Answer :
Answer:
A=[tex]8.7\times 10^{-7}m^{2}[/tex]
Explanation:
We have given power =40 W but only 10% of this power is used so actual power [tex]P=\frac{40\times 10}{100}=4 W[/tex]
We know that from Stefan's law [tex]p=\sigma AT^4[/tex] where [tex]\sigma[/tex] is Boltzmann constant which value is [tex]5.6\times 10^{-8}Wm^{-2}K^{-4}[/tex]
So [tex]4=5.67\times 10^{-8}\times A\times 3000^4[/tex]
A=[tex]8.7\times 10^{-7}m^{2}[/tex]
The surface area of the 40-watt lightbulb filamen is about 8.7 × 10⁻⁷ m²
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Further explanation
Let's recall Heat Transfer Rate of Radiation as follows:
[tex]\boxed{P = \sigma A T^4}[/tex]
where:
P = Heat Transfer Rate ( W )
σ = Stefan-Boltzman Constant ( 5.67 × 10⁻⁸ W/m²K⁴ )
A = Surface Area ( m² )
T = Temperature ( K )
Let us now tackle the problem!
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Given:
temperature = T = 3000 K
power of lightbulb filament = P_total = 40 W
efficiency of light bulb = η = 10%
heat transfer rate = P = 10% × ( 40 W ) = 4 W
voltage = V = 120 V
Asked:
surface area = A = ?
Solution:
[tex]P = \sigma A T^4[/tex]
[tex]A = P \div ( \sigma T^4 )[/tex]
[tex]A = 4 \div ( 5.67 \times 10^{-8} \times 3000^4 )[/tex]
[tex]\boxed{A \approx 8.7 \times 10^{-7} \texttt{ m}^2}[/tex]
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Conclusion:
The surface area of the 40-watt lightbulb filamen is about 8.7 × 10⁻⁷ m²
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Learn more
- Efficiency of Engine : https://brainly.com/question/5597682
- Flow of Heat : https://brainly.com/question/3010079
- Difference Between Temperature and Heat : https://brainly.com/question/3821712
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Answer details
Grade: College
Subject: Physics
Chapter: Thermal Physics
