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 2 500 K?

Question

Quizplus · Accepted Answer

Using the Stefan-Boltzmann’s law:
P = σ x A x T⁴ x ε
where
σ = Stefan-Boltzmann constant = 5.67 x 10^-8 W/m^2K^4
A = Surface area of the filament (since it remains constant) = 1
ε = emissivity (since it remains constant)
T1 = 2000 K, P1 = 40 W
T2 = 2500 K, P2 = ?
So, P2 = P1 x (T2/T1)⁴
P2 = 40W x (2500K/2000K)⁴
P2 = 98W
Therefore, the answer is C.

The Filament Temperature of a Light Bulb Is 2 000