The tungsten filament of a light bulb has an operating temperature of about 2100 K.If the emitting area of the filament is 0.80 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) 100 W B) 75 W C) 60 W D) 40 W

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Quizplus · Accepted Answer

The power output can be calculated using the Stefan-Boltzmann law: P = εσAT⁴. Substituting the given values: P = 0.68 * 5.67 × 10⁻⁸ W/m²·K⁴ * 0.80 × 10⁻⁴ m² * (2100 K)⁴ ≈ 60 W.

The Tungsten Filament of a Light Bulb Has an Operating