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you turn up the dimmer switch so that the temperature of the filament reaches 2600 degrees c. the light bulb filament has an area of 6.45 x 10⁻⁴ m² (0.1 square inches) and an emissivity of 0.8. how much electrical power must it be using?

User Ross
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1 Answer

4 votes

Final answer:

To find the power usage of the light bulb filament at 2600°C, the Stefan-Boltzmann law is used, incorporating the area of the filament, its emissivity, temperature in Kelvin, and the Stefan-Boltzmann constant, to calculate the power radiated.

Step-by-step explanation:

To calculate the electrical power being used by the light bulb filament at a temperature of 2600 degrees Celsius, we need to apply the Stefan-Boltzmann law for black-body radiation, which states that the power radiated per unit area of a black body is directly proportional to the fourth power of the black body's thermodynamic temperature.

The formula for the Stefan-Boltzmann law is:

P = ε·A·σ·T^4

Where:

P is the power radiated,

ε is the emissivity of the material (0.8 for the filament),

A is the area of the emitting body in square meters (6.45 x 10⁻⁴ m² for the filament),

σ is the Stefan-Boltzmann constant (≈ 5.67 x 10⁻⁸ W/m²K⁴),

T is the absolute temperature in Kelvins (2600°C = 2873K).

We can rearrange this formula to solve for P (the power) and substitute in the values:

P = 0.8 · 6.45 x 10⁻⁴ m² · 5.67 x 10⁻⁸ W/m²K⁴ · (2873K)⁴

Upon calculating, we find that the filament's power usage is significant, which is typical for an incandescent bulb operating at such high temperatures.

User Prateek Narendra
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9.2k points
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