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Chapter 19: Problem 7

A spherical body of \(2.0 \mathrm{~cm}\) diameter is maintained at \(600{ }^{\circ} \mathrm{C}\). Assuming that it radiates as if it were a blackbody, at what rate (in watts) is energy radiated from the sphere? $$ \begin{array}{l} A=\text { Surface area }=4 \pi r^{2}=4 \pi(0.01 \mathrm{~m})^{2}=1.26 \times 10^{-3} \mathrm{~m}^{2} \\ \mathrm{P}=A \sigma T^{4}=\left(1.26 \times 10^{-3} \mathrm{~m}^{2}\right)\left(5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)(873 \mathrm{~K})^{4}=41 \mathrm{~W} \end{array} $$

Short Answer

Expert verified
The energy radiated from the sphere is approximately 41 watts.

Step by step solution

01

Understand the Context

We have a spherical object with a given diameter, which we need to convert to radius to calculate the surface area. The object behaves as a perfect blackbody radiator, meaning it emits thermal radiation according to the Stefan-Boltzmann Law. We need to find the rate at which this body emits energy (power), given its temperature.
02

Convert Diameter to Radius

The sphere has a diameter of 2.0 cm. To use this in our calculations, we first convert it to meters (since the SI unit is meters) and then find the radius. Diameter = 2.0 cm = 0.02 m Radius ( r ) = Diameter / 2 = 0.02 m / 2 = 0.01 m.
03

Calculate the Surface Area

The formula for the surface area ( A ) of a sphere is given by: \[ A = 4 \pi r^2 \]Substitute the radius: \[ A = 4 \pi (0.01 \, \text{m})^2 = 1.26 \times 10^{-3} \, \text{m}^2 \]
04

Convert Temperature to Kelvin

Temperature is given in Celsius, so we need to convert it to Kelvin (K) to apply the Stefan-Boltzmann Law. The conversion formula is: \[ T(\text{in K}) = T(\text{in } ^{\circ}\text{C}) + 273.15 \] \[ T = 600^{\circ}\text{C} + 273.15 = 873 \text{ K} \]
05

Use Stefan-Boltzmann Law to Calculate Power

The Stefan-Boltzmann Law relates the power radiated by a blackbody to its temperature and surface area: \[ P = A \sigma T^4 \]Where \( \sigma \) is the Stefan-Boltzmann constant \( 5.67 \times 10^{-8} \, \text{W} / \text{m}^2 \cdot \text{K}^4 \).Substitute the known values: \[ P = (1.26 \times 10^{-3} \, \text{m}^2) \times (5.67 \times 10^{-8} \, \text{W} / \text{m}^2 \cdot \text{K}^4) \times (873 \, \text{K})^4 \]
06

Calculate the Result

Carrying out the multiplication: \[ P \approx 1.26 \times 10^{-3} \times 5.67 \times 10^{-8} \times 5.76 \times 10^{10} = 41 \text{ W} \]
07

Conclusion: Power Radiated

The sphere radiates energy at a rate of approximately 41 watts.

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Key Concepts

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

Blackbody Radiation
Blackbody radiation refers to the thermal electromagnetic radiation emitted by a body in thermal equilibrium. The concept of a blackbody is an idealized model used in physics to describe an object that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. Such a body also emits radiation uniformly in all directions, following Planck's law for blackbody radiation. A perfect blackbody is an efficient radiator and will emit energy at a rate directly proportional to the fourth power of the temperature of its surface, as stated by the Stefan-Boltzmann Law. This means that even slight temperature increases can lead to significant increases in energy emission. Understanding blackbody radiation is crucial in areas like astrophysics, where celestial objects are frequently modeled as blackbodies to estimate their temperature or emitted power by observing their radiation.
Surface Area of a Sphere
The calculation of the surface area of a sphere is essential in determining how much radiation a spherical object emits. The formula is given as \[ A = 4 \pi r^2 \] where \( r \) is the radius of the sphere. This formula arises from the geometric understanding that a sphere's surface area is related to the number of square units that can cover the entire outer surface of the spherical shape.In practical applications like this exercise, one needs to ensure that the units remain consistent. For instance, if the diameter of the sphere is given in centimeters, it must be converted to meters before calculating the surface area, as the Stefan-Boltzmann constant is typically expressed in meters.With the radius calculated from the diameter, you can effectively find the surface area, which forms a critical part of subsequent power calculations.
Temperature Conversion
Temperature conversion is a vital step when working with formulas that require a particular standard unit system. In scientific calculations, temperature is often required in Kelvin (K) rather than Celsius (掳C) since Kelvin is the SI unit for thermodynamic temperature. To convert from Celsius to Kelvin, the following formula applies:\[ T(\text{in K}) = T(\text{in } ^{\circ}\text{C}) + 273.15 \]This formula shifts the Celsius scale to make it absolute, beginning at absolute zero, which is 0 K, equivalent to -273.15 掳C. In this exercise, converting the sphere's temperature from 600掳C to Kelvin provides the correct dimensions to apply the Stefan-Boltzmann Law. The converted temperature of 873 K ensures the correct application in power calculations.
Power Calculation
Power calculation in blackbody radiation problems involves applying the Stefan-Boltzmann Law. The law is expressed as:\[ P = A \sigma T^4 \] Where:
  • \( P \) represents the power radiated in watts,
  • \( A \) is the surface area in square meters,
  • \( \sigma \) is the Stefan-Boltzmann constant \( 5.67 \times 10^{-8} \, \text{W} / \text{m}^2 \cdot \text{K}^4 \),
  • \( T \) is the absolute temperature in Kelvin.
Each of these values must be used in their correct scientific units to ensure an accurate calculation. In this specific problem, after calculating the surface area and converting the temperature, substituting into the formula provides the power. The step includes raising the temperature in Kelvin to the fourth power, which highlights the sensitivity of radiated power to temperature changes.By following these careful steps, we've determined that our sphere radiates energy at a rate of approximately 41 watts.

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Most popular questions from this chapter

An unclothed person whose body has a surface area of \(1.40 \mathrm{~m}^{2}\) with an emissivity of \(0.85\) has a skin temperature of \(37^{\circ} \mathrm{C}\) and stands in a \(20^{\circ} \mathrm{C}\) room. How much energy does the person lose through radiation per minute? Energy is power (P) multiplied by time \((\Delta t)\). From \(\mathrm{P}=\varepsilon A \sigma\left(T^{4}\right.\) \(-T_{e}^{4}\) ), the energy loss is $$ \varepsilon A \sigma\left(T^{4}-T_{e}^{4}\right) \Delta t=(0.85)\left(1.40 \mathrm{~m}^{2}\right)(\sigma)\left(T^{4}-T_{e}^{4}\right)(60 \mathrm{~s}) $$ Using \(\sigma=5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}^{4}, T=273+37=310 \mathrm{~K}\), and \(T_{e}=\) \(273+20=293 \mathrm{~K}\) results in an energy loss of $$ 7.6 \mathrm{~kJ}=1.8 \mathrm{kcal} $$

A copper tube (length, \(3.0 \mathrm{~m}\); inner diameter, \(1.500 \mathrm{~cm}\); outer diameter, \(1.700 \mathrm{~cm}\) ) extends across a 3.0-m-long vat of rapidly circulating water maintained at \(20^{\circ} \mathrm{C}\). Live steam at \(100^{\circ} \mathrm{C}\) passes through the tube. ( \(a\) ) What is the heat flow rate from the steam into the vat? ( \(b\) ) How much steam is condensed each minute? For copper, \(k_{L}=1.0 \mathrm{cal} / \mathrm{s} \cdot \mathrm{cm} \cdot{ }^{\circ} \mathrm{C}\) To determine the rate at which heat flows through the tube wall, approximate it as a flat sheet. Because the thickness of the tube is much smaller than its radius, the inner surface area of the tube, $$ 2 \pi r_{i} L=2 \pi(0.750 \mathrm{~cm})(300 \mathrm{~cm})=1410 \mathrm{~cm}^{2} $$ nearly equals its outer surface area, $$ 2 \pi r_{0} L=2 \pi(0.850 \mathrm{~cm})(300 \mathrm{~cm})=1600 \mathrm{~cm}^{2} $$ As an approximation, consider the tube to be a plate of thickness \(0.100 \mathrm{~cm}\) and area given by $$ A=\frac{1}{2}\left(1410 \mathrm{~cm}^{2}+1600 \mathrm{~cm}^{2}\right)=1500 \mathrm{~cm}^{2} $$ (a) \(\frac{\Delta Q}{\Delta t}=k_{T} A \frac{\Delta T}{L}=\left(1.0 \frac{\mathrm{cal}}{\mathrm{s} \cdot \mathrm{cm} \cdot{ }^{\circ} \mathrm{C}}\right) \frac{\left(1500 \mathrm{~cm}^{2}\right)\left(80^{\circ} \mathrm{C}\right)}{(0.100 \mathrm{~cm})}=1.2 \times 10^{6} \mathrm{cal} \mathrm{s} / \mathrm{s}\) (b) In one minute, the heat conducted from the tube is $$ \Delta Q=\left(1.2 \times 10^{6} \mathrm{cal} / \mathrm{s}\right)(60 \mathrm{~s})=72 \times 10^{6} \mathrm{cal} $$ It takes 540 cal to condense \(1.0 \mathrm{~g}\) of steam at \(100{ }^{\circ} \mathrm{C}\). Therefore, Steam condensed per min \(=\frac{72 \times 10^{6} \mathrm{cal}}{540 \mathrm{cal} / \mathrm{g}}=13.3 \times 10^{4} \mathrm{~g}=1.3 \times 10^{2} \mathrm{~kg}\) In practice, various factors would greatly reduce this theoretical value.

Suppose the area of a window on a passenger jet is doubled in size and all else is kept constant. Compared with the original window, what will happen to the rate at which heat passes out through the new window from the cabin?

A single-thickness glass window on a house actually has layers of stagnant air on its two surfaces. But if it did not, how much heat would flow out of an \(80-\mathrm{cm} \times 40-\mathrm{cm} \times 3.0\) -mm window each hour on a day when the outside temperature was precisely \(0{ }^{\circ} \mathrm{C}\) and the inside temperature was \(18^{\circ} \mathrm{C}\) ? For glass, \(k_{T}\) is \(0.84 \mathrm{~W} / \mathrm{k} \cdot \mathrm{m}\).

A beverage cooler is in the shape of a cube, \(42 \mathrm{~cm}\) on each inside edge. Its 3.0-cm-thick walls are made of plastic \(\left(k_{T}=0.050\right.\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) ). When the outside temperature is \(20^{\circ} \mathrm{C}\), how much ice will melt inside the cooler each hour? We have to determine the amount of heat conducted into the box. The cubical box has six sides, each with an area of about \((0.42\) \(\mathrm{m})^{2}\). From \(\Delta Q / \Delta t=k_{T} A \Delta T / L\), we have, with the ice inside at \(0{ }^{\circ} \mathrm{C}\), $$ \frac{\Delta Q}{\Delta t}=(0.050 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})(0.42 \mathrm{~m})^{2}(6)\left(\frac{20^{\circ} \mathrm{C}}{0.030 \mathrm{~m}}\right)=35.3 \mathrm{~J} / \mathrm{s}=8.43 \mathrm{cal} / \mathrm{s} $$ In one hour, \(\Delta Q=(60)^{2}(8.43)=30350\) cal. To melt \(1.0 \mathrm{~g}\) of ice requires 80 cal, so the mass of ice melted in one hour is $$ m=\frac{30350 \mathrm{cal}}{80 \mathrm{cal} / \mathrm{g}}=0.38 \mathrm{~kg} $$
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