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Chapter 11: Problem 13

A copper sphere is suspended in an evacuated chamber maintained at \(300 \mathrm{~K}\). The sphere is maintained at constant temperature of \(900 \mathrm{~K}\) by heating electrically. A total of \(300 \mathrm{~W}\) electric power is needed to do this. When half of the surface of the copper sphere is completely blackened, \(600 \mathrm{~W}\) is needed to maintain the same temperature of sphere. The emissivity of copper is (A) \(1 / 4\) (B) \(1 / 3\) (C) \(1 / 2\) (D) 1

Short Answer

Expert verified
The emissivity of copper is \(\frac{1}{3}\).

Step by step solution

01

Write down the Stefan-Boltzmann Law for the power radiated

The power radiated by an object is given by the Stefan-Boltzmann Law: \[P = \epsilon A \sigma T^4\] Where: - \(P\) is the power (in Watts) - \(\epsilon\) is the emissivity (dimensionless) - \(A\) is the area of the object (in m虏) - \(\sigma\) is the Stefan-Boltzmann constant, approximately \(5.67 \times 10^{-8} \mathrm{W \cdot m^{-2} \cdot K^{-4}}\) - \(T\) is the temperature of the object (in K)
02

Set up the equations for the two situations

Initially, with the entire sphere non-blackened, we need 300 W to maintain the sphere's temperature. Let's denote the initial emissivity as \(\epsilon_c\), for the initial copper surface. The equation becomes: \[P_1 = 300 = \epsilon_c A\sigma T^4\] When half of the sphere is blackened, we need 600 W to maintain the temperature. The blackened part has an emissivity of 1. Let's denote the black and non-black areas as \(A_b\) and \(A_c\), respectively, and the total area as \(A\), we get: \[P_2 = 600 = (1 \cdot A_b + \epsilon_c A_c) \sigma T^4\]
03

Simplify the equations and solve for the emissivity of copper

First, we know that half the sphere is blackened, so \(A_b = \frac{A}{2}\) and \(A_c = \frac{A}{2}\). We can plug that into the second equation: \[600 = \left(1 \cdot \frac{A}{2} + \epsilon_c \cdot \frac{A}{2}\right) \sigma T^4\] Now, divide the second equation by the first equation: \[\frac{600}{300} = \frac{\left(1 \cdot \frac{A}{2} + \epsilon_c \cdot \frac{A}{2}\right) \sigma T^4}{\epsilon_c A\sigma T^4}\] Simplify the equation: \[\frac{600}{300} = \frac{1 + \epsilon_c}{2\epsilon_c}\] Solve for \(\epsilon_c\): \[\epsilon_c = \frac{1}{3}\] So the correct answer is (B). The emissivity of copper is \(\frac{1}{3}\).

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Most popular questions from this chapter

On the surface of lake when the atmospheric temperature is \(-15^{\circ} \mathrm{C}, 1.5 \mathrm{~cm}\) thick layer of ice is formed in \(20 \mathrm{~min}\), time taken to change its thickness from \(1.5 \mathrm{~cm}\) to \(3 \mathrm{~cm}\) will be (A) \(20 \mathrm{~min}\) (B) less than \(20 \mathrm{~min}\) (C) greater than \(40 \mathrm{~min}\) (D) less than \(40 \mathrm{~min}\) and greater than \(20 \mathrm{~min}\)

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The rate of flow of thermal current depends on the nature of material (thermal conductivity), cross-sectional area \(A\), and temperature gradient. More the temperature difference, higher the thermal current flow. This fact identifies the thermal resistance offered by the material while conducting heat. One can find by equivalent resistance in heat flow using the same principles as for current. A body may transfer energy better by radiation. The nature of radiating surfaces play a role in the power radiated from it. On covering a surface by non-conducting/radiating media, the loss of the heat energy can be controlled. A rod of length \(l\) and conductivity \(k\) is placed between two reservoirs maintained at \(0^{\circ} \mathrm{C}\) and \(100^{\circ} \mathrm{C}\). At a distance \(\frac{l}{3}\) and \(\frac{l}{5}\) from \(100^{\circ} \mathrm{C}\) reservoir the rate of flow of thermal current are \(I_{1}\) and \(I_{2} .\) The temperatures at \(\frac{l}{5}\) and \(\frac{l}{3}\) are \(\theta_{1}\) and \(\theta_{2}\) then (A) \(I_{1}=I_{2}, 2 \theta_{1}=3 \theta_{2}\) (B) \(5 \theta_{1}-3 \theta_{2}=200\) (C) \(I_{1}=I_{2}, 5 \theta_{1}-3 \theta_{2}=200\) (D) \(I_{1} \neq I_{2}, \quad 5 \theta_{2}=3 \theta_{1}\)
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