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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 2 f f (or 43.7%) 11-25 A glass window transmits 90% of the radiation in a specified wavelength range and is opaque for radiation at other wavelengths. The rate of radiation transmitted through this window is to be determined
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