wavelengths. The peak wavelength corresponds to the maximum emissive power (maximum intensity of sunlight). This exercise helps to understand the behavior of the sun as a blackbody and how its temperature is related to the emitted radiation. Moreover, it gives insight into the characteristics of sunlight, which is essential for various applications such as solar energy harvesting.

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Most popular questions from this chapter

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 \(300 \mathrm{~K}\) ?

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