Question

g Incandescent bulbs generate visible light by heating up a thin metal filament to a very high temperature so that the thermal radiation from the filament becomes visible. One bulb filament has a surface area of 30 mm2 and emits 60 W when operating. If the bulb filament has an emissivity of 0.8, what is the operating temperature of the filament

Answer 1

The operating temperature of the incandescent bulb filament can be determined using the Stefan-Boltzmann law. By rearranging the formula and substituting the given values, we find that the temperature of the filament is approximately 2428 K (Kelvin).

Step-by-step explanation:

An incandescent bulb filament emits visible light by heating up a thin metal filament to a very high temperature. The operating temperature of the filament can be determined using the Stefan-Boltzmann law, which relates the power radiated by an object to its surface area and temperature.

Given that the filament has a surface area of 30 mm2, emits 60 W of power, and has an emissivity of 0.8, we can use the formula:

Power emitted = emissivity x Stefan-Boltzmann constant x surface area x temperature^4

By rearranging the equation and substituting the given values, we can solve for the temperature of the filament:

Temperature^4 = Power emitted / (emissivity x Stefan-Boltzmann constant x surface area)

Temperature = (Power emitted / (emissivity x Stefan-Boltzmann constant x surface area))^(1/4)

Plugging in the given values, we find that the operating temperature of the filament is approximately 2428 K (Kelvin).

Answer 2

2577 K

Step-by-step explanation:

Power radiated , P = σεAT⁴ where σ = Stefan-Boltzmann constant = 5.6704 × 10⁻⁸ W/m²K⁴, ε = emissivity of bulb filament = 0.8, A = surface area of bulb = 30 mm² = 30 × 10⁻⁶ m² and T = operating temperature of filament.

So, T = ⁴√(P/σεA)

Since P = 60 W, we substitute the vales of the variables into T. So,

T = ⁴√(P/σεA)

= ⁴√(60 W/(5.6704 × 10⁻⁸ W/m²K⁴ × 0.8 × 30 × 10⁻⁶ m²)

= ⁴√(60 W/(136.0896 × 10⁻¹⁴ W/K⁴)

= ⁴√(60 W/(13608.96 × 10⁻¹⁶ W/K⁴)

= ⁴√(0.00441 × 10¹⁶K⁴)

= 0.2577 × 10⁴ K

= 2577 K