Homework Help is Here – Start Your Trial Now!
Science
Physics
Working It Out 6.3 Wien's Law Look again at Figure 6.17. Notice where the peak of each curve lines up along the horizontal axis. As the temperature, T, increases, the peak of the spectrum shifts toward shorter wave- lengths. This is an inverse proportion. Translating this into math, and inserting the constant of 2,900 um-K (derived from measure- ments of this relationship) to fix the units, we find: Suppose we want to calculate the peak wavelength at which Earth radiates. If we insert Earth's average temperature of 288 K into Wien's law, we get 2,900 pm -K 2.900 um-K 288 K 2,900 pum - K A 10.1 pum This is Wien's law (pronounced "Voen's law"), named for Ger- man physicist Wilhelm Wien (1864-1928), who discovered the relationship. In this equation, Xyusa (pronounced "lambda peak") is the wavelength where the spectrum is at its peak, where the electromagnetic radiation from an object is greatest. or slightly more than 10 pm. Earth's radiation peaks in the infra- red region of the spectrum. The average surface temperature of Venus is 773 Ke.

Working It Out 6.3 Wien's Law Look again at Figure 6.17. Notice where the peak of each curve lines up along the horizontal axis. As the temperature, T, increases, the peak of the spectrum shifts toward shorter wave- lengths. This is an inverse proportion. Translating this into math, and inserting the constant of 2,900 um-K (derived from measure- ments of this relationship) to fix the units, we find: Suppose we want to calculate the peak wavelength at which Earth radiates. If we insert Earth's average temperature of 288 K into Wien's law, we get 2,900 pm -K 2.900 um-K 288 K 2,900 pum - K A 10.1 pum This is Wien's law (pronounced "Voen's law"), named for Ger- man physicist Wilhelm Wien (1864-1928), who discovered the relationship. In this equation, Xyusa (pronounced "lambda peak") is the wavelength where the spectrum is at its peak, where the electromagnetic radiation from an object is greatest. or slightly more than 10 pm. Earth's radiation peaks in the infra- red region of the spectrum. The average surface temperature of Venus is 773 Ke.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
See similar textbooks
icon
Related questions
Question
100%
P X %23 in a 20 poster wll Sp Sp famu.instructure.com/courses/9823/assignments/177283 M Update : or 16. If the temperature triples, then the flux goes up by a factor of 34, or 81. FLORIDA MECHANICAL AGRICULTURA HEAD Suppose we want to find the flux and luminosity of Earth. Earth's average temperature is 288 K, so the flux from its surface 0.5 1.0 1.5 2.0 2.5 3.0 HEART HAND Wavelength, i (um) is FIELD FIGURE 6.17 Blackbody spectra emitted by sources with temperatures F = oT* F = (5.67 x 10- W/(m²K*)) (288 K)* F = 390 W/m? ranging from 2000 K to 6000 K. At higher temperatures the peak of the spectrum shifts toward shorter wavelengths, and the amount of energy radiated per second from each square meter of the source increases. Account Working It Out 6.3 Wien's Law Look again at Figure 6.17. Notice where the peak of each curve lines up along the horizontal axis. As the temperature, T, increases, the peak of the spectrum shifts toward shorter wave- lengths. This is an inverse proportion. Translating this into math, and inserting the constant of 2,900 pm-K (derived from measure- ments of this relationship) to fix the units, we find: Suppose we want to calculate the peak wavelength at which Earth radiates. If we insert Earth's average temperature of 288 K into Wien's law, we get Dashboard Apeak 2,900 um K T. %3D Courses 2,900 um K Apaak 2,900 pm K 288 K Apask Apaak = 10.1 pm or slightly more than 10 um. Earth's radiation peaks in the infra- red region of the spectrum. This is Wien's law (pronounced "Veen's law"), named for Ger- man physicist Wilhelm Wien (1864-1928), who discovered the relationship. In this equation, Xpask (pronounced "lambda peak") is the wavelength where the spectrum is at its peak, where the electromagnetic radiation from an object is greatest. Groups Calendar The average surface temperature of Venus is 773 Ke. 34 а. Inbox What is the peak wavelength of radiation from a blackbody e with the same temperature as Venus? History b. Studio What is the radiation flux 2 from the surface of Venus? 1:17 AM e Type here to search 4 2/20/2021 | + (3) II UNIVERSITY
Transcribed Image Text:P X %23 in a 20 poster wll Sp Sp famu.instructure.com/courses/9823/assignments/177283 M Update : or 16. If the temperature triples, then the flux goes up by a factor of 34, or 81. FLORIDA MECHANICAL AGRICULTURA HEAD Suppose we want to find the flux and luminosity of Earth. Earth's average temperature is 288 K, so the flux from its surface 0.5 1.0 1.5 2.0 2.5 3.0 HEART HAND Wavelength, i (um) is FIELD FIGURE 6.17 Blackbody spectra emitted by sources with temperatures F = oT* F = (5.67 x 10- W/(m²K*)) (288 K)* F = 390 W/m? ranging from 2000 K to 6000 K. At higher temperatures the peak of the spectrum shifts toward shorter wavelengths, and the amount of energy radiated per second from each square meter of the source increases. Account Working It Out 6.3 Wien's Law Look again at Figure 6.17. Notice where the peak of each curve lines up along the horizontal axis. As the temperature, T, increases, the peak of the spectrum shifts toward shorter wave- lengths. This is an inverse proportion. Translating this into math, and inserting the constant of 2,900 pm-K (derived from measure- ments of this relationship) to fix the units, we find: Suppose we want to calculate the peak wavelength at which Earth radiates. If we insert Earth's average temperature of 288 K into Wien's law, we get Dashboard Apeak 2,900 um K T. %3D Courses 2,900 um K Apaak 2,900 pm K 288 K Apask Apaak = 10.1 pm or slightly more than 10 um. Earth's radiation peaks in the infra- red region of the spectrum. This is Wien's law (pronounced "Veen's law"), named for Ger- man physicist Wilhelm Wien (1864-1928), who discovered the relationship. In this equation, Xpask (pronounced "lambda peak") is the wavelength where the spectrum is at its peak, where the electromagnetic radiation from an object is greatest. Groups Calendar The average surface temperature of Venus is 773 Ke. 34 а. Inbox What is the peak wavelength of radiation from a blackbody e with the same temperature as Venus? History b. Studio What is the radiation flux 2 from the surface of Venus? 1:17 AM e Type here to search 4 2/20/2021 | + (3) II UNIVERSITY
P X %23 in a 20 poster wal Sp Sp famu.instructure.com/courses/9823/assignments/177283 M Update : THERMAL RADIATION 2 FLORIDA MECHANICAL AGRICULTURA HEAD HEART HAND Problem 4. Planetary Temperatures: Radiation of Heat to Space (Palen, et. al. 1st Ed. Chapter 6 Problem 63 ) FIELD Working It Out 6.2 The Stefan-Boltzmann Law Account Look at Figure 6.17, which shows the spectra of a light source at several different temperatures. This source is assumed to emit electromagnetic radiation only because of its temperature, not its composition. This kind of source is called a blackbody, and if we graph the intensity of its emitted radiation across all wave- lengths (as in Figure 6.17), we obtain a characteristic curve called a blackbody spectrum. As the object's temperature increases, it emits more radiation at every wavelength, so each increase in temperature raises the curve. The luminosity of the object (the total amount of light emitted) increases. In fact, it increases quite fast as the temperature rises. Adding up all the energy in a spec- trum shows that the increase in luminosity, L, is proportional to the fourth power of the temperature, T But if we want to know the total energy radiated every second, we must multiply by the surface area (A) of Earth. Surface area is given by 4xR, and the radius of Earth is 6,378,000 meters, or 6.378 x 10° meters. So the luminosity is L= F.A Dashboard L= F.4nR L= (390 W/m")-(4T-(6.378 x 10" m)7) L 2 x 1017 W This means that Earth emits the equivalent of 2,000,000,000,000,000 hundred-watt light bulbs. This is an enor- mous amount of energy, but not anywhere close to the amount emitted by the Sun. Courses Lx T+ This result is known as the Stefan-Boltzmann law because it Groups was discovered in the laboratory by Austrian physicist Josef Ste- fan (1835–1893) and derived by his student, Ludwig Boltzmann (1844-1906). The amount of energy radiated by each square meter of the surface of an object each second is called the flux, F. This is pro- portional to the luminosity, but easier to measure. (Think about trying to catch all the photons emitted by the Earth in all direc- tions, versus just the ones from the particular square meter under your feet. You can find the luminosity from the flux by multiply- ing by the total surface area.) We can relate the flux to the temper- ature by using the Stefan-Boltzmann constant, o (the Greek letter sigma). The value of o is 5.67 x 1o W/(m2K*), where W stands for watts. Expressing all this in math, we find: T = 6000 K, peak =0.48 um Calendar Making an object hotter makes it more luminous... 34 ...and shifts the peak of its blackbody spectrum to shorter wavelengths. Inbox T 5000 K, Apaak =0.58 um F = oT* T = 4000 K, k =0.73 um Even modest changes in temperature can result in large changes in the amount of power radiated by an object. If the tem- perature of an object doubles, the flux increases by a factor of 24, or 16. If the temperature triples, then the flux goes up by a factor of 3*, or 81. T = 3000 K, Aeak = 0.97 um -T = 2000 K, peak =1.45 um History Suppose we want to find the flux and luminosity of Earth. Earth's average temperature is 288 K, so the flux from its surface 0.5 1.0 1.5 2.0 2.5 3.0 Wavelength, i (um) is Studio FIGURE 6.17 Blackbody spectra emitted by sources with temperatures F = oT F = (5.67 x 10- W/(m²K*)) (288 K)* F = 390 W/m? ranging from 2000 K to 6000 K. At higher temperatures the peak of the spectrum shifts toward shorter wavelengths, and the amount of energy radiated per second from each square meter of the source increases. 1:16 AM e Type here to search 日i 4 2/20/2021 | + (3) Intensity UNIVERSITY
Transcribed Image Text:P X %23 in a 20 poster wal Sp Sp famu.instructure.com/courses/9823/assignments/177283 M Update : THERMAL RADIATION 2 FLORIDA MECHANICAL AGRICULTURA HEAD HEART HAND Problem 4. Planetary Temperatures: Radiation of Heat to Space (Palen, et. al. 1st Ed. Chapter 6 Problem 63 ) FIELD Working It Out 6.2 The Stefan-Boltzmann Law Account Look at Figure 6.17, which shows the spectra of a light source at several different temperatures. This source is assumed to emit electromagnetic radiation only because of its temperature, not its composition. This kind of source is called a blackbody, and if we graph the intensity of its emitted radiation across all wave- lengths (as in Figure 6.17), we obtain a characteristic curve called a blackbody spectrum. As the object's temperature increases, it emits more radiation at every wavelength, so each increase in temperature raises the curve. The luminosity of the object (the total amount of light emitted) increases. In fact, it increases quite fast as the temperature rises. Adding up all the energy in a spec- trum shows that the increase in luminosity, L, is proportional to the fourth power of the temperature, T But if we want to know the total energy radiated every second, we must multiply by the surface area (A) of Earth. Surface area is given by 4xR, and the radius of Earth is 6,378,000 meters, or 6.378 x 10° meters. So the luminosity is L= F.A Dashboard L= F.4nR L= (390 W/m")-(4T-(6.378 x 10" m)7) L 2 x 1017 W This means that Earth emits the equivalent of 2,000,000,000,000,000 hundred-watt light bulbs. This is an enor- mous amount of energy, but not anywhere close to the amount emitted by the Sun. Courses Lx T+ This result is known as the Stefan-Boltzmann law because it Groups was discovered in the laboratory by Austrian physicist Josef Ste- fan (1835–1893) and derived by his student, Ludwig Boltzmann (1844-1906). The amount of energy radiated by each square meter of the surface of an object each second is called the flux, F. This is pro- portional to the luminosity, but easier to measure. (Think about trying to catch all the photons emitted by the Earth in all direc- tions, versus just the ones from the particular square meter under your feet. You can find the luminosity from the flux by multiply- ing by the total surface area.) We can relate the flux to the temper- ature by using the Stefan-Boltzmann constant, o (the Greek letter sigma). The value of o is 5.67 x 1o W/(m2K*), where W stands for watts. Expressing all this in math, we find: T = 6000 K, peak =0.48 um Calendar Making an object hotter makes it more luminous... 34 ...and shifts the peak of its blackbody spectrum to shorter wavelengths. Inbox T 5000 K, Apaak =0.58 um F = oT* T = 4000 K, k =0.73 um Even modest changes in temperature can result in large changes in the amount of power radiated by an object. If the tem- perature of an object doubles, the flux increases by a factor of 24, or 16. If the temperature triples, then the flux goes up by a factor of 3*, or 81. T = 3000 K, Aeak = 0.97 um -T = 2000 K, peak =1.45 um History Suppose we want to find the flux and luminosity of Earth. Earth's average temperature is 288 K, so the flux from its surface 0.5 1.0 1.5 2.0 2.5 3.0 Wavelength, i (um) is Studio FIGURE 6.17 Blackbody spectra emitted by sources with temperatures F = oT F = (5.67 x 10- W/(m²K*)) (288 K)* F = 390 W/m? ranging from 2000 K to 6000 K. At higher temperatures the peak of the spectrum shifts toward shorter wavelengths, and the amount of energy radiated per second from each square meter of the source increases. 1:16 AM e Type here to search 日i 4 2/20/2021 | + (3) Intensity UNIVERSITY
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 3 steps with 3 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
The boltzmann Factor
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
Recommended textbooks for you
College Physics
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
University Physics (14th Edition)
University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON
Introduction To Quantum Mechanics
Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press
Physics for Scientists and Engineers
Physics for Scientists and Engineers
Physics
ISBN:
9781337553278
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Lecture- Tutorials for Introductory Astronomy
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:
9780321820464
Author:
Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:
Addison-Wesley
College Physics: A Strategic Approach (4th Editio…
College Physics: A Strategic Approach (4th Editio…
Physics
ISBN:
9780134609034
Author:
Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:
PEARSON
SEE MORE TEXTBOOKS