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Define emissivity and absorptivity. What is Kirchhoff's law of radiation?

Short Answer

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Answer: Emissivity measures an object's ability to emit thermal radiation compared to a perfect black body, while absorptivity describes the effectiveness of a material in absorbing incident radiation. Kirchhoff's law of radiation states that, at thermal equilibrium and for a given wavelength, the emissivity of a body is equal to its absorptivity. This implies that good absorbers are also good emitters, and poor absorbers are poor emitters, maintaining a balance between the energy being absorbed and the energy being emitted to keep a constant temperature.

Step by step solution

01

Define Emissivity

Emissivity (\(ε\)) is a dimensionless quantity (ranges between 0 and 1) that measures an object's ability to emit thermal radiation compared to that of a perfect black body, which absorbs all incident radiation and emits radiation at the maximum possible intensity for a given temperature. Emissivity depends on factors such as the object's material, surface properties, and temperature. In mathematical terms, it can be represented as: \(ε = \frac{\text{Radiative power emitted by the object}}{\text{Radiative power emitted by a black body at the same temperature}}\)
02

Define Absorptivity

Absorptivity (\(α\)) is a dimensionless quantity (ranges between 0 and 1) that describes the effectiveness of a material in absorbing incident radiation. It represents the ratio of the radiation energy absorbed by a material to the total incident radiation energy at a specific wavelength. A higher absorptivity means a material is more efficient in absorbing radiation. Mathematically, it can be expressed as: \(α = \frac{\text{Radiative energy absorbed by the object}}{\text{Incident radiative energy}}\)
03

Explain Kirchhoff's Law of Radiation

Kirchhoff's law of radiation states that, at thermal equilibrium and for a given wavelength, the emissivity of a body is equal to its absorptivity. In other words: \(ε = α\) This implies that good absorbers are also good emitters, and poor absorbers are poor emitters. The main idea behind this law is that a body in thermal equilibrium must have a balance between the energy being absorbed and the energy being emitted to maintain a constant temperature.

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Most popular questions from this chapter

A cold bottled drink ( $\left.m=2.5 \mathrm{~kg}, c_{p}=4200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\( at \)5^{\circ} \mathrm{C}$ is left on a table in a room. The average temperature of the drink is observed to rise to \(15^{\circ} \mathrm{C}\) in \(30 \mathrm{~min}\). The average rate of heat transfer to the drink is (a) \(23 \mathrm{~W}\) (b) \(29 \mathrm{~W}\) (c) \(58 \mathrm{~W}\) (d) \(88 \mathrm{~W}\) (e) \(122 \mathrm{~W}\)

An ice skating rink is located in a building where the air is at $T_{\text {air }}=20^{\circ} \mathrm{C}\( and the walls are at \)T_{w}=25^{\circ} \mathrm{C}$. The convection heat transfer coefficient between the ice and the surrounding air is \(h=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The emissivity of ice is \(\varepsilon=0.95\). The latent heat of fusion of ice is \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\), and its density is $920 \mathrm{~kg} / \mathrm{m}^{3}$. (a) Calculate the refrigeration load of the system necessary to maintain the ice at \(T_{s}=0^{\circ} \mathrm{C}\) for an ice rink of \(12 \mathrm{~m}\) by \(40 \mathrm{~m}\). (b) How long would it take to melt \(\delta=3 \mathrm{~mm}\) of ice from the surface of the rink if no cooling is supplied and the surface is considered insulated on the back side?

Can all three modes of heat transfer occur simultaneously (in parallel) in a medium?

A \(0.3\)-cm-thick, \(12-\mathrm{cm}\)-high, and \(18-\mathrm{cm}\)-long circuit board houses 80 closely spaced logic chips on one side, each dissipating $0.06 \mathrm{~W}$. The board is impregnated with copper fillings and has an effective thermal conductivity of $16 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. All the heat generated in the chips is conducted across the circuit board and is dissipated from the back side of the board to the ambient air. Determine the temperature difference between the two sides of the circuit board. Answer: \(0.042^{\circ} \mathrm{C}\)

It is well known that at the same outdoor air temperature a person is cooled at a faster rate under windy conditions than under calm conditions due to the higher convection heat transfer coefficients associated with windy air. The phrase wind-chill is used to relate the rate of heat loss from people under windy conditions to an equivalent air temperature for calm conditions (considered to be a wind or walking speed of \(3 \mathrm{mph}\) or $5 \mathrm{~km} / \mathrm{h}$ ). The hypothetical wind-chill temperature (WCT), called the wind-chill temperature index (WCTI), is an air temperature equivalent to the air temperature needed to produce the same cooling effect under calm conditions. A 2003 report on wind-chill temperature by the U.S. National Weather Service gives the WCTI in metric units as WCTI $\left({ }^{\circ} \mathrm{C}\right)=13.12+0.6215 T-11.37 \mathrm{~V}^{0.16}+0.3965 T V^{0.16}$ where \(T\) is the air temperature in \({ }^{\circ} \mathrm{C}\) and \(V\) the wind speed in \(\mathrm{km} / \mathrm{h}\) at \(10 \mathrm{~m}\) elevation. Show that this relation can be expressed in English units as WCTI $\left({ }^{\circ} \mathrm{F}\right)=35.74+0.6215 T-35.75 V^{0.16}+0.4275 T V^{0.16}$ where \(T\) is the air temperature in \({ }^{\circ} \mathrm{F}\) and \(V\) the wind speed in \(\mathrm{mph}\) at \(33 \mathrm{ft}\) elevation. Also, prepare a table for WCTI for air temperatures ranging from 10 to \(-60^{\circ} \mathrm{C}\) and wind speeds ranging from 10 to \(80 \mathrm{~km} / \mathrm{h}\). Comment on the magnitude of the cooling effect of the wind and the danger of frostbite

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