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Chapter 2: Problem 54

A body whose surface area is \(0.5 \mathrm{~m}^{2}\), emissivity is \(0.8\), and temperature is \(150^{\circ} \mathrm{C}\) is placed in a large, evacuated chamber whose walls are at \(25^{\circ} \mathrm{C}\). What is the rate at which radiation is emitted by the surface, in W? What is the net rate at which radiation is exchanged between the surface and the chamber walls, in W?

Short Answer

Expert verified
The rate at which radiation is emitted by the surface is 72.52 W. The net rate at which radiation is exchanged between the surface and the chamber walls is 50 W.

Step by step solution

01

- Convert temperatures to Kelvin

Convert the given temperatures from degrees Celsius to Kelvin. Use the formula: \text{temperature in Kelvin} = \text{temperature in Celsius} + 273.15 For the body: \(150^{\circ} \mathrm{C} = 150 + 273.15 = 423.15 \mathrm{K}\) For the chamber walls: \(25^{\circ} \mathrm{C} = 25 + 273.15 = 298.15 \mathrm{K}\)
02

- Use Stefan-Boltzmann law to find rate of radiation emission

The Stefan-Boltzmann law is given by: \(E = \epsilon \sigma A T^4\) Where: \(E\): Power radiated per unit area (\(W\)) \(\epsilon\): Emissivity of the surface (0.8) \(\sigma\): Stefan-Boltzmann constant \((5.67 \times 10^{-8} \mathrm{W} \mathrm{m}^{-2} \mathrm{K}^{-4})\) \(A\): Surface area \((0.5 \mathrm{~m}^2)\) \(T\): Temperature in Kelvin (423.15 K) Plug in the values: \(E = 0.8 \times 5.67 \times 10^{-8} \times 0.5 \times (423.15)^4\) Calculate the power: \(E \approx 0.8 \times 5.67 \times 10^{-8} \times 0.5 \times 3.19 \times 10^9 \approx 72.52 \mathrm{W}\)
03

- Use Stefan-Boltzmann law for the chamber walls

Calculate the power radiated by the chamber walls: \(E_{walls} = \sigma A T_{walls}^4\) \(E_{walls} = 1 \times 5.67 \times 10^{-8} \times 0.5 \times (298.15)^4\) \(E_{walls} \approx 5.67 \times 10^{-8} \times 0.5 \times 7.94 \times 10^8 \approx 22.52 \mathrm{W}\)
04

- Compute the net radiation exchange rate

To find the net rate of radiation exchange between the surface and the chamber walls: \(E_{net} = E_{body} - E_{walls}\) \(E_{net} = 72.52 \mathrm{W} - 22.52 \mathrm{W} = 50 \mathrm{W}\)

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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 thermal radiation, describing how much energy a body emits based on its temperature. The formula is given by: \[ E = \epsilon \sigma A T^4 \]Here, \(E\) represents the power emitted per unit area in watts (W). The symbol \(\epsilon\) stands for emissivity, which indicates how effective the surface is at emitting energy compared to a perfect blackbody. \(\sigma\) is the Stefan-Boltzmann constant with a value of \(5.67 \times 10^{-8} \mathrm{W} \mathrm{m}^{-2} \mathrm{K}^{-4}\). \(A\) is the surface area, and \(T\) is the temperature in Kelvin. This law plays a critical role in understanding energy transfer by radiation, allowing us to quantify the power emitted based on temperature, surface area, and material properties.
emissivity
Emissivity \(\left(\epsilon\right\) is a material property that ranges from 0 to 1 and indicates how efficiently a surface emits thermal radiation compared to an ideal blackbody, which has an emissivity of 1.
  • An emissivity of 0 means the body does not emit any thermal radiation.
  • An emissivity of 1 means the body is a perfect emitter of thermal radiation.
For example, in the exercise, the body has an emissivity of 0.8. This means it emits 80% of the thermal radiation that an ideal blackbody would emit at the same temperature.
thermal radiation
Thermal radiation is the emission of electromagnetic waves from a body due to its temperature. All objects with a temperature above absolute zero emit thermal radiation. This process allows energy to be transferred through space without needing a medium, unlike conduction or convection.
  • Higher temperatures result in more emitted radiation.
  • The nature of the surface (e.g., color, texture) also affects the emission.
Radiation is crucial in various scientific fields, such as understanding Earth's energy balance, designing thermal management systems in engineering, and studying astronomical phenomena.
temperature conversion
Converting temperatures between different scales is essential in many scientific calculations. The two commonly used temperature scales are Celsius (掳C) and Kelvin (K). The relationships between these scales are straightforward: \[ \text{Temperature in Kelvin} = \text{Temperature in Celsius} + 273.15 \]For example, in our problem, we converted 150掳C to Kelvin: \[ 150 \degree \mathrm{C} + 273.15 = 423.15 \mathrm{K} \]Similarly, we converted 25掳C to Kelvin: \[ 25 \degree \mathrm{C} + 273.15 = 298.15 \mathrm{K} \]This conversion is necessary because the Stefan-Boltzmann law and most thermodynamic equations require temperature inputs in Kelvin.

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Most popular questions from this chapter

Beginning from rest, an object of mass \(200 \mathrm{~kg}\) slides down a 10-m-long ramp. The ramp is inclined at an angle of \(40^{\circ}\) from the horizontal. If air resistance and friction between the object and the ramp are negligible, determine the velocity of the object, in \(\mathrm{m} / \mathrm{s}\), at the bottom of the ramp. Let \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\).

At steady state, a spherical interplanetary electronicsladen probe having a diameter of \(0.5 \mathrm{~m}\) transfers energy by radiation from its outer surface at a rate of \(150 \mathrm{~W}\). If the probe does not receive radiation from the sun or deep space, what is the surface temperature, in \(\mathrm{K}\) ? Let \(\varepsilon=0.8\).

A composite plane wall consists of a 12 -in.-thick layer of insulating concrete block \(\left(\kappa_{\mathrm{c}}=0.27 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{R}\right)\) and a \(0.625\)-in.-thick layer of gypsum board \(\left(\kappa_{\mathrm{b}}=1.11 \mathrm{Btu} / \mathrm{h}\right.\) \(\mathrm{ft} \cdot{ }^{\circ} \mathrm{R}\) ). The outer surface temperature of the concrete block and gypsum board are \(460^{\circ} \mathrm{R}\) and \(560^{\circ} \mathrm{R}\), respectively, and there is perfect contact at the interface between the two layers. Determine at steady state the instantaneous rate of heat transfer, in \(\mathrm{Btu} / \mathrm{h}\) per \(\mathrm{ft}^{2}\) of surface area, and the temperature, in \({ }^{\circ} \mathrm{R}\), at the interface between the concrete block and gypsum board.

A gas undergoes a process in a piston-cylinder assembly during which the pressure-specific volume relation is \(p v^{1.2}=\) constant. The mass of the gas is \(0.4 \mathrm{lb}\) and the following data are known: \(p_{1}=160 \mathrm{lbf} / \mathrm{in}^{2}, V_{1}=1 \mathrm{ft}^{3}\), and \(p_{2}=\) \(390 \mathrm{lbf} /\) in. \({ }^{2}\) During the process, heat transfer from the gas is \(2.1\) Btu. Kinetic and potential energy effects are negligible. Determine the change in specific internal energy of the gas, in Btu/lb.

A heat pump cycle operating at steady state receives energy by heat transfer from well water at \(10^{\circ} \mathrm{C}\) and discharges energy by heat transfer to a building at the rate of \(1.2 \times 10^{5} \mathrm{~kJ} / \mathrm{h}\). Over a period of 14 days, an electric meter records that \(1490 \mathrm{~kW}\) - \(\mathrm{h}\) of electricity is provided to the heat pump. These are the only energy transfers involved. Determine (a) the amount of energy that the heat pump receives over the 14-day period from the well water by heat transfer, in \(\mathrm{kJ}\), and (b) the heat pump's coefficient of performance.
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