Thermodynamics HW Solutions 872

Thermodynamics HW Solutions 872 - Chapter 11 Fundamentals...

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Chapter 11 Fundamentals of Thermal Radiation 11-24 Radiation emitted by a light source is maximum in the blue range. The temperature of this light source and the fraction of radiation it emits in the visible range are to be determined. Assumptions The light source behaves as a black body. Analysis The temperature of this light source is T = ? K 6166 = μ μ = ⎯→ μ = λ m 47 . 0 K m 8 . 2897 K m 8 . 2897 ) ( power max T T The visible range of the electromagnetic spectrum extends from . Noting that = 6166 K, the blackbody radiation functions corresponding to m 76 . 0 to m 40 . 0 2 1 μ = λ μ = λ T λ λ 1 T and 2 T are determined from Table 11-2 to be 59141 . 0 f mK 4686 = K) m)(6166 76 . 0 ( 15444 . 0 f mK 2466 = K) m)(6166 40 . 0 ( 2 1 2 1 = ⎯→ μ μ = λ = ⎯→ μ μ = λ λ λ T T Then the fraction of radiation emitted between these two wavelengths becomes 0.437 = λ λ 15444 . 0 59141 . 0 1
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This note was uploaded on 01/24/2012 for the course PHY 4803 taught by Professor Dr.danielarenas during the Fall '10 term at UNF.

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