P X%23ina20posterwllSpSpfamu.instructure.com/courses/9823/assignments/177283MUpdate :or 16. If the temperature triples, then the flux goes up by a factorof 34, or 81.FLORIDAMECHANICALAGRICULTURAHEADSuppose we want to find the flux and luminosity of Earth.Earth's average temperature is 288 K, so the flux from its surface0.51.01.52.02.53.0HEARTHANDWavelength, i (um)isFIELDFIGURE 6.17 Blackbody spectra emitted by sources with temperaturesF = 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 thespectrum shifts toward shorter wavelengths, and the amount of energyradiated per second from each square meter of the source increases.AccountWorking It Out 6.3 Wien's LawLook again at Figure 6.17. Notice where the peak of each curvelines 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 whichEarth radiates. If we insert Earth's average temperature of 288 Kinto Wien's law, we getDashboardApeak2,900 um KT.%3DCourses2,900 um KApaak2,900 pm K288 KApaskApaak= 10.1 pmor 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 therelationship. In this equation, Xpask (pronounced "lambda peak")is the wavelength where the spectrum is at its peak, where theelectromagnetic radiation from an object is greatest.GroupsCalendarThe average surface temperature of Venus is 773 Ke.34а.InboxWhat is the peak wavelength of radiation from a blackbody e with the sametemperature as Venus?Historyb.StudioWhat is the radiation flux 2 from the surface of Venus?1:17 AMe Type here to search42/20/2021|+(3)IIUNIVERSITY P X%23ina20posterwalSpSpfamu.instructure.com/courses/9823/assignments/177283MUpdate :THERMAL RADIATION 2FLORIDAMECHANICALAGRICULTURAHEADHEARTHANDProblem 4. Planetary Temperatures: Radiation of Heat to Space (Palen, et. al. 1stEd. Chapter 6 Problem 63 )FIELDWorking It Out 6.2 The Stefan-Boltzmann LawAccountLook at Figure 6.17, which shows the spectra of a light source atseveral different temperatures. This source is assumed to emitelectromagnetic radiation only because of its temperature, notits composition. This kind of source is called a blackbody, and ifwe graph the intensity of its emitted radiation across all wave-lengths (as in Figure 6.17), we obtain a characteristic curve calleda blackbody spectrum. As the object's temperature increases, itemits more radiation at every wavelength, so each increase intemperature raises the curve. The luminosity of the object (thetotal amount of light emitted) increases. In fact, it increases quitefast as the temperature rises. Adding up all the energy in a spec-trum shows that the increase in luminosity, L, is proportional tothe fourth power of the temperature, TBut if we want to know the total energy radiated every second,we must multiply by the surface area (A) of Earth. Surface areais given by 4xR, and the radius of Earth is 6,378,000 meters, or6.378 x 10° meters. So the luminosity isL= F.ADashboardL= F.4nRL= (390 W/m")-(4T-(6.378 x 10" m)7)L 2 x 1017 WThis means that Earth emits the equivalent of2,000,000,000,000,000 hundred-watt light bulbs. This is an enor-mous amount of energy, but not anywhere close to the amountemitted by the Sun.CoursesLx T+This result is known as the Stefan-Boltzmann law because itGroupswas 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 thesurface of an object each second is called the flux, F. This is pro-portional to the luminosity, but easier to measure. (Think abouttrying to catch all the photons emitted by the Earth in all direc-tions, versus just the ones from the particular square meter underyour 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 lettersigma). The value of o is 5.67 x 1o W/(m2K*), where W stands forwatts. Expressing all this in math, we find:T = 6000 K, peak =0.48 umCalendarMaking an objecthotter makes itmore luminous...34...and shifts the peak ofits blackbody spectrumto shorter wavelengths.InboxT 5000 K, Apaak =0.58 umF = oT*T = 4000 K, k =0.73 umEven modest changes in temperature can result in largechanges 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 factorof 3*, or 81.T = 3000 K, Aeak = 0.97 um-T = 2000 K, peak =1.45 umHistorySuppose we want to find the flux and luminosity of Earth.Earth's average temperature is 288 K, so the flux from its surface0.51.01.52.02.53.0Wavelength, i (um)isStudioFIGURE 6.17 Blackbody spectra emitted by sources with temperaturesF = oTF = (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 thespectrum shifts toward shorter wavelengths, and the amount of energyradiated per second from each square meter of the source increases.1:16 AMe Type here to search日i42/20/2021|+(3)IntensityUNIVERSITY

Question

P X%23ina20posterwllSpSpfamu.instructure.com/courses/9823/assignments/177283MUpdate :or 16. If the temperature triples, then the flux goes up by a factorof 34, or 81.FLORIDAMECHANICALAGRICULTURAHEADSuppose we want to find the flux and luminosity of Earth.Earth's average temperature is 288 K, so the flux from its surface0.51.01.52.02.53.0HEARTHANDWavelength, i (um)isFIELDFIGURE 6.17 Blackbody spectra emitted by sources with temperaturesF = 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 thespectrum shifts toward shorter wavelengths, and the amount of energyradiated per second from each square meter of the source increases.AccountWorking It Out 6.3 Wien's LawLook again at Figure 6.17. Notice where the peak of each curvelines 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 whichEarth radiates. If we insert Earth's average temperature of 288 Kinto Wien's law, we getDashboardApeak2,900 um KT.%3DCourses2,900 um KApaak2,900 pm K288 KApaskApaak= 10.1 pmor slightly more than 10 um. Earth's radiation peaks in the infra-red region of the spectrum.This is Wien's law (pronounced &#34;Veen's law&#34;), named for Ger-man physicist Wilhelm Wien (1864-1928), who discovered therelationship. In this equation, Xpask (pronounced &#34;lambda peak&#34;)is the wavelength where the spectrum is at its peak, where theelectromagnetic radiation from an object is greatest.GroupsCalendarThe average surface temperature of Venus is 773 Ke.34а.InboxWhat is the peak wavelength of radiation from a blackbody e with the sametemperature as Venus?Historyb.StudioWhat is the radiation flux 2 from the surface of Venus?1:17 AMe Type here to search42/20/2021|+(3)IIUNIVERSITY P X%23ina20posterwalSpSpfamu.instructure.com/courses/9823/assignments/177283MUpdate :THERMAL RADIATION 2FLORIDAMECHANICALAGRICULTURAHEADHEARTHANDProblem 4. Planetary Temperatures: Radiation of Heat to Space (Palen, et. al. 1stEd. Chapter 6 Problem 63 )FIELDWorking It Out 6.2 The Stefan-Boltzmann LawAccountLook at Figure 6.17, which shows the spectra of a light source atseveral different temperatures. This source is assumed to emitelectromagnetic radiation only because of its temperature, notits composition. This kind of source is called a blackbody, and ifwe graph the intensity of its emitted radiation across all wave-lengths (as in Figure 6.17), we obtain a characteristic curve calleda blackbody spectrum. As the object's temperature increases, itemits more radiation at every wavelength, so each increase intemperature raises the curve. The luminosity of the object (thetotal amount of light emitted) increases. In fact, it increases quitefast as the temperature rises. Adding up all the energy in a spec-trum shows that the increase in luminosity, L, is proportional tothe fourth power of the temperature, TBut if we want to know the total energy radiated every second,we must multiply by the surface area (A) of Earth. Surface areais given by 4xR, and the radius of Earth is 6,378,000 meters, or6.378 x 10° meters. So the luminosity isL= F.ADashboardL= F.4nRL= (390 W/m&#34;)-(4T-(6.378 x 10&#34; m)7)L 2 x 1017 WThis means that Earth emits the equivalent of2,000,000,000,000,000 hundred-watt light bulbs. This is an enor-mous amount of energy, but not anywhere close to the amountemitted by the Sun.CoursesLx T+This result is known as the Stefan-Boltzmann law because itGroupswas 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 thesurface of an object each second is called the flux, F. This is pro-portional to the luminosity, but easier to measure. (Think abouttrying to catch all the photons emitted by the Earth in all direc-tions, versus just the ones from the particular square meter underyour 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 lettersigma). The value of o is 5.67 x 1o W/(m2K*), where W stands forwatts. Expressing all this in math, we find:T = 6000 K, peak =0.48 umCalendarMaking an objecthotter makes itmore luminous...34...and shifts the peak ofits blackbody spectrumto shorter wavelengths.InboxT 5000 K, Apaak =0.58 umF = oT*T = 4000 K, k =0.73 umEven modest changes in temperature can result in largechanges 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 factorof 3*, or 81.T = 3000 K, Aeak = 0.97 um-T = 2000 K, peak =1.45 umHistorySuppose we want to find the flux and luminosity of Earth.Earth's average temperature is 288 K, so the flux from its surface0.51.01.52.02.53.0Wavelength, i (um)isStudioFIGURE 6.17 Blackbody spectra emitted by sources with temperaturesF = oTF = (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 thespectrum shifts toward shorter wavelengths, and the amount of energyradiated per second from each square meter of the source increases.1:16 AMe Type here to search日i42/20/2021|+(3)IntensityUNIVERSITY

Accepted Answer

Given,