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Chapter 12: Problem 25

At a wavelength of \(0.7 \mu \mathrm{m}\), the black body emissive power is equal to \(10^{8} \mathrm{~W} / \mathrm{m}^{3}\). Determine \((a)\) the temperature of the blackbody and \((b)\) the total emissive power at this temperature.

Short Answer

Expert verified
Question: A black body has a maximum emissive power at a wavelength of 0.7 micrometers. Calculate (a) the temperature of the black body, and (b) the total emissive power at that temperature. Answer: (a) The temperature of the black body is approximately 4139.08 K. (b) The total emissive power at this temperature is approximately \(1.813 \times 10^7 \mathrm{W/m^2}\).

Step by step solution

01

Determine the temperature using Wien's Displacement Law

Wien's Displacement Law is given by the equation: $$ \lambda_{max}T = b $$ where \(\lambda_{max}\) is the wavelength of maximum emissive power (in this case, \(0.7 \mu \mathrm{m}\)), \(T\) is the temperature of the black body (in Kelvin), and \(b\) is Wien's displacement constant, which is approximately \(2.898\times10^{-3} \mathrm{m \cdot K}\). We are given \(\lambda_{max}=0.7 \times 10^{-6} \mathrm{m}\) and want to find \(T\). Rearranging and solving for \(T\) we get: $$ T = \frac{b}{\lambda_{max}} $$
02

Calculate the temperature of the black body

Using the values provided, we can calculate the temperature as follows: $$ T = \frac{2.898\times10^{-3} \mathrm{m \cdot K}}{0.7 \times 10^{-6} \mathrm{m}} \approx 4139.08 \mathrm{K} $$ So, the temperature of the black body is approximately 4139.08 K.
03

Determine the total emissive power using the Stefan-Boltzmann Law

The Stefan-Boltzmann Law states that the total emissive power of a black body is given by: $$ E = \sigma T^{4} $$ Where \(E\) is the total emissive power, \(T\) is the temperature of the black body, and \(\sigma\) is the Stefan-Boltzmann constant, which is approximately \(5.67\times10^{-8} \mathrm{W/m^2 \cdot K^4}\). We already know the temperature (\(T \approx 4139.08 \mathrm{K}\)), so we can calculate the total emissive power as follows:
04

Calculate the total emissive power at the given temperature

Using the values provided, calculate the total emissive power: $$ E = (5.67\times10^{-8} \mathrm{W/m^2 \cdot K^4})(4139.08 \mathrm{K})^4 \approx 1.813 \times 10^7 \mathrm{W/m^2} $$ So, the total emissive power at the given temperature is approximately \(1.813 \times 10^7 \mathrm{W/m^2}\). In summary, the temperature of the black body is approximately 4139.08 K and the total emissive power at this temperature is approximately \(1.813 \times 10^7 \mathrm{W/m^2}\).

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Most popular questions from this chapter

The temperature of the filament of an incandescent lightbulb is \(2500 \mathrm{~K}\). Assuming the filament to be a blackbody, determine the fraction of the radiant energy emitted by the filament that falls in the visible range. Also, determine the wavelength at which the emission of radiation from the filament peaks.

Define the total and spectral blackbody emissive powers. How are they related to each other? How do they differ?

A surface is exposed to solar radiation. The direct and diffuse components of solar radiation are 350 and \(250 \mathrm{~W} / \mathrm{m}^{2}\), and the direct radiation makes a \(35^{\circ}\) angle with the normal of the surface. The solar absorptivity and the emissivity of the surface are \(0.24\) and \(0.41\), respectively. If the surface is observed to be at \(315 \mathrm{~K}\) and the effective sky temperature is \(256 \mathrm{~K}\), the net rate of radiation heat transfer to the surface is (a) \(-129 \mathrm{~W} / \mathrm{m}^{2}\) (b) \(-44 \mathrm{~W} / \mathrm{m}^{2}\) (c) \(0 \mathrm{~W} / \mathrm{m}^{2}\) (d) \(129 \mathrm{~W} / \mathrm{m}^{2}\) (e) \(537 \mathrm{~W} / \mathrm{m}^{2}\)

An opaque horizontal plate is well insulated on the edges and the lower surface. The irradiation on the plate is \(3000 \mathrm{~W} / \mathrm{m}^{2}\), of which \(500 \mathrm{~W} / \mathrm{m}^{2}\) is reflected. The plate has a uniform temperature of \(700 \mathrm{~K}\) and has an emissive power of \(5000 \mathrm{~W} / \mathrm{m}^{2}\). Determine the total emissivity and absorptivity of the plate.

A small circular plate has a diameter of \(2 \mathrm{~cm}\) and can be approximated as a blackbody. To determine the radiation from the plate, a radiometer is placed normal to the direction of viewing from the plate at a distance of \(50 \mathrm{~cm}\). If the radiometer measured an irradiation of \(85 \mathrm{~W} / \mathrm{m}^{2}\) from the plate, determine the temperature of the plate.
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