The tungsten filament of a light bulb has an operating temperature of about 2 050 K.If the emitting area of the filament is 0.70 cm²,and its emissivity is 0.68,what is the power output of the light bulb? (σ = 5.67 × 10−⁸ W/m²⋅K⁴) A)70.1 W B)47.7 W C)477 W D)42.7 W E)39.7 W

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Quizplus · Accepted Answer

The power output can be calculated using the Stefan-Boltzmann law: P = εσAT⁴. Convert the area from cm² to m², then substitute the values: P = 0.68 * 5.67 × 10⁻⁸ W/m²⋅K⁴ * 0.70 × 10⁻⁴ m² * (2050 K)⁴ ≈ 47.7 W.

The Tungsten Filament of a Light Bulb Has an Operating