   Chapter 11, Problem 54P

Chapter
Section
Textbook Problem

The filament of a 75-W light bulb is at a temperature of 3300 K. Assuming the filament has an emissivity e = 1.0, find its surface area.

To determine
The surface area of a 75 W light bulb filament.

Explanation

Given info: The power of the filament is 75 W, temperature of the filament is 3300 K, and the emissivity of the filament is 1.0.

Formula to calculate the power radiated by the bulb is,

P=σAeT2

• P is the power radiated by the bulb,
• σ is Stefan’s – Boltzmann constant,
• A is the surface area of the filament,
• e is emissivity of the Star,
• T is the temperature of filament

Rewrite the above expression in terms of A.

A=PσeT4

Substitute 75 W for P, 5.66696×108W/m2K4 for σ , 1.0 for e, 3300 K for T to find A.

A=75W(5

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