The tungsten filament of a light bulb has an operating temperature of about 2 100 k. if the emitting area of the filament is 1.0 cm2, and its emissivity is 0.68, what is the power output of the light bulb? (σ = 5.67 × 10−8w/m2⋅k4)
a. 100 wb. 75 wc. 60 wd. 40 w

75 W

Explanation:

= temperature of the filament = 2100 K

= Emitting area of the filament = 1 cm² = 10⁻⁴ m²

= Emissivity = 0.68

= Stefan's constant = 5.67 x 10⁻⁸ Wm⁻²K⁻⁴

Using Stefan's law, Power output of the light bulb is given as

b would be the answer

do you have a formula possibly