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Chapter 18: Problem 51

A sphere of radius \(0.500 \mathrm{~m}\), temperature \(27.0^{\circ} \mathrm{C}\), and emissivity \(0.850\) is located in an environment of temperature \(77.0^{\circ} \mathrm{C}\). At what rate does the sphere (a) emit and (b) absorb thermal radiation? (c) What is the sphere's net rate of energy exchange?

Short Answer

Expert verified
The sphere emits at 459 W, absorbs 886.9 W, and the net exchange is 427.9 W.

Step by step solution

01

Gather Known Values

We begin by identifying the known values in the problem:- Radius of the sphere, \( r = 0.500 \, \text{m} \).- Temperature of the sphere, \( T_s = 27.0^{\circ} \text{C} = 300.15 \, \text{K} \).- Emissivity of the sphere, \( \epsilon = 0.850 \).- Temperature of the environment, \( T_e = 77.0^{\circ} \text{C} = 350.15 \, \text{K} \).- Stefan-Boltzmann constant, \( \sigma = 5.67 \times 10^{-8} \, \text{W/m}^2 \cdot \text{K}^4 \).
02

Calculate Surface Area of the Sphere

The surface area \( A \) of a sphere is calculated using the formula:\[ A = 4 \pi r^2 \]Substituting the radius:\[ A = 4 \pi (0.500)^2 = 3.1416 \, \text{m}^2 \]
03

Calculate the Rate of Emission

The rate of emission \( P_{emit} \) by the sphere is given by the formula:\[ P_{emit} = \epsilon \cdot \sigma \cdot A \cdot T_s^4 \]Substitute the known values:\[ P_{emit} = 0.850 \times 5.67 \times 10^{-8} \times 3.1416 \times (300.15)^4 \]Calculate the expression to find \( P_{emit} \approx 459.0 \, \text{W} \).
04

Calculate the Rate of Absorption

The rate of absorption \( P_{absorb} \) by the sphere from the environment is:\[ P_{absorb} = \epsilon \cdot \sigma \cdot A \cdot T_e^4 \]Substitute the values:\[ P_{absorb} = 0.850 \times 5.67 \times 10^{-8} \times 3.1416 \times (350.15)^4 \]Calculate \( P_{absorb} \approx 886.9 \, \text{W} \).
05

Calculate the Net Rate of Energy Exchange

The net rate of energy exchange \( P_{net} \) is the difference between the rate of absorption and emission:\[ P_{net} = P_{absorb} - P_{emit} \]Substitute the calculated rates:\[ P_{net} = 886.9 - 459.0 = 427.9 \, \text{W} \]
06

State the Final Results

- The rate at which the sphere emits thermal radiation is approximately \( 459.0 \, \text{W} \).- The rate at which the sphere absorbs thermal radiation is approximately \( 886.9 \, \text{W} \).- The net rate of energy exchange is approximately \( 427.9 \, \text{W} \).

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Key Concepts

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

Emissivity
Emissivity is a measure of how efficiently a surface emits thermal radiation compared to an ideal black body. It ranges from 0 to 1, with 1 being a perfect emitter like a black body, and 0 indicating no emission at all. A surface with high emissivity is better at radiating heat. In our case, the sphere has an emissivity of 0.850, which means it is quite efficient at emitting radiation, though not perfectly so.

Understanding emissivity is essential in calculations involving thermal radiation. The concept plays a crucial role in determining how much energy an object emits or absorbs from its surroundings, leading us to the next principle, the Stefan-Boltzmann Law.
Stefan-Boltzmann Law
The Stefan-Boltzmann Law describes how much thermal radiation is emitted by a black body in terms of its temperature. The law states that the power radiated by an object is proportional to the fourth power of its temperature, given by the formula:
\[ P = \ \ \ epsilon \ \ \ \ sigma \ \ \ \ A \ \ \ \ T^4 \]Where:
  • \( P \) is the power radiated (in Watts).
  • \( epsilon \) is the object's emissivity.
  • \( sigma \) is the Stefan-Boltzmann constant \( (5.67 \ \times \ 10^{-8} \ \text{W/m}^2 \ \cdot \ \text{K}^4) \).
  • \( A \) is the surface area of the object.
  • \( T \) is the absolute temperature in Kelvin.
In our exercise, the sphere's ability to emit radiation (emission rate) was calculated using this law. By inserting the known values, we determined both the emission and absorption rate of the thermal radiation using the sphere's temperature and the temperature of its surrounding environment.
Net Energy Exchange
Net energy exchange considers both the energy emitted by the object and the energy absorbed from its surroundings. It is calculated by determining the difference between the power absorbed and emitted. This difference reveals whether the object is gaining or losing energy over time.

In the sphere's case, we calculated that it absorbs approximately 886.9 Watts from the environment while it emits about 459.0 Watts. To find the net rate of energy exchange, we subtract the emission rate from the absorption rate:
\[ P_{net} = P_{absorb} - P_{emit} \] This results in a net energy exchange rate of 427.9 Watts, showing that the sphere is absorbing more heat than it emits. Over time, this will affect the sphere's temperature unless it reaches equilibrium with its environment. Understanding the concept of net energy exchange is pivotal for recognizing an object's thermal dynamics and how it interacts with its environment.

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Most popular questions from this chapter

co Ethyl alcohol has a boiling point of \(78.0^{\circ} \mathrm{C}\), a freezing point of \(-114^{\circ} \mathrm{C}\), a heat of vaporization of \(879 \mathrm{~kJ} / \mathrm{kg}\), a heat of fusion of \(109 \mathrm{~kJ} / \mathrm{kg}\), and a specific heat of \(2.43 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\). How much energy must be removed from \(0.510 \mathrm{~kg}\) of ethyl alcohol that is initially a gas at \(78.0^{\circ} \mathrm{C}\) so that it becomes a solid at \(-114^{\circ} \mathrm{C} ?\)

\(.\) Penguin huddling. To withstand the harsh weather of the Antarctic, emperor penguins huddle in groups (Fig. \(18-49)\). Assume that a penguin is a circular cylinder with a top surface area \(a=0.34 \mathrm{~m}^{2}\) and height \(h=1.1 \mathrm{~m}\). Let \(P_{r}\) be the rate at which an individual penguin radiates energy to the environment (through the top and the sides); thus \(N P_{r}\) is the rate at which \(N\) identical, well-separated penguins radiate. If the penguins huddle closely to form a huddled cylinder with top surface area \(N a\) and height \(h\), the cylinder radiates at the rate \(P_{h}\). If \(N=1000,(\) a) what is the value of the fraction \(P_{h} / N P_{r}\) and (b) by what percentage does huddling reduce the total radiation loss?

(a) Calculate the rate at which body heat is conducted through the clothing of a skier in a steady-state process, given the following data: the body surface area is \(1.8 \mathrm{~m}^{2}\), and the clothing is \(1.0 \mathrm{~cm}\) thick; the skin surface temperature is \(33^{\circ} \mathrm{C}\) and the outer surface of the clothing is at \(1.0^{\circ} \mathrm{C} ;\) the thermal conductivity of the clothing is \(0.040\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K} .\) (b) If, after a fall, the skier's clothes became soaked with water of thermal conductivity \(0.60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), by how much is the rate of conduction multiplied?

\(\$$ An insulated Thermos contains \)130 \mathrm{~cm}^{3}\( of hot coffee at \)80.0^{\circ} \mathrm{C}\(. You put in a \)12.0 \mathrm{~g}$ ice cube at its melting point to cool the coffee. By how many degrees has your coffee cooled once the ice has melted and equilibrium is reached? Treat the coffee as though it were pure water and neglect energy exchanges with the environment.

(c) Evaporative cooling of beverages. A cold beverage can be kept cold even on a warm day if it is slipped into a porous ceramic container that has been soaked in water. Assume that energy lost to evaporation matches the net energy gained via the radiation exchange through the top and side surfaces. The container and beverage have temperature \(T=15^{\circ} \mathrm{C}\), the environment has temperature \(T_{\text {env }}=32^{\circ} \mathrm{C}\), and the container is a cylinder with radius \(r=2.2 \mathrm{~cm}\) and height \(10 \mathrm{~cm}\). Approximate the emissivity as \(\varepsilon=1\), and neglect other energy exchanges. At what rate \(d m / d t\) is the container losing water mass?
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