Question

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

Answer 1

75 W

Step-by-step explanation:

`T` = temperature of the filament = 2100 K

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

`e` = Emissivity = 0.68

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

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

`P = \sigma e AT^(4) \\P = (5.67*10^(-8)) (0.68) (10^(-4)) (2100)^(4)\\P = 75 W`