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Chapter 17: Problem 86

The heat one feels when sitting near the fire in a fireplace or at a campfire is due almost entirely to thermal radiation. (a) Estimate the diameter and length of an average campfire log. (b) Compute the surface area of such a log. (c) Use the StefanBoltzmann law to determine the power emitted by thermal radiation by such a log when it burns at a typical temperature of \(700^{\circ} \mathrm{C}\) in a surrounding air temperature of \(20.0^{\circ} \mathrm{C}\). The emissivity of a burning log is close to unity.

Short Answer

Expert verified
The estimation of the log size is subjective, but let's assume it to be a cylinder with a diameter of 0.1 m and a length of 1 m. The surface area of the log is \(0.628 m^{2}\). The power emitted, computed by using the Stefan-Boltzmann law, is approximately \(9.34 kW\).

Step by step solution

01

Estimation of log size

Estimate the size of a typical campfire log. This estimation is subjective but, for purpose of mathematically modeling, let's assume the log to be a cylinder with a diameter of 0.1 m and a length of 1 m.
02

Calculation of surface area

Compute the surface area of the cylindrical log. The formula for the surface area of a cylinder (excluding the bases) is \(2 \pi r h\), where \(r\) is the radius and \(h\) is the height/length of the cylinder. Insert the previously estimated values for diameter and length into this equation to obtain the surface area.
03

Understand Stefan-Boltzmann law

Understand the concepts behind the Stefan-Boltzmann law, which states that the power radiated per unit area of a black body is proportional to the fourth power of its thermodynamic temperature. The law can be mathematically represented as \( P = \sigma e A (T_{\text{log}}^4 - T_{\text{air}}^4)\), where \( \sigma \) is the Stefan-Boltzmann constant, \(e\) is the emissivity, \(A\) is the area, and \(T_{\text{log}}\) and \(T_{\text{air}}\) are the temperatures of the log and the surrounding air respectively.
04

Calculation of power emitted

After understanding the law, calculate the power emitted by plugging in the values into the Stefan-Boltzmann law equation. The temperatures have to be in Kelvin for the formula to work. The temperature in Celsius is converted to Kelvin by adding 273.15. Also, given that the emissivity of the log is close to unity, it can be considered as 1 for the calculation. The value of the Stefan-Boltzmann constant (\( \sigma \)) is \(5.67 \times 10^{-8} W m^{-2} K^{-4}\).

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Key Concepts

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

Stefan-Boltzmann Law
The Stefan-Boltzmann Law is essential for understanding how objects emit thermal radiation. This law describes the power radiated from a black body in terms of its temperature. It states that the power radiated per unit area is proportional to the fourth power of the object's thermodynamic temperature. This is mathematically represented by the equation:\[P = \sigma e A (T_{\text{log}}^4 - T_{\text{air}}^4)\]Here,
  • \( P \) is the power emitted.
  • \( \sigma \) is the Stefan-Boltzmann constant, valued at \( 5.67 \times 10^{-8} \, W m^{-2} K^{-4} \).
  • \( e \) represents the emissivity, which measures how efficiently an object radiates energy compared to a perfect black body.
  • \( A \) is the surface area of the object.
  • \( T_{\text{log}} \) and \( T_{\text{air}} \) are the temperatures of the log and the surrounding air, respectively, in Kelvin.
This law is powerful for calculating the thermal power output of objects like a burning campfire log, helping us understand how much energy it radiates into its surroundings.
Surface Area Calculation
To determine the surface area of a cylindrical object such as a campfire log, you need to use the formula for the lateral surface area of a cylinder. A cylinder is characterized by its radius \( r \) and height \( h \) (or length in this context), and its lateral surface area is given by:\[2 \pi r h\]This formula calculates only the curved surface area, excluding the circular ends of the cylinder. For practical purposes, especially when focusing on thermal radiation emitted over the side of a log, this part is often the most relevant.
  • For a log with a diameter of 0.1 m, the radius \( r \) would be half of that, or 0.05 m.
  • If the log's length (which serves as \( h \)) is 1 m, the surface area can be computed as:\[ 2 \pi \times 0.05 \times 1 = 0.314 \, m^2 \]
Understanding this calculation is fundamental, as the surface area directly influences the amount of thermal radiation the log can emit.
Emissivity
Emissivity is a measure of an object's ability to emit thermal radiation relative to a perfect black body. It ranges between 0 and 1, with 1 corresponding to an ideal black body, which emits the maximum radiation possible for its temperature.
  • An emissivity of 0 means the object does not emit thermal radiation at all.
  • An emissivity of 1 indicates maximum emission efficiency.
  • In practical scenarios, materials have varying emissivity values depending on their surface properties and temperature.
For a burning log, the emissivity is close to unity (1), indicating it emits almost all the thermal radiation possible for its temperature. When calculating the power emitted through the Stefan-Boltzmann Law, this high emissivity allows for a straightforward multiplication of the surface area with the Stefan-Boltzmann constant and the temperature difference in the formula, maximizing the accuracy of the obtained power emission rate.
Temperature Conversion
Temperature conversion is crucial when using the Stefan-Boltzmann Law, as the temperatures need to be in Kelvin, the SI unit for thermodynamic temperature. This is because the formula requires absolute temperatures and because the Kelvin scale allows for a direct application to the equation.
  • To convert from Celsius to Kelvin, you simply add 273.15 to the Celsius temperature.
  • For example, the typical burning temperature of the log is given as \(700^{\circ}C\), which converts to \[T_{\text{log}} = 700 + 273.15 = 973.15 \, K\]
  • Similarly, if the surrounding air temperature is \(20^{\circ}C\), it converts to \[T_{\text{air}} = 20 + 273.15 = 293.15 \, K\]
Understanding these conversions enables one to correctly apply the Stefan-Boltzmann Law without errors resulting from temperature unit discrepancies.

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Most popular questions from this chapter

A carpenter builds a solid wood door with dimensions \(2.00 \mathrm{~m} \times 0.95 \mathrm{~m} \times 5.0 \mathrm{~cm} .\) Its thermal conductivity is \(k=0.120 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The air films on the inner and outer surfaces of the door have the same combined thermal resistance as an additional \(1.8 \mathrm{~cm}\) thickness of solid wood. The inside air temperature is \(20.0^{\circ} \mathrm{C},\) and the outside air temperature is \(-8.0^{\circ} \mathrm{C}\). (a) What is the rate of heat flow through the door? (b) By what factor is the heat flow increased if a window \(0.500 \mathrm{~m}\) on a side is inserted in the door? The glass is \(0.450 \mathrm{~cm}\) thick, and the glass has a thermal conductivity of \(0.80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The air films on the two sides of the glass have a total thermal resistance that is the same as an additional \(12.0 \mathrm{~cm}\) of glass.

You have \(750 \mathrm{~g}\) of water at \(10.0^{\circ} \mathrm{C}\) in a large insulated beaker. How much boiling water at \(100.0^{\circ} \mathrm{C}\) must you add to this beaker so that the final temperature of the mixture will be \(75^{\circ} \mathrm{C}\) ?

There is \(0.050 \mathrm{~kg}\) of an unknown liquid in a plastic container of negligible mass. The liquid has a temperature of \(90.0^{\circ} \mathrm{C}\). To measure the specific heat capacity of the unknown liquid, you add a mass \(m_{w}\) of water that has a temperature of \(0.0^{\circ} \mathrm{C}\) to the liquid and measure the final temperature \(T\) after the system has reached thermal equilibrium. You repeat this measurement for several values of \(m_{\mathrm{w}},\) with the initial temperature of the unknown liquid always equal to \(90.0^{\circ} \mathrm{C}\). The plastic container is insulated, so no heat is exchanged with the surroundings. You plot your data as \(m_{\mathrm{w}}\) versus \(T^{-1}\), the inverse of the final temperature \(T\). Your data points lie close to a straight line that has slope \(2.15 \mathrm{~kg} \cdot \mathrm{C}^{\circ} .\) What does this result give for the value of the specific heat capacity of the unknown liquid?

A copper sphere with density \(8900 \mathrm{~kg} / \mathrm{m}^{3},\) radius \(5.00 \mathrm{~cm}\) and emissivity \(e=1.00\) sits on an insulated stand. The initial temperature of the sphere is \(300 \mathrm{~K}\). The surroundings are very cold, so the rate of absorption of heat by the sphere can be neglected. (a) How long does it take the sphere to cool by \(1.00 \mathrm{~K}\) due to its radiation of heat energy? Neglect the change in heat current as the temperature decreases. (b) To assess the accuracy of the approximation used in part (a), what is the fractional change in the heat current \(H\) when the temperature changes from \(300 \mathrm{~K}\) to \(299 \mathrm{~K} ?\)

A typical doughnut contains \(2.0 \mathrm{~g}\) of protein, \(17.0 \mathrm{~g}\) of carbohydrates, and \(7.0 \mathrm{~g}\) of fat. Average food energy values are \(4.0 \mathrm{kcal} / \mathrm{g}\) for protein and carbohydrates and \(9.0 \mathrm{kcal} / \mathrm{g}\) for fat. (a) During heavy exercise, an average person uses energy at a rate of \(510 \mathrm{kcal} / \mathrm{h}\). How long would you have to exercise to "work off" one doughnut? (b) If the energy in the doughnut could somehow be converted into the kinetic energy of your body as a whole, how fast could you move after eating the doughnut? Take your mass to be \(60 \mathrm{~kg}\), and express your answer in \(\mathrm{m} / \mathrm{s}\) and in \(\mathrm{km} / \mathrm{h}\).
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