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Chapter 14: Problem 61

An incandescent lightbulb has a tungsten filament that is heated to a temperature of \(3.00 \times 10^{3} \mathrm{K}\) when an electric current passes through it. If the surface area of the filament is approximately $1.00 \times 10^{-4} \mathrm{m}^{2}\( and it has an emissivity of \)0.32,$ what is the power radiated by the bulb?

Short Answer

Expert verified
Answer: The power radiated by the incandescent lightbulb is approximately \(155.84 \mathrm{W}\).

Step by step solution

01

Define the given parameters

Temperature (T): \(3.00 \times 10^{3} \mathrm{K}\) Surface area (A): \(1.00 \times 10^{-4} \mathrm{m}^{2}\) Emissivity (ε): \(0.32\)
02

Define the Stefan-Boltzmann constant

Stefan-Boltzmann constant (σ): \(5.67 \times 10^{-8} \mathrm{Wm}^{-2}\mathrm{K}^{-4}\)
03

Apply the Stefan-Boltzmann law

The power radiated per unit area by a black body (P) can be calculated using the Stefan-Boltzmann law: \(P = \sigma T^4\)
04

Calculate the power radiated by the filament

First, we need to find the power radiated per unit area: \(P = \sigma T^4 = (5.67 \times 10^{-8} \mathrm{Wm}^{-2}\mathrm{K}^{-4}) (3.00 \times 10^{3} \mathrm{K})^4 = 4.87 \times 10^6 \mathrm{Wm}^{-2}\) Now, we can multiply the power per unit area by the total surface area to find the total power radiated by the filament as a black body: \(P_\text{black body} = P \times A = (4.87 \times 10^6 \mathrm{Wm}^{-2}) (1.00 \times 10^{-4} \mathrm{m}^{2}) = 487 \mathrm{W}\)
05

Adjust for emissivity

Finally, to find the actual power radiated by the filament, we need to adjust the black body power by multiplying by the emissivity: \(P_\text{radiated} = ε \times P_\text{black body} = 0.32 \times 487 \mathrm{W} = 155.84 \mathrm{W}\)
06

Provide the final answer

The power radiated by the incandescent lightbulb is approximately \(155.84 \mathrm{W}\).

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