/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 An enclosure has an inside area ... [FREE SOLUTION] | 91Ó°ÊÓ

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An enclosure has an inside area of \(100 \mathrm{~m}^{2}\), and its inside surface is black and is maintained at a constant temperature. A small opening in the enclosure has an area of \(0.02 \mathrm{~m}^{2}\). The radiant power emitted from this opening is \(70 \mathrm{~W}\). What is the temperature of the interior enclosure wall? If the interior surfice is maintained at this temperature, but is now polished. what will be the value of the radiant power emitted from the opening?

Short Answer

Expert verified
The interior wall temperature is approximately 368.5 K; polished surface emits about 7 W.

Step by step solution

01

Understand the Problem

We need to find the temperature of the enclosure's interior wall using given data: the radiant power from a small opening, the area of the opening, and the Stefan-Boltzmann law of black bodies. Then, determine the radiant power when the surface becomes polished.
02

Apply the Stefan-Boltzmann Law

The Stefan-Boltzmann law states that the power per unit area emitted by a black body is given by \( P = \sigma A T^4 \), where \( \sigma \) is the Stefan-Boltzmann constant \((5.67 \times 10^{-8} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~K}^{-4}) \), \( A \) is the area, and \( T \) is the temperature in Kelvins. With \( P = 70 \mathrm{~W} \) and \( A = 0.02 \mathrm{~m}^{2} \), we solve for \( T \):
03

Calculate for Black Surface

Substitute known values into the Stefan-Boltzmann formula: \( 70 = (5.67 \times 10^{-8}) \times 0.02 \times T^4 \). Solve for \( T^4 \): \( T^4 = \frac{70}{(5.67 \times 10^{-8}) \times 0.02} \). Calculate \( T \):
04

Compute the Temperature

Perform the calculation: \( T^4 = 6172839.8373 \). Thus, \( T = \sqrt[4]{6172839.8373} \), which approximately gives \( T \approx 368.5 \mathrm{~K} \).
05

Consider Polished Surface

When the surface is polished, its emissivity changes. For a polished surface, assume the emissivity \( \varepsilon \) is significantly smaller, like 0.1. The radiant power now becomes \( P = \varepsilon \times \sigma A T^4 \).
06

Calculate Emitted Power for Polished Surface

Substitute \( \varepsilon = 0.1 \), \( A = 0.02 \mathrm{~m}^{2} \), \( T = 368.5 \mathrm{~K} \): \( P = 0.1 \times (5.67 \times 10^{-8}) \times 0.02 \times (368.5)^4 \). Calculate the new \( P \): \( P \approx 7 \mathrm{~W} \).

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Key Concepts

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

Black Body Radiation
Black body radiation is a fundamental concept in physics, referring to an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle. These ideal bodies then emit radiation in a characteristic spectrum determined solely by their temperature.
It follows the principle that perfect absorbers at equilibrium also serve as perfect emitters. This means a black body does not reflect or transmit any energy, making it an important model for understanding thermal emission in real objects.
  • The radiation emitted is referred to as black body radiation.
  • The intensity of emission relies on the temperature and surface area of the body.
In practical scenarios where a perfect black body isn't realizable, materials with very high emissivities, approaching 1, are used to simulate this condition.
Emissivity
Emissivity is a measure of an object's ability to emit thermal radiation, corresponding to how closely a real object's emission resembles that of a black body. It quantifies the efficiency with which an object radiates energy as thermal radiation compared to the maximum possible emission.
  • Values range from 0 to 1, with 1 being perfect emission like a black body.
  • Real-world objects have varying emissivities based on their surfaces. Rough, dark surfaces tend to have high emissivity, while smooth, shiny surfaces have low emissivity.
For example, in the original exercise, a black interior surface was considered to have an emissivity close to 1, while a polished surface significantly lowers the emissivity, altering the thermal emission characteristics.
Thermal Radiation
Thermal radiation is energy emitted by materials due to their finite temperature, stemming from the thermal motion of charged particles. It is one of the primary methods through which heat energy moves from one place to another, not requiring a medium to transfer through, unlike conduction and convection.
Understanding thermal radiation involves analyzing the wavelengths and frequencies of the emitted radiation, which are closely tied to the object's temperature.
  • As the temperature of an object increases, the peak of the emitted radiation shifts to shorter wavelengths.
  • This mode of heat transfer is especially crucial in systems like the one described in the exercise where radiant energy needs to be managed efficiently.
Heat Transfer Calculations
Heat transfer calculations using the Stefan-Boltzmann Law allow the determination of emitted power from a surface based on its temperature and emissivity. The law is expressed as \( P = \varepsilon \sigma A T^4 \), where \( P \) is the power emitted, \( \varepsilon \) is the emissivity, \( \sigma \) is the Stefan-Boltzmann constant, \( A \) is the surface area, and \( T \) is the temperature in Kelvin.
In exercises like this, the calculation begins by identifying known quantities such as the area of the opening and the radiant power emitted.
  • For a black body (emissivity near 1), solve for temperature with the given power and area.
  • Adjust the formula for cases when the surface finish changes, influencing emissivity and thus affecting the emitted power.
These calculations are crucial for determining the temperature and power changes when varying surface properties, as seen in the original problem solution.

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Most popular questions from this chapter

A sphere is suspended in air in a dark room and mair tained at a uniform incandescent temperature. Whet firs viewed with the naked eye, the sphere appears s be brighter around the rim. After several hours, however, it appears to be brighter in the center. Of what type material would you reason the sphere is nude? Give plasible reasons for the nonunifornity of brightness of the sphere and for the changing appes ance with time.

Estimate the wavelength corresponding to maximum emission from each of the following surfaces: the sun, a tungsten filament at \(2500 \mathrm{~K}\), a heated metal at \(1500 \mathrm{~K}\), human skin at \(305 \mathrm{~K}\). and a cryogenically cooled metal surface at \(60 \mathrm{~K}\). Estimate the fraction of the solar emission that is in the following spectral regions: the ultraviolet, the visible, and the infrared.

A temperature sensor imbedded in the tip of a small tube having a diffuse, gray surface with an emissivity of \(0.8\) is centrally positioned within a large air-conditioned toom whose walls and air temperature are 30 and \(20^{\circ} \mathrm{C}\), respectively. (a) What temperature will the sensor indicate if the convection coefficient between the sensot tube and the air is \(5 \mathrm{~W} / \mathrm{m}^{2}\) - K? (b) What would be the effect of using a fan to induce airflow over the tube? Plot the sensor temperature as a function of the convection cocfficient for \(2 \leq h \leq 25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and values of \(\mathrm{e}=0.2,0.5\). and \(0.8\).

A horizontal, opaque surface at a steady-sate temperature of \(77^{\circ} \mathrm{C}\) is exposed in an airflow having a free stream temperature of \(27^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(28 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The emissive power of the surface is \(628 \mathrm{~W} / \mathrm{m}^{2}\), the irradiation is \(1380 \mathrm{~W} / \mathrm{m}^{2}\), and the reflectivity is \(0.40\). Determine the absorptivity of the surface. Determine the act ractiation heat transfer rate for this surface. Is this heat transfer to the wurface or from the surface? Determine the combined heat transfer rate for the surface. Is this heat transfer to the surface or from the surface?

It is not uncommon for the night sky temperature in desert regions to drop to \(-40^{\circ} \mathrm{C}\). If the ambient air temperature is \(20^{\circ} \mathrm{C}\) and the convection coefficient for still air conditions is approximately \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). can a shallow pan of water freeze?

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