Question

The filament of an evacuated light bulb has a
length 10 cm, diameter 0.2 mm and emissivity 0.2,
calculate the power it radiates at 2000 K.
(σ = 5.67 × 10⁻⁸ W/m² K⁴)
(a) 21.5 W (b) 15.5 W
(c) 8.9 W (d) 11.4 W

Answer 1

Given:

• The length of the filament (l) = 0.1m
• The diameter of the filament (d) = 2*10⁻⁴m
• The emissivity of the filament (e) = 0.2
• The temperature of the filament (T) = 2000K
• Stefan's constant (σ) = 5.67*10⁻⁸ Wm⁻²K⁻⁴

To find:

The power radiated by the filament.

Solution:

• The area of the filament (A) = πdl = 6.28*10⁻⁵m²

• According to Stefan-Boltzmann law, the power radiated equals σeAT⁴ = 11.4 W

Answer:

The power radiated by the filament is equal to 11.4 W.