The filament temperature of a light bulb is 2,000 K when the bulb delivers 40 W of power.If its emissivity remains constant,what power is delivered when the filament temperature is 2,400 K?

Question

The filament temperature of a light bulb is 2,000 K when the bulb delivers 40 W of power.If its emissivity remains constant,what power is delivered when the filament temperature is 2,400 K?
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Quizplus · Accepted Answer

The power radiated by an object is given by the Stefan-Boltzmann law:

P = εσAT^4

where P is the power radiated, ε is the emissivity, σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/m^2/K^4), A is the surface area of the object, and T is the temperature in Kelvin.

Since the emissivity remains constant, we can write:

P1/P2 = (T1/T2)^4

where P1 is the power when T1 = 2,000 K, and P2 is the power when T2 = 2,400 K.

Plugging in the values and solving for P2, we get:

P2 = P1 (T2/T1)^4 
    = 40 W (2.4/2)^4 
    ≈ 83 W

Therefore, the answer is C.

The Filament Temperature of a Light Bulb Is 2,000 K