The tungsten filament of an incandescent light bulb has atemperature of approxim
ID: 1667658 • Letter: T
Question
The tungsten filament of an incandescent light bulb has atemperature of approximately 3000 K. The emissivity of tungsten isapproximately 1/3, and you may assume that it is independent ofwavelength.(a). If the bulb gives off a total of 100 watts, what is thesurface area of its filament in square millimeters?
(b). At what value of the photon energy does the peak in the bulb'sspectrum occur? What is the wavelength corresponding to this photonenergy?
(c). Sketch the spectrum of light given off by the filament.Indicate the region on the graph that corresponds to visiblewavelengths, between 400 and 700 nm.
(d). Calculate the fraction of the bulb's energy that comes out asvisible light. (Do the integral numerically on a calculator orcomputer.) Check your result qualitatively from the graph of part(c).
(e). To increase the efficiency of an incandescent bulb, would youwant to raise or lower the temperature?
(f). Estimate the maximum possible efficiency (i.e. fraction ofenergy in the visible spectrum) of an incandescent bulb, and thecorresponding filament temperature. Neglect the fact that tungstenmelts at 3695 K.
Explanation / Answer
Given temperature of the bulb is T =3000 K emissivity of tungesten e = 1/3 = 0.333 a) power of the bulb is 100 W From stefan's boltzmann constant P = A e T 4 here is stefans boltzmann constant ( 5.67 *10 -8 J s -1 m -2 k -4 ) on substituting the numerical values 100 W = A ( 0.33)( 5.67 *10 -8) (3000) 4 thesurface area of its filament A = 6.5980*10-5 m 2 = 0.6598 mm2 b) from weins displacement law the relation between wave length and temperature is max = b / T = 2.897*10-3 m k / 3000 k = 9.6566*10^ -7 m photon energy is E = h c / = ( 6.635* 10 -34 Js ) ( 3*10 8 m/s ) / ( 9.6566*10^ -7 = 2.061*10^ -19 J = 1.2883 eV = 1.2883 eVRelated Questions
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