Assuming a filament in a 100 W light bulb acts like a perfect blackbody,what is the temperature of the hottest portion of the filament if it has a surface area of 6.3 × 10^-⁵ m²? The Stefan-Boltzmann constant is 5.67 × 10^-⁸ W/(m² · K²). A)130 K B)1100 K C)2300 K D)5800 K E)30 000 K

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Quizplus · Accepted Answer

The power emitted by a blackbody is given by the Stefan-Boltzmann law: $P = \sigma A T^4$, where $P$ is the power, $\sigma$ is the Stefan-Boltzmann constant, $A$ is the surface area, and $T$ is the temperature in Kelvin. Rearranging for $T$, we get $T = \left(\frac{P}{\sigma A}\right)^{1/4}$. Substituting the given values: $T = \left(\frac{100}{5.67 \times 10^{-8} \times 6.3 \times 10^{-5}}\right)^{1/4}$, we find that $T$ is approximately 2300 K.

Assuming a Filament in a 100 W Light Bulb Acts