Question

A 60 W filament lamp loses all its energy by radiation from its surface. The emissivity of the filament surface is 0.5 and the surface area is 5 × 10⁻⁵ m². Calculate the temperature of the filament. (Given: σ = 5.67 × 10⁻⁸ Jm⁻² s⁻¹ K⁻⁴)

Answer 1

Given: σ = 5.67 × 10⁻⁸ Jm⁻² s⁻¹ K⁻⁴

A = 5 × 10⁻⁵ m²

e = 0.5 

 `(dQ)/dt` (P) = 60 W

To find: Temperature (T) = ?

Formula: `(dQ)/dt` (P) = eAσT⁴

∴ 60 = 0.5 × 5 × 10⁻⁵ ×  5.67 × 10⁻⁸  × T⁴

∴ 60 = 14.175 × 10⁻¹³ × T⁴

∴ T⁴ = `60/(14.175 xx 10^-13`

∴ T⁴ = `60/(14.175) xx 10^13`

∴ T⁴ = 4.233 × 10¹³

∴ T⁴ = 42.33 × 10¹²

∴ `T^4 = (42.33 xx 10^12)^(1/4)` 

∴ T = 2.551 × 10³ K

∴ T = 2551 K