Question

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 1

Answer:

The value is   $A = 2.80 *10^{-4} \ m^2$

Explanation:

From the question we are told that

The  operating temperature is  $T = 2450 \ K$

The emissivity is  $e = 0.350$

 The  power rating is  $P = 200 \ W$

Generally the area is mathematically represented as

      $A = \frac{P}{ e * \sigma * T^2}$

Where  $\sigma$ is the Stefan Boltzmann constant  with value  

      $\sigma = 5.67 *10^{-8} \ W/m^2\cdot K^4$

So

     $A = \frac{200}{0.350 * 5.67*10^{-8} * 2450^{4}}$

     $A = 2.80 *10^{-4} \ m^2$