A 100W electric light bulb has diameter of (8)10^-5m.Estimate working temperature of solar surface Ts=6000K,earth sun distance d=(1.5)(100^11)m (1)1900K (2)2018K (3)2800K (4)2946K

To estimate the working temperature of a 100W electric light bulb based on the given parameters, we can use the Stefan-Boltzmann Law, which relates the power radiated by a black body to its temperature. This law states that the power emitted per unit area of a black body is proportional to the fourth power of its absolute temperature. Let's break this down step by step.

Understanding the Stefan-Boltzmann Law

The Stefan-Boltzmann Law is expressed mathematically as:

P = σAT^4

Where:

• P is the power emitted (in watts)
• σ is the Stefan-Boltzmann constant (approximately 5.67 x 10^-8 W/m²K⁴)
• A is the surface area of the emitting body (in m²)
• T is the absolute temperature (in Kelvin)

Calculating the Surface Area of the Bulb

First, we need to calculate the surface area of the light bulb. Since the bulb is approximately spherical, we can use the formula for the surface area of a sphere:

A = 4πr²

Given the diameter of the bulb is 8 x 10^-5 m, the radius (r) will be half of that:

r = (8 x 10^-5 m) / 2 = 4 x 10^-5 m

Now, substituting the radius into the surface area formula:

A = 4π(4 x 10^-5 m)²

A ≈ 4π(16 x 10^-10 m²) ≈ 2.01 x 10^-9 m²

Applying the Stefan-Boltzmann Law

Now that we have the surface area, we can rearrange the Stefan-Boltzmann Law to solve for temperature:

T = (P / (σA))^(1/4)

Substituting the values:

• P = 100 W
• σ = 5.67 x 10^-8 W/m²K⁴
• A ≈ 2.01 x 10^-9 m²

Now plug these values into the equation:

T = (100 W / (5.67 x 10^-8 W/m²K⁴ * 2.01 x 10^-9 m²))^(1/4)

Calculating the denominator:

5.67 x 10^-8 W/m²K⁴ * 2.01 x 10^-9 m² ≈ 1.14 x 10^-16 W/K⁴

Now substituting back into the equation:

T = (100 W / 1.14 x 10^-16 W/K⁴)^(1/4)

Calculating the fraction:

T ≈ (8.77 x 10^16 K⁴)^(1/4)

Taking the fourth root:

T ≈ 2946 K

Final Result

Based on our calculations, the estimated working temperature of the 100W electric light bulb is approximately 2946 K. Therefore, the correct answer from the options provided is (4) 2946K.