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Chapter 1: Problem 50

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

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Question: Define emissivity, absorptivity, and explain Kirchhoff's law of radiation. Answer: Emissivity (\(\epsilon\)) is a measure of how effectively a material emits thermal radiation compared to a perfect blackbody radiator. Absorptivity (\(\alpha\)) represents the fraction of incident electromagnetic radiation absorbed by a material compared to the total incident radiation. Kirchhoff's law of radiation states that, for an object in thermal equilibrium, the emissivity and absorptivity are equal at the same temperature and wavelength (i.e., \(\epsilon = \alpha\)). This law implies that good absorbers are also good emitters, while poor absorbers are poor emitters.

Step by step solution

01

Definition of Emissivity

Emissivity, represented by the symbol \(\epsilon\), is a measure of how effectively a material emits thermal radiation as compared to a perfect blackbody radiator. It is a dimensionless quantity with values ranging from 0 to 1. A higher value of emissivity indicates more effective emission of thermal radiation by the material.
02

Definition of Absorptivity

Absorptivity, represented by the symbol \(\alpha\), is the fraction of incident electromagnetic radiation absorbed by a material compared to the total incident radiation. It is also a dimensionless quantity, with values ranging from 0 to 1. A higher value of absorptivity indicates more effective absorption of incident radiation by the material.
03

Kirchhoff's Law of Radiation

Kirchhoff's law of radiation states that, for an object in thermal equilibrium, the emissivity and absorptivity are equal at the same temperature and wavelength. Mathematically, it is expressed as: $$\epsilon = \alpha$$ This law implies that good absorbers are also good emitters of radiation and that poor absorbers are poor emitters. This principle is essential in understanding the behavior of materials in various applications such as thermal insulation, solar panels, and radiative cooling systems.

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Most popular questions from this chapter

A series of experiments were conducted by passing \(40^{\circ} \mathrm{C}\) air over a long \(25 \mathrm{~mm}\) diameter cylinder with an embedded electrical heater. The objective of these experiments was to determine the power per unit length required \((\dot{W} / L)\) to maintain the surface temperature of the cylinder at \(300^{\circ} \mathrm{C}\) for different air velocities \((V)\). The results of these experiments are given in the following table: $$ \begin{array}{lccccc} \hline V(\mathrm{~m} / \mathrm{s}) & 1 & 2 & 4 & 8 & 12 \\ \dot{W} / L(\mathrm{~W} / \mathrm{m}) & 450 & 658 & 983 & 1507 & 1963 \\ \hline \end{array} $$ (a) Assuming a uniform temperature over the cylinder, negligible radiation between the cylinder surface and surroundings, and steady state conditions, determine the convection heat transfer coefficient \((h)\) for each velocity \((V)\). Plot the results in terms of \(h\left(\mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)\) vs. \(V(\mathrm{~m} / \mathrm{s})\). Provide a computer generated graph for the display of your results and tabulate the data used for the graph. (b) Assume that the heat transfer coefficient and velocity can be expressed in the form of \(h=C V^{m}\). Determine the values of the constants \(C\) and \(n\) from the results of part (a) by plotting \(h\) vs. \(V\) on log-log coordinates and choosing a \(C\) value that assures a match at \(V=1 \mathrm{~m} / \mathrm{s}\) and then varying \(n\) to get the best fit.

The north wall of an electrically heated home is 20 \(\mathrm{ft}\) long, \(10 \mathrm{ft}\) high, and \(1 \mathrm{ft}\) thick, and is made of brick whose thermal conductivity is \(k=0.42 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}{ }^{\circ} \mathrm{F}\). On a certain winter night, the temperatures of the inner and the outer surfaces of the wall are measured to be at about \(62^{\circ} \mathrm{F}\) and \(25^{\circ} \mathrm{F}\), respectively, for a period of \(8 \mathrm{~h}\). Determine \((a)\) the rate of heat loss through the wall that night and \((b)\) the cost of that heat loss to the home owner if the cost of electricity is \(\$ 0.07 / \mathrm{kWh}\).

A hollow spherical iron container with outer diameter \(20 \mathrm{~cm}\) and thickness \(0.2 \mathrm{~cm}\) is filled with iced water at \(0^{\circ} \mathrm{C}\). If the outer surface temperature is \(5^{\circ} \mathrm{C}\), determine the approximate rate of heat loss from the sphere, in \(\mathrm{kW}\), and the rate at which ice melts in the container. The heat of fusion of water is \(333.7 \mathrm{~kJ} / \mathrm{kg}\).

A 40-cm-long, 0.4-cm-diameter electric resistance wire submerged in water is used to determine the convection heat transfer coefficient in water during boiling at \(1 \mathrm{~atm}\) pressure. The surface temperature of the wire is measured to be \(114^{\circ} \mathrm{C}\) when a wattmeter indicates the electric power consumption to be \(7.6 \mathrm{~kW}\). The heat transfer coefficient is (a) \(108 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(13.3 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(68.1 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(0.76 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(256 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Conduct this experiment to determine the combined heat transfer coefficient between an incandescent lightbulb and the surrounding air and surfaces using a \(60-\mathrm{W}\) lightbulb. You will need a thermometer, which can be purchased in a hardware store, and a metal glue. You will also need a piece of string and a ruler to calculate the surface area of the lightbulb. First, measure the air temperature in the room, and then glue the tip of the thermocouple wire of the thermometer to the glass of the lightbulb. Turn the light on and wait until the temperature reading stabilizes. The temperature reading will give the surface temperature of the lightbulb. Assuming 10 percent of the rated power of the bulb is converted to light and is transmitted by the glass, calculate the heat transfer coefficient from Newton's law of cooling.
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