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Chapter 12: Problem 130

Consider a surface at \(500 \mathrm{~K}\). The spectral blackbody emissive power at a wavelength of \(50 \mu \mathrm{m}\) is (a) \(1.54 \mathrm{~W} / \mathrm{m}^{2} \cdot \mu \mathrm{m}\) (b) \(26.3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mu \mathrm{m}\) (c) \(108.4 \mathrm{~W} / \mathrm{m}^{2} \cdot \mu \mathrm{m}(d) 2750 \mathrm{~W} / \mathrm{m}^{2} \cdot \mu \mathrm{m}(e) 8392 \mathrm{~W} / \mathrm{m}^{2} \cdot \mu \mathrm{m}\)

Short Answer

Expert verified
Answer: The spectral blackbody emissive power at a wavelength of 50 µm for a surface at 500 K is approximately 1.54 W/m²·µm.

Step by step solution

01

Write down Planck's Law formula

Planck's Law describes the spectral distribution of blackbody radiation and can be written as: $$ E_\lambda(\lambda, T) = \frac{2 \pi h c^2}{\lambda^5} \frac{1}{e^{\frac{hc}{\lambda k_BT}} - 1} $$ Where: \(E_\lambda\) is the spectral blackbody emissive power (\(\mathrm{W} / \mathrm{m}^{2} \cdot \mu \mathrm{m}\)), \(\lambda\) is the wavelength (\(\mathrm{m}\)), \(h\) is the Planck constant (\(6.626 \times 10^{-34} \mathrm{~Js}\)), \(c\) is the speed of light (\(2.998 \times 10^{8} \mathrm{~m/s}\)), \(T\) is the temperature in Kelvin (K), \(kB\) is the Boltzmann constant (\(1.381 \times 10^{-23} \mathrm{~J/K}\)).
02

Convert given wavelength to meters

The given wavelength is 50 µm, which we need to convert to meters for the calculation: $$ \lambda = 50 \times 10^{-6} \mathrm{~m} $$
03

Calculate the spectral blackbody emissive power

Now, we can plug in the given data into the Planck's Law formula to determine the spectral blackbody emissive power \(E_\lambda\): $$ E_\lambda = \frac{2 \pi (6.626 \times 10^{-34}) (2.998 \times 10^{8})^2}{(50 \times 10^{-6})^5} \frac{1}{e^{\frac{(6.626 \times 10^{-34})(2.998 \times 10^{8})}{(50 \times 10^{-6})(1.381 \times 10^{-23})(500)}} - 1} $$ After performing the calculations, we get: $$ E_\lambda \approx 1.54 \mathrm{~W} / \mathrm{m}^{2} \cdot \mu \mathrm{m} $$ The spectral blackbody emissive power at a wavelength of 50 µm for a surface at 500 K is approximately 1.54 \(\mathrm{W} / \mathrm{m}^{2} \cdot \mu \mathrm{m}\), which corresponds to option (a).

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Most popular questions from this chapter

What changes would you notice if the sun emitted radiation at an effective temperature of \(2000 \mathrm{~K}\) instead of \(5762 \mathrm{~K}\) ?

A long metal bar \(\left(c_{p}=450 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \rho=\right.\) \(7900 \mathrm{~kg} / \mathrm{m}^{3}\) ) is being conveyed through a water bath to be quenched. The metal bar has a cross section of \(30 \mathrm{~mm} \times 15 \mathrm{~mm}\), and it enters the water bath at \(700^{\circ} \mathrm{C}\). During the quenching process, \(500 \mathrm{~kW}\) of heat is released from the bar in the water bath. In order to prevent thermal burn on people handling the metal bar, it must exit the water bath at a temperature below \(45^{\circ} \mathrm{C}\). A radiometer is placed normal to and at a distance of \(1 \mathrm{~m}\) from the bar to monitor the exit temperature. The radiometer receives radiation from a target area of \(1 \mathrm{~cm}^{2}\) of the bar surface. Irradiation signal detected by the radiometer is used to control the speed of the bar being conveyed through the water bath so that the exit temperature is safe for handing. If the radiometer detects an irradiation of \(0.015 \mathrm{~W} / \mathrm{m}^{2}\), determine the speed of the bar bar can be approximated as a blackbody.

By what properties is an electromagnetic wave characterized? How are these properties related to each other?

A radiometer is employed to monitor the temperature of manufactured parts \(\left(A_{1}=10 \mathrm{~cm}^{2}\right)\) on a conveyor. The radiometer is placed at a distance of \(1 \mathrm{~m}\) from and normal to the manufactured parts. When a part moves to the position normal to the radiometer, the sensor measures the radiation emitted from the part. In order to prevent thermal burn on people handling the manufactured parts at the end of the conveyor, the temperature of the parts should be below \(45^{\circ} \mathrm{C}\). An array of spray heads is programmed to discharge mist to cool the parts when the radiometer detects a temperature of \(45^{\circ} \mathrm{C}\) or higher on a part. If the manufactured parts can be approximated as blackbody, determine the irradiation on the radiometer that should trigger the spray heads to release cooling mist when the temperature is not below \(45^{\circ} \mathrm{C}\).

Solar radiation is incident on a semi-transparent body at a rate of \(500 \mathrm{~W} / \mathrm{m}^{2}\). If \(150 \mathrm{~W} / \mathrm{m}^{2}\) of this incident radiation is reflected back and \(225 \mathrm{~W} / \mathrm{m}^{2}\) is transmitted across the body, the absorptivity of the body is (a) 0 (b) \(0.25\) (c) \(0.30\) (d) \(0.45\) (e) 1
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