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antoniya [11.8K]
3 years ago
13

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.
Physics
1 answer:
natulia [17]3 years ago
8 0

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

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