Question

Assume that the surface temperature of the lightbulb filament when it is plugged into an outlet of 120 V is about 3000 K and the power rating of the bulb is the electric energy it consumes (not what it radiates). Incandescent lightbulbs usually radiate in visible light about 10% of the electric energy that they consume. Estimate the surface area of a 40-watt lightbulb filament. Express your answer with the appropriate units. A = 8.7×10−7 m2

Answer 1

Answer:

A=$8.7\times 10^{-7}m^{2}$

Explanation:

We have given power =40 W but only 10% of this power is used so actual power $P=\frac{40\times 10}{100}=4 W$

We know that from Stefan's law $p=\sigma AT^4$ where $\sigma$ is Boltzmann constant which value is $5.6\times 10^{-8}Wm^{-2}K^{-4}$

So $4=5.67\times 10^{-8}\times A\times 3000^4$

A=$8.7\times 10^{-7}m^{2}$

Answer 2

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:

$\boxed{P = \sigma A T^4}$

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:

$P = \sigma A T^4$

$A = P \div ( \sigma T^4 )$

$A = 4 \div ( 5.67 \times 10^{-8} \times 3000^4 )$

$\boxed{A \approx 8.7 \times 10^{-7} \texttt{ m}^2}$

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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 : brainly.com/question/5597682
• Flow of Heat : brainly.com/question/3010079
• Difference Between Temperature and Heat : brainly.com/question/3821712

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Answer details

Grade: College

Subject: Physics

Chapter: Thermal Physics