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Chapter 18: Problem 93

Suppose that you intercept \(5.0 \times 10^{-3}\) of the energy radiated by a hot sphere that has a radius of \(0.020 \mathrm{~m}\), an emissivity of \(0.80\), and a surface temperature of \(500 \mathrm{~K}\). How much energy do you intercept in \(2.0 \mathrm{~min} ?\)

Short Answer

Expert verified
The energy intercepted in 2 minutes is 6.822 J.

Step by step solution

01

Understand the Problem

We need to calculate the energy intercepted over 2 minutes from a sphere with a given radius, emissivity, and temperature, while accounting for only 0.005 of the radiated energy being intercepted.
02

Use the Stefan-Boltzmann Law

The power radiated by a blackbody is given by the Stefan-Boltzmann law: \( P = e imes \sigma imes A imes T^4 \), where \( e \) is the 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 sphere, and \( T \) is the temperature in Kelvin.
03

Calculate the Surface Area of the Sphere

The surface area \( A \) of a sphere is given by \( A = 4 \pi r^2 \), where \( r \) is the radius. Substituting \( r = 0.020 \, \text{m} \), we get: \( A = 4 \pi (0.020)^2 = 5.0265493 \times 10^{-3} \, \text{m}^2 \).
04

Calculate the Total Power Radiated

Substitute the known values into the Stefan-Boltzmann equation: \( P = 0.80 \times 5.67 \times 10^{-8} \times 5.0265493 \times 10^{-3} \times 500^4 \). Simplifying, we get: \( P = 11.37 \, \text{W} \).
05

Determine the Energy Radiated in 2 Minutes

Power is energy per unit time. Thus, the energy \( E \) radiated in 2 minutes \( (120 \, ext{s}) \) is \( E = P \times t = 11.37 \, \text{W} \times 120 \, ext{s} = 1364.4 \, \text{J} \).
06

Calculate the Intercepted Energy

Since only \( 5.0 \times 10^{-3} \) of the energy is intercepted, the intercepted energy \( E_{\text{intercepted}} \) is \( E_{\text{intercepted}} = 5.0 \times 10^{-3} \times 1364.4 \, \text{J} \). So, \( E_{\text{intercepted}} = 6.822 \, \text{J} \).

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Key Concepts

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

Understanding Emissivity
Emissivity is a measure of how efficiently a surface emits thermal radiation compared to an ideal blackbody, which has an emissivity of 1. It ranges from 0 to 1, where 1 indicates perfect thermal emission, while 0 means no emission. For the sphere in our problem, the emissivity is 0.80. This suggests that the sphere emits 80% of the thermal radiation that a perfect blackbody would emit at the same temperature.

In essence, emissivity affects the amount of radiation a body can emit. The formula used in the Stefan-Boltzmann Law incorporates emissivity to adjust the theoretical maximum emission rate (of a blackbody) to real-life applications. Therefore, even if two objects have the same temperature and surface area, their emitted energies can differ due to variations in emissivity. This is crucial for accurately estimating energy outputs in physics and engineering applications.
Surface Area Calculation
To understand the amount of energy a surface emits, we must first know the surface area of the object in question. This is because the Stefan-Boltzmann law includes surface area as a key factor in determining total power emission.

For our problem, the sphere's surface area is calculated using the formula for the surface area of a sphere: \[ A = 4 \pi r^2 \] Given the radius \( r = 0.020 \text{m} \), we calculate: \[ A = 4 \pi (0.020)^2 \approx 5.03 \times 10^{-3} \text{ m}^2 \] This tells us how much surface is available to emit radiation. Understanding surface area is crucial because it directly influences the rate of energy emission. A larger surface area, all else being equal, results in more energy being emitted. Knowing how to calculate it ensures accurate estimations in problems involving thermal radiation.
Energy Interception
Energy interception is the process of capturing a portion of the energy emitted by a radiating body. In many real-world scenarios, we do not harness the entirety of emitted energy due to various constraints like system inefficiencies or geometric positioning.

In our exercise, only \( 5.0 \times 10^{-3} \) or 0.5% of the sphere's energy is intercepted. This means for every 1000 units of energy the sphere emits, only 5 units are actually captured. To find the intercepted energy:\[ E_{\text{intercepted}} = \text{fraction captured} \times E_{\text{total emitted}} \]When the total emitted energy over 2 minutes is 1364.4 J, the intercepted energy is:\[ E_{\text{intercepted}} = 0.005 \times 1364.4 = 6.822 \, \text{J} \] This concept is vital as it affects energy efficiency calculations in devices that rely on energy capture from radiation sources such as solar panels or radiative heaters.

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Most popular questions from this chapter

On finding your stove out of order, you decide to boil the water for a cup of tea by shaking it in a thermos flask. Suppose that you use tap water at \(19^{\circ} \mathrm{C}\), the water falls \(32 \mathrm{~cm}\) each shake, and you make 27 shakes each minute. Neglecting any loss of thermal energy by the flask, how long (in minutes) must you shake the flask until the water reaches \(100^{\circ} \mathrm{C}\) ?

Suppose the temperature of a gas is \(373.15 \mathrm{~K}\) when it is at the boiling point of water. What then is the limiting value of the ratio of the pressure of the gas at that boiling point to its pressure at the triple point of water? (Assume the volume of the gas is the same at both temperatures.)

Leidenfrost effect. \(\mathrm{A}\) water drop will last about \(1 \mathrm{~s}\) on a hot skillet with a temperature between \(100^{\circ} \mathrm{C}\) and about \(200^{\circ} \mathrm{C}\). However, if the skillet is much hotter, the drop can last several minutes, an effect named after an early investigator. The longer lifetime is due to the support of a thin layer of air and water vapor that separates the drop from the metal (by distance \(L\) in Fig. \(18-48)\). Let \(L=\) \(0.100 \mathrm{~mm}\), and assume that the drop is flat with height \(h=1.50 \mathrm{~mm}\) and bottom face area \(A=4.00 \times 10^{-6} \mathrm{~m}^{2}\). Also assume that the skillet has a constant temperature \(T_{s}=300^{\circ} \mathrm{C}\) and the drop has a temperature of \(100^{\circ} \mathrm{C}\). Water has density \(\rho=1000 \mathrm{~kg} / \mathrm{m}^{3}\), and the supporting layer has thermal conductivity \(k=0.026 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} .\) (a) At what rate is energy conducted from the skillet to the drop through the drop's bottom surface? (b) If conduction is the primary way energy moves from the skillet to the drop, how long will the drop last?

One way to keep the contents of a garage from becoming too cold on a night when a severe subfreezing temperature is forecast is to put a tub of water in the garage. If the mass of the water is \(125 \mathrm{~kg}\) and its initial temperature is \(20^{\circ} \mathrm{C},(\mathrm{a})\) how much energy must the water transfer to its surroundings in order to freeze completely and (b) what is the lowest possible temperature of the water and its surroundings until that happens?

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 \(T_{\mathrm{env}}=32^{\circ} \mathrm{C}\), 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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