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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

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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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Key Concepts

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

Wien's Displacement Law
Wien's Displacement Law is crucial for understanding blackbody radiation and helps us determine the temperature of black bodies. This law states that the wavelength where a blackbody emits most intensely is inversely proportional to its temperature. Mathematically, it is expressed as:\[\lambda_{max} T = b\]Here, \(\lambda_{max}\) represents the wavelength at maximum emissive power, \(T\) is the temperature in Kelvin, and \(b\) is Wien's constant \(\approx 2.898 \times 10^{-3} \text{m} \cdot \text{K}\). This relationship indicates that as temperature increases, the peak wavelength of emitted radiation shifts to shorter wavelengths.For example, hotter objects like stars emit more of their energy at shorter, visible wavelengths, while cooler objects radiate primarily in the infrared. In the problem, we use this law to calculate the temperature of a blackbody when given its peak emission wavelength \(\lambda_{max} = 0.7 \, \mu\text{m}\). Rearranging the formula gives:\[T = \frac{b}{\lambda_{max}}\]Plugging in the values:\[T = \frac{2.898 \times 10^{-3} \, \text{m} \cdot \text{K}}{0.7 \times 10^{-6} \, \text{m}} \approx 4139.08 \text{K}\]This result shows how Wien's Displacement Law helps us extract important thermal information from electromagnetic measurements.
Stefan-Boltzmann Law
The Stefan-Boltzmann Law provides the total power radiated per unit surface area of a blackbody across all wavelengths. It's an essential tool for understanding how temperature affects radiation emissions. The equation is given by:\[E = \sigma T^4\]Where \(E\) is the emissive power, \(T\) is the absolute temperature of the body in Kelvin, and \(\sigma\) is the Stefan-Boltzmann constant \(\approx 5.67 \times 10^{-8} \, \text{W/m}^2 \cdot \text{K}^4\). This law illustrates that the radiant energy from a body increases dramatically with temperature, specifically proportional to the fourth power.This means a slight increase in temperature results in a much larger increase in radiant energy output. In our problem, using the temperature \(T \approx 4139.08 \text{K}\), the total emissive power can be calculated:\[E = (5.67 \times 10^{-8} \, \text{W/m}^2 \cdot \text{K}^4)(4139.08 \text{K})^4 \approx 1.813 \times 10^7 \, \text{W/m}^2\]This calculation demonstrates how powerful thermal radiation can be, just by increasing a body's temperature. Understanding the Stefan-Boltzmann Law is vital for many fields, from astrophysics to climate science.
Emissive Power Calculation
In the context of blackbody radiation, emissive power is critical. It refers to the energy emitted per unit area from the surface, across all wavelengths. Calculating emissive power involves using laws such as Wien's displacement and Stefan-Boltzmann's to derive broader thermal characteristics. Emissive power calculations help determine vital properties of stars, the Earth's atmosphere, and even technological surfaces like solar panels. For instance:- In our exercise, given a temperature derived from Wien's Law, the Stefan-Boltzmann formula helps calculate the precise power output at that temperature.\[E = \sigma T^4\]This formula gives a sense of how much energy a surface emits due to its temperature. It underscores how Object's temperature exponentially affects the energy radiated, providing insights into thermal efficiencies and heat losses.Understanding this confluence of physics in calculating emissive power allows engineers and scientists to optimize energy usage, assess environmental heat exchanges, and design more efficient thermal management systems.

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Most popular questions from this chapter

What is a graybody? How does it differ from a blackbody? What is a diffuse gray surface?

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.

Consider a black spherical ball, with a diameter of \(25 \mathrm{~cm}\), is being suspended in air. Determine the surface temperature of the ball that should be maintained in order to heat \(10 \mathrm{~kg}\) of air from 20 to \(30^{\circ} \mathrm{C}\) in the duration of 5 minutes.

Consider an opaque horizontal plate that is well insulated on the edges and the lower surface. The plate is uniformly irradiated from above while air at \(T_{\infty}=300 \mathrm{~K}\) flows over the surface providing a uniform convection heat transfer coefficient of \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Under steady state conditions the surface has a radiosity of \(4000 \mathrm{~W} / \mathrm{m}^{2}\), and the plate temperature is maintained uniformly at \(350 \mathrm{~K}\). If the total absorptivity of the plate is \(0.40\), determine \((a)\) the irradiation on the plate, \((b)\) the total reflectivity of the plate, \((c)\) the emissive power of the plate, and \((d)\) the total emissivity of the plate.

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) \(537 \mathrm{~W} / \mathrm{m}^{2}\)

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