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Chapter 1: Problem 85

Consider a person whose exposed surface area is \(1.7 \mathrm{~m}^{2}\), emissivity is \(0.5\), and surface temperature is \(32^{\circ} \mathrm{C}\). Determine the rate of heat loss from that person by radiation in a large room having walls at a temperature of \((a) 300 \mathrm{~K}\) and (b) \(280 \mathrm{~K}\). Answers: (a) \(26.7 \mathrm{~W}\), (b) \(121 \mathrm{~W}\)

Short Answer

Expert verified
Question: Calculate the rate of heat loss from a person by radiation for two different temperatures of surrounding room walls, with given exposed surface area, emissivity, and surface temperature. Answer: The rate of heat loss from the person by radiation in a large room having walls at a temperature of (a) 300 K is 26.7 W, and (b) 280 K is 121 W.

Step by step solution

01

Calculate surface temperature in Kelvin

We are given the surface temperature \(32^{\circ} \mathrm{C}\). First, convert this temperature to Kelvin by adding \(273.15\): \(T_{surface} = 32 + 273.15 = 305.15 \mathrm{~K}\)
02

Calculate the heat loss rate due to radiation using Stefan-Boltzmann Law

Stefan-Boltzmann Law states that the rate of heat transfer by radiation is: \(Q = e \cdot s \cdot A (T_{surface}^4 - T_{room}^4)\) where \(e\) is the emissivity, \(s\) is the Stefan-Boltzmann constant (\(5.67 \cdot 10^{-8} \mathrm{~W/m^2K^4}\)), \(A\) is the person's exposed surface area, \(T_{surface}\) is the surface temperature in Kelvin, and \(T_{room}\) is the room walls' temperature in Kelvin. Now, we will apply this formula to calculate the rate of heat loss for both cases: (a) For a room with walls' temperature \(300 \mathrm{~K}\): \(Q_{a} = (0.5) \cdot (5.67 \cdot 10^{-8}) \cdot (1.7) \cdot (305.15^4 - 300^4)\) \(Q_{a} \approx 26.7 \mathrm{~W}\) (b) For a room with walls' temperature \(280 \mathrm{~K}\): \(Q_{b} = (0.5) \cdot (5.67 \cdot 10^{-8}) \cdot (1.7) \cdot (305.15^4 - 280^4)\) \(Q_{b} \approx 121 \mathrm{~W}\) Therefore, the rate of heat loss from the person by radiation in a large room having walls at a temperature of (a) \(300 \mathrm{~K}\) is \(26.7 \mathrm{~W}\), and (b) \(280 \mathrm{~K}\) is \(121 \mathrm{~W}\).

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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 a fundamental principle in thermodynamics which describes the power radiated from a black body in terms of its temperature. Specifically, it establishes that the total energy radiated per unit surface area of a black body across all wavelengths per unit time (also known as the black-body irradiance or emissive power) is directly proportional to the fourth power of the black body's temperature.

The law is mathematically represented by the equation: \[ Q = e \cdot s \cdot A \cdot (T_{surface}^4 - T_{room}^4) \] where:\
  • \( Q \) is the total heat loss due to radiation,
  • \( e \) is the emissivity of the material (a dimensionless coefficient),
  • \( s \) is the Stefan-Boltzmann constant (\( 5.67 \times 10^{-8} \mathrm{~W/m^2K^4} \)),
  • \( A \) is the surface area of the radiating body,
  • \( T_{surface} \) is the surface temperature in Kelvin,
  • \( T_{room} \) is the ambient temperature around the object, also in Kelvin.
The Stefan-Boltzmann Law helps us understand that objects at a higher temperature emit more thermal radiation compared to objects at a lower temperature.
Surface Temperature
Surface temperature refers to the temperature of an object's surface and plays a crucial role in calculations regarding thermal radiation. To apply the Stefan-Boltzmann Law, surface temperature must be measured in Kelvin, a thermodynamic temperature scale.

In the context of the given problem, the individual's skin temperature was initially measured in degrees Celsius, \(32^\circ \mathrm{C}\), and converted to Kelvin by adding \(273.15\), resulting in \(305.15 \mathrm{~K}\).

This conversion is essential for applying the Stefan-Boltzmann Law because the law is formulated in terms of absolute temperature, which is best represented on the Kelvin scale where zero denotes the absence of thermal energy.
Emissivity
Emissivity is a measure of the efficiency with which a surface emits thermal radiation and is defined as the ratio of radiation emitted by a surface to the radiation emitted by a black body at the same temperature. It鈥檚 a unitless value that ranges from 0 to 1 - a perfect black body, which is an ideal emitter and absorber of radiation, has an emissivity of 1, while real-world objects have emissivities less than 1.

In our exercise, the person鈥檚 skin has an emissivity of \(0.5\), indicating that the skin radiates heat at half the efficiency of a perfect black body. This factor significantly affects the rate of heat loss through radiation, as seen in the calculation steps for determining the heat lost by the person in various room temperatures.
Thermal Radiation
Thermal radiation is the process by which energy is emitted from a surface in the form of electromagnetic waves due to the surface鈥檚 temperature. This form of heat transfer does not require a medium; it can occur through a vacuum, making it distinct from conduction and convection which require a medium to transfer heat.

Objects at all temperatures emit thermal radiation and, as described by the Stefan-Boltzmann Law, the amount of this radiation depends on the fourth power of the temperature of the object. This principle plays an instrumental role in understanding how objects cool down or warm up through energy emission, including the human body, as illustrated in the sample problem where a person's heat loss due to thermal radiation is calculated.

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Most popular questions from this chapter

Solar radiation is incident on a \(5 \mathrm{~m}^{2}\) solar absorber plate surface at a rate of \(800 \mathrm{~W} / \mathrm{m}^{2}\). Ninety-three percent of the solar radiation is absorbed by the absorber plate, while the remaining 7 percent is reflected away. The solar absorber plate has a surface temperature of \(40^{\circ} \mathrm{C}\) with an emissivity of \(0.9\) that experiences radiation exchange with the surrounding temperature of \(-5^{\circ} \mathrm{C}\). In addition, convective heat transfer occurs between the absorber plate surface and the ambient air of \(20^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(7 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the efficiency of the solar absorber, which is defined as the ratio of the usable heat collected by the absorber to the incident solar radiation on the absorber.

In a power plant, pipes transporting superheated vapor are very common. Superheated vapor is flowing at a rate of \(0.3 \mathrm{~kg} / \mathrm{s}\) inside a pipe with \(5 \mathrm{~cm}\) in diameter and \(10 \mathrm{~m}\) in length. The pipe is located in a power plant at \(20^{\circ} \mathrm{C}\), and has a uniform pipe surface temperature of \(100^{\circ} \mathrm{C}\). If the temperature drop between the inlet and exit of the pipe is \(30^{\circ} \mathrm{C}\), and the specific heat of the vapor is \(2190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), determine the heat transfer coefficient as a result of convection between the pipe surface and the surrounding.

One way of measuring the thermal conductivity of a material is to sandwich an electric thermofoil heater between two identical rectangular samples of the material and to heavily insulate the four outer edges, as shown in the figure. Thermocouples attached to the inner and outer surfaces of the samples record the temperatures. During an experiment, two \(0.5-\mathrm{cm}\) thick samples \(10 \mathrm{~cm} \times\) \(10 \mathrm{~cm}\) in size are used. When steady operation is reached, the heater is observed to draw \(25 \mathrm{~W}\) of electric power, and the temperature of each sample is observed to drop from \(82^{\circ} \mathrm{C}\) at the inner surface to \(74^{\circ} \mathrm{C}\) at the outer surface. Determine the thermal conductivity of the material at the average temperature.

A concrete wall with a surface area of \(20 \mathrm{~m}^{2}\) and a thickness of \(0.30 \mathrm{~m}\) separates conditioned room air from ambient air. The temperature of the inner surface of the wall \(\left(T_{1}\right)\) is maintained at \(25^{\circ} \mathrm{C}\). (a) Determine the heat loss \(\dot{Q}(\mathrm{~W})\) through the concrete wall for three thermal conductivity values of \((0.75,1\), and \(1.25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and outer wall surface temperatures of \(T_{2}=-15,-10,-5,0,5,10,15,20,25,30\), and \(38^{\circ} \mathrm{C}\) (a total of 11 data points for each thermal conductivity value). Tabulate the results for all three cases in one table. Also provide a computer generated graph [Heat loss, \(\dot{Q}(\mathrm{~W})\) vs. Outside wall temperature, \(\left.T_{2}\left({ }^{\circ} \mathrm{C}\right)\right]\) for the display of your results. The results for all three cases should be plotted on the same graph. (b) Discuss your results for the three cases.

A 2-kW electric resistance heater submerged in 30-kg water is turned on and kept on for \(10 \mathrm{~min}\). During the process, \(500 \mathrm{~kJ}\) of heat is lost from the water. The temperature rise of water is (a) \(5.6^{\circ} \mathrm{C}\) (b) \(9.6^{\circ} \mathrm{C}\) (c) \(13.6^{\circ} \mathrm{C}\) (d) \(23.3^{\circ} \mathrm{C}\) (e) \(42.5^{\circ} \mathrm{C}\)
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