Question

The operating temperature of a tungsten filament in an incandescent light bulb is 2400 K, and its emissivity is 0.350.

Find the surface area of the filament of a 150 W bulb if all the electrical energy consumed by the bulb is radiated by the filament as electromagnetic waves. (Only a fraction of the radiation appears as visible light.) Express your answer in meters squared.

Answer 1

By applying the Stefan-Boltzmann law, we find that the surface area of the filament in a 150 W bulb, assuming all electrical energy is radiated as electromagnetic waves, is approximately 0.000086 m^2.

Step-by-step explanation:

This problem involves a concept in Physics known as the Stefan-Boltzmann law. The power or energy radiated by a black body (in this case, the tungsten filament) is given by the equation P = εσAT^4. Here, P is the power (150W), ε is the emissivity (0.350), σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/m^2K^4), A is the surface area we want to find, and T is the temperature (2400K).

Substituting the known values into the equation, we get 150W = 0.350 * (5.67 x 10^-8 W/m^2K^4) * A * (2400K)^4. Solving for A, we find that the surface area of the filament is approximately 0.000086 m^2 or 8.6 x 10^-5 m^2.

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