Final answer:
Using the Stefan-Boltzmann law and assuming an emissivity of 1, the surface area of a filament in a 100 W lightbulb at 2.5 KK is found to be approximately 45.17 mm².
Step-by-step explanation:
To find the surface area of the filament of a 100 W lightbulb when it's at 2.5 KK (or 2500 Kelvin), we can use the Stefan-Boltzmann law, which relates the power emitted per unit area of a black body to its temperature.
The Stefan-Boltzmann equation is:
P = eσAT^4
Where:
P = Power emitted (Watts)
e = Emissivity (dimensionless)
σ = Stefan-Boltzmann constant (5.67 x 10^-8 W/m^2K^4)
A = Surface area (m^2)
T = Absolute temperature (Kelvin)
Given that P = 100 W, e = 1 (for black bodies), and T = 2.5 KK, we can rearrange to solve for the surface area A:
A = P / (eσT^4)
A = 100 W / (1 × 5.67 x 10^-8 W/m^2K^4 × (2500 K)^4)
After calculating, we then convert the area from square meters to square millimeters by multiplying the result by 1,000,000 (since 1 m^2 = 1,000,000 mm^2).
Substituting the values and performing the calculation:
A = 100 / (1 × 5.67 x 10^-8 × 2500^4)
A = 100 / (1 × 5.67 x 10^-8 × 3.90625 x 10^13)
A = 100 / (5.67 x 10^-8 × 3.90625 x 10^13)
A = 100 / (2.2140625 x 10^6)
A = 4.517456 x 10^-5 m^2
A in mm^2 = 4.517456 x 10^-5 m^2 × 1,000,000 mm^2/m^2
A in mm^2 = 45.17456 mm^2
The surface area of the filament is approximately 45.17 mm^2.