Calculate the electric power which must be supplied to the filament of of a light bulb operating at 3000K. The total surface area of the filament is

8×10−6m2
and its emissivity is 0.92.

R=kσT⁴ = 0.92•5.67•10⁻⁸•3000⁴=… W/m²

R=P/A =>
P=AR= A•kσT⁴ =
=8•10⁻⁶•0.92•5.67•10⁻⁸•3000⁴= 33.8 W

To calculate the electric power supplied to the filament of a light bulb, you can use the Stefan-Boltzmann law and the equation for electric power.

The Stefan-Boltzmann law states that the total radiated power from a black body (or an idealized filament) is proportional to the fourth power of its absolute temperature:

P = σ * A * ε * T^4

Where:
P = Power radiated (in watts)
σ = Stefan-Boltzmann constant (5.67 × 10^-8 W/m^2K^4)
A = Surface area of the filament (in square meters)
ε = Emissivity of the filament
T = Absolute temperature of the filament (in Kelvin)

Let's plug in the given values:
σ = 5.67 × 10^-8 W/m^2K^4
A = 8 × 10^-6 m^2
ε = 0.92
T = 3000K

P = (5.67 × 10^-8) * (8 × 10^-6) * 0.92 * (3000)^4

Now let's calculate the power:

P = 1.14 * 10^-3 W

Therefore, the electric power that must be supplied to the filament of the light bulb is approximately 1.14 milliwatts (1.14 mW).

To calculate the electric power required to supply the filament of a light bulb, we need to use the Stefan-Boltzmann Law, which relates the power radiated by a black body to its temperature and emissivity.

The Stefan-Boltzmann Law states that the power radiated by a black body is proportional to the fourth power of its temperature and its emissivity. Mathematically, it can be written as:

P = εσAT⁴

Where:
P is the power radiated (in watts)
ε is the emissivity of the filament
σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/(m^2K^4))
A is the total surface area of the filament (in square meters)
T is the temperature of the filament (in kelvin)

In this case, we are given:
ε = 0.92
A = 8 x 10^-6 m^2
T = 3000 K

Let's substitute these values into the equation and calculate the power radiated:

P = (0.92)(5.67 x 10^-8 W/(m^2K^4))(8 x 10^-6 m^2)(3000 K)^4

P = (0.92)(5.67 x 10^-8 W/m^2K^4)(8 x 10^-6 m^2)(81 x 10^16 K^4)

P = (0.92)(4.5928 x 10^10 W)(8 x 81 x 10^10)

P ≈ 0.35 x 10^11 W

Therefore, the electric power that must be supplied to the filament of the light bulb operating at 3000 K is approximately 0.35 x 10^11 W.