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What is the rate of energy radiation per unit area of a blackbody at (a) \(273 \mathrm{~K}\) and (b) \(2730 \mathrm{~K} ?\)

Short Answer

Expert verified
The rate of radiation of a blackbody at (a) 273 K is approximately 315.09 W/m^2 and at (b) 2730 K is approximately 315.09*10^4 W/m^2.

Step by step solution

01

Understanding the Stefan-Boltzmann Law

The Stefan-Boltzmann law is represented by the equation \( E = \sigma T^4 \), where \( E \) is the energy radiation per unit area, \( \sigma \) is the Stefan-Boltzmann constant equal to \( 5.67x10^{-8} Wm^{-2}K^{-4} \), and \( T \) is the temperature in Kelvin.
02

Calculating energy radiation for (a) 273 K

Plug the given temperature into the equation: \( E = 5.67x10^{-8} * (273)^4 \) and calculate the result, which will be the rate of energy radiation per unit area at 273 K.
03

Calculating energy radiation for (b) 2730 K

Repeat the process for the second temperature: \( E = 5.67x10^{-8} * (2730)^4 \). The result will be the rate of energy radiation per unit area at 2730 K.

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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 an idealized object, known as a blackbody, which absorbs all incident radiation, regardless of frequency or angle of incidence. This type of radiation is a key concept in the field of physics as it provides insight into how energy is emitted by objects based on their temperatures. Unlike real-world objects that can reflect, refract, or transmit photons, a perfect blackbody is a theoretical construct that represents an object which absorbs all incoming light and emits radiation at all wavelengths in a predictable pattern. The wavelength distribution, and thus the intensity of the radiated energy, depends solely on the object’s temperature. Blackbody radiation is characterized by the spectrum of radiation it would emit, which shifts towards higher frequencies as the temperature increases.
Energy Radiation
Energy radiation, in the context of blackbodies, refers to the process by which energy is emitted in the form of electromagnetic waves due to the thermal motion of particles within the object. This emission occurs at all temperatures and increases with rising temperature. The rate of energy radiation per unit area from a blackbody is determined by the Stefan-Boltzmann Law. As temperature increases, energy radiation becomes more intense and shifts to shorter wavelengths. Understanding the rate at which energy is radiated is essential for applications across various scientific fields such as astrophysics, climate science, and engineering. This includes theories explaining the brightness of stars and understanding radiation from the Earth's surface.
Temperature in Kelvin
Temperature in Kelvin is a key factor in calculating the amount of radiation emitted by a blackbody using the Stefan-Boltzmann Law. The Kelvin scale is an absolute temperature scale, starting from absolute zero, where thermal motion ceases. In blackbody radiation calculations, the temperature must be in Kelvin to accurately apply the Stefan-Boltzmann equation: \[ E = \sigma T^4 \]where \(T\) represents the object's temperature in Kelvin. The Kelvin scale allows scientists to unambiguously measure thermal energy, as it directly relates temperature to energy, removing the negative values found in other scales, such as Celsius. Using Kelvin is vital for precision and consistency in scientific computations and hypotheses.
Stefan-Boltzmann Constant
The Stefan-Boltzmann constant, denoted as \(\sigma\), is a fundamental constant in physics that relates the energy radiated by a blackbody in terms of its temperature. Its value is approximately \(5.67 \times 10^{-8} \, \mathrm{Wm^{-2}K^{-4}}\).This constant is used in the Stefan-Boltzmann Law, which articulates that the total energy emission from a blackbody per unit area is directly proportional to the fourth power of the absolute temperature: \[ E = \sigma T^4 \]The constant ensures that the relationship between temperature and emitted energy is accurately quantified. It plays a crucial role not only in theoretical applications but also in practical endeavors like calculating the radiative properties of stars, planets, and other celestial bodies, as well as determining thermal radiation in engineering systems.

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Most popular questions from this chapter

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 \(299 \mathrm{~K} ?\)

You have probably seen people jogging in extremely hot weather. There are good reasons not to do this! When jogging strenuously, an average runner of mass \(68 \mathrm{~kg}\) and surface area \(1.85 \mathrm{~m}^{2}\) produces energy at a rate of up to \(1300 \mathrm{~W}, 80 \%\) of which is converted to heat. The jogger radiates heat but actually absorbs more from the hot air than he radiates away. At such high levels of activity, the skin's temperature can be elevated to around \(33^{\circ} \mathrm{C}\) instead of the usual \(30^{\circ} \mathrm{C}\). (Ignore conduction, which would bring even more heat into his body.) The only way for the body to get rid of this extra heat is by evaporating water (sweating). (a) How much heat per second is produced just by the act of jogging? (b) How much net heat per second does the runner gain just from radiation if the air temperature is \(40.0^{\circ} \mathrm{C}\left(104^{\circ} \mathrm{F}\right) ?\) (Remember: He radiates out, but the environment radiates back in.) (c) What is the total amount of excess heat this runner's body must get rid of per second? (d) How much water must his body evaporate every minute due to his activity? The heat of vaporization of water at body temperature is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\) (e) How many \(750 \mathrm{~mL}\) bottles of water must he drink after (or preferably before!) jogging for a half hour? Recall that a liter of water has a mass of \(1.0 \mathrm{~kg}\).

The basal metabolic rate is the rate at which energy is produced in the body when a person is at rest. A \(75 \mathrm{~kg}(165 \mathrm{lb})\) person of height \(1.83 \mathrm{~m}(6 \mathrm{ft})\) has a body surface area of approximately \(2.0 \mathrm{~m}^{2}\). (a) What is the net amount of heat this person could radiate per second into a room at \(18^{\circ} \mathrm{C}\) (about \(65^{\circ} \mathrm{F}\) ) if his skin's surface temperature is \(30^{\circ} \mathrm{C}\) ? (At such temperatures, nearly all the heat is infrared radiation, for which the body's emissivity is \(1.0,\) regardless of the amount of pigment.) (b) Normally, \(80 \%\) of the energy produced by metabolism goes into heat, while the rest goes into things like pumping blood and repairing cells. Also normally, a person at rest can get rid of this excess heat just through radiation. Use your answer to part (a) to find this person's basal metabolic rate.

A steel wire has density \(7800 \mathrm{~kg} / \mathrm{m}^{3}\) and mass \(2.50 \mathrm{~g}\). It is stretched between two rigid supports separated by \(0.400 \mathrm{~m}\). (a) When the temperature of the wire is \(20.0^{\circ} \mathrm{C}\), the frequency of the fundamental standing wave for the wire is \(440 \mathrm{~Hz}\). What is the tension in the wire? (b) What is the temperature of the wire if its fundamental standing wave has frequency \(460 \mathrm{~Hz}\) ? For steel the coefficient of linear expansion is \(1.2 \times 10^{-5} \mathrm{~K}^{-1}\) and Young's modulus is \(20 \times 10^{10} \mathrm{~Pa}\)

An unknown liquid has density \(\rho\) and coefficient of volume expansion \(\beta\). A quantity of heat \(Q\) is added to a volume \(V\) of the liquid, and the volume of the liquid increases by an amount \(\Delta V\). There is no phase change. In terms of these quantities, what is the specific heat capacity \(c\) of the liquid?

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