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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.

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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]

fichoh

Using the relation between surface area, temperature and power, the surface area of the filament is [tex] A = 2.80 \times 10^{-4} m^{2} [/tex]

Using the relation, we can obtain the surface area of the filament :

[tex] A = \frac{P}{σ \times \epsilon \times T^{4}[/tex]

T = temperature = 2450 k

P = Power = 200 W

e = Emissivity = 0.350

σ = Boltzmann constant = [tex]5.67 \times 10^{-8}[/tex]

Substituting the values into the relation :

[tex] A = \frac{P}{σ \times \epsilon \times T^{4}}[/tex]

[tex] A = \frac{200}{5.67 \times 10^{-8} \times 0.350 \times 2450^{4}}[/tex]

[tex] A = \frac{200}{715015.47403125} [/tex]

[tex] A = 2.80 \times 10^{-4} [/tex]

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