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

the moon affects the earth gravitation ally, causing certain motions of that body and also producing tides in its ocean and atmosphere. the tidal wave in our atmosphere is, however, far too minute to be an effective cause in any known meteorological phenomenon.

explanation:

total distance/total time

(85 + 82)/(5+3)

167/8

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