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

A=

Explanation:

We have given power =40 W but only 10% of this power is used so actual power

We know that from Stefan's law where is Boltzmann constant which value is

So

A=

The surface area of the 40-watt lightbulb filamen is about 8.7 × 10⁻⁷ m²

Let's recall Heat Transfer Rate of Radiation as follows:

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!

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:

The surface area of the 40-watt lightbulb filamen is about 8.7 × 10⁻⁷ m²

Grade: College

Subject: Physics

Chapter: Thermal Physics

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