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? ($\sigma$ = 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: ε = 0.68, σ = 5.67 × 10⁻⁸ W/m²·K⁴, A = 0.80 cm² = 0.00008 m², and T = 2100 K, we find P ≈ 60 W.

The Tungsten Filament of a Light Bulb Has an Operating σ\sigmaσ

The Tungsten Filament of a Light Bulb Has an Operating $\sigma$