The operating temperature of a tungsten filament in an incandescent light bulb is 2450 K, and its emissivity is 0.350. Find the surface area of the filament of a 200 W bulb if all the electrical energy consumed by the bulb is radiated by the filament as electromagnetic waves.
Answer:
The value is [tex]A = 2.80 *10^{-4} \ m^2[/tex]
Explanation:
From the question we are told that
The operating temperature is [tex]T = 2450 \ K[/tex]
The emissivity is [tex]e = 0.350[/tex]
The power rating is [tex]P = 200 \ W[/tex]
Generally the area is mathematically represented as
[tex]A = \frac{P}{ e * \sigma * T^2}[/tex]
Where [tex]\sigma[/tex] is the Stefan Boltzmann constant with value
[tex]\sigma = 5.67 *10^{-8} \ W/m^2\cdot K^4[/tex]
So
[tex]A = \frac{200}{0.350 * 5.67*10^{-8} * 2450^{4}}[/tex]
[tex]A = 2.80 *10^{-4} \ m^2[/tex]