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Chapter 14: Problem 62

A tungsten filament in a lamp is heated to a temperature of $2.6 \times 10^{3} \mathrm{K}\( by an electric current. The tungsten has an emissivity of \)0.32 .$ What is the surface area of the filament if the lamp delivers $40.0 \mathrm{W}$ of power?

Short Answer

Expert verified
Answer: The approximate surface area of the tungsten filament is \(4.24 \times 10^{-6} m^2\).

Step by step solution

01

Write down the given information and the Stefan-Boltzmann Law formula

We are given: \(T = 2.6 \times 10^{3} K\) \(ε = 0.32\) \(P = 40.0 W\) \(σ = 5.67 \times 10^{-8} Wm^{-2}K^{-4}.\) The Stefan-Boltzmann Law formula is: \(P = ε × σ × A × T^4\)
02

Rearrange the formula to solve for the surface area (A)

We need to find \(A\). Rearrange the formula: \(A = \frac{P}{ε × σ × T^4}\)
03

Plug in the given values and calculate the surface area (A)

Now, we can plug in the given values to the formula: \(A = \frac{40.0 W}{0.32 \times (5.67 \times 10^{-8} Wm^{-2}K^{-4}) \times (2.6 \times 10^{3} K)^4}\)
04

Perform the calculations

After plugging in the values, perform the calculations to find the surface area: \(A = \frac{40.0}{0.32 \times 5.67 \times 10^{-8} \times (2.6 \times 10^{3})^4} \approx 4.24 \times 10^{-6} m^2\) So, the surface area of the tungsten filament is approximately \(4.24 \times 10^{-6} m^2\).

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