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​

Respuesta :

Answer:

T = 2398 K

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 = [tex]\sqrt[4]{\frac{P}{ \sigma A} }[/tex]

    let's calculate

           T =[tex]\sqrt[4]{\frac{120}{5.67 \ 10^{-8} \ 6.4 \ 10^{-5}} }[/tex]

           T = [tex]\sqrt[4]{33.06878 \ 10^{12} }[/tex]

           T = 2,398 10³ K

           T = 2398 K