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A 100 W light bulb has a tungsten filament (emissivity = 0.30) which is required to operate at 2780 K. If the bulb is completely evacuated, calculate the minimum surface area of the tungsten filament
- a)0.98 * 10 -4 m2
- b)1.98 * 10 -4 m2
- c)2.98 * 10 -4 m2
- d)3.98 * 10 -4 m2
Correct answer is option 'A'. Can you explain this answer?
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A 100 W light bulb has a tungsten filament (emissivity = 0.30) which i...
E = (Emissivity) σ A T4. So, A = 0.98 * 10 -4 m2.
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A 100 W light bulb has a tungsten filament (emissivity = 0.30) which i...
Calculating the minimum surface area of the tungsten filament:
To calculate the minimum surface area of the tungsten filament, we can use the Stefan-Boltzmann Law, which relates the power radiated by a blackbody to its temperature and surface area.
Stefan-Boltzmann Law:
The power radiated by a blackbody is given by the equation:
\[ P = \varepsilon \sigma A T^4 \]
Where:
- \( P \) = Power radiated (100 W)
- \( \varepsilon \) = Emissivity of tungsten filament (0.30)
- \( \sigma \) = Stefan-Boltzmann constant (\( 5.67 \times 10^{-8} W/m^2K^4 \))
- \( A \) = Surface area of the tungsten filament
- \( T \) = Temperature (2780 K)
Calculating the surface area:
Substitute the given values into the Stefan-Boltzmann Law equation and solve for the surface area:
\[ 100 = 0.30 \times 5.67 \times 10^{-8} \times A \times 2780^4 \]
\[ A = \frac{100}{0.30 \times 5.67 \times 10^{-8} \times 2780^4} \]
Calculating the value of A, we get:
\[ A = 0.98 \times 10^{-4} m^2 \]
Therefore, the minimum surface area of the tungsten filament required to operate at 2780 K is approximately \( 0.98 \times 10^{-4} m^2 \). Hence, option 'A' is the correct answer.
To calculate the minimum surface area of the tungsten filament, we can use the Stefan-Boltzmann Law, which relates the power radiated by a blackbody to its temperature and surface area.
Stefan-Boltzmann Law:
The power radiated by a blackbody is given by the equation:
\[ P = \varepsilon \sigma A T^4 \]
Where:
- \( P \) = Power radiated (100 W)
- \( \varepsilon \) = Emissivity of tungsten filament (0.30)
- \( \sigma \) = Stefan-Boltzmann constant (\( 5.67 \times 10^{-8} W/m^2K^4 \))
- \( A \) = Surface area of the tungsten filament
- \( T \) = Temperature (2780 K)
Calculating the surface area:
Substitute the given values into the Stefan-Boltzmann Law equation and solve for the surface area:
\[ 100 = 0.30 \times 5.67 \times 10^{-8} \times A \times 2780^4 \]
\[ A = \frac{100}{0.30 \times 5.67 \times 10^{-8} \times 2780^4} \]
Calculating the value of A, we get:
\[ A = 0.98 \times 10^{-4} m^2 \]
Therefore, the minimum surface area of the tungsten filament required to operate at 2780 K is approximately \( 0.98 \times 10^{-4} m^2 \). Hence, option 'A' is the correct answer.
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