Question

assuming a filament in a 120W light bulb acts like a prefect blackbody, what is the temperature of the hottest portion of the filament if it has a surface area of 6.4×10^_5m^2. The stefan- boltzmann constant is 5.67×10^-8W/(m2.k2) A. 12OOk B. 2400K C. 2100K​

Answer 1

T = 2398 K

Step-by-step explanation:

To calculate the emission of the light bulb we use the law is Stefan

P = σ A e T⁴

as they indicate that the filament is a black body, the emissivity is equal to 1 (e = 1)

T =
`\sqrt[4]{(P)/( \sigma A) }`

let's calculate

T =
`\sqrt[4]{(120)/(5.67 \ 10^(-8) \ 6.4 \ 10^(-5)) }`

T =
`\sqrt[4]{33.06878 \ 10^(12) }`

T = 2,398 10³ K

T = 2398 K