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Chapter 14: Problem 65

An incandescent light bulb radiates at a rate of \(60.0 \mathrm{W}\) when the temperature of its filament is \(2820 \mathrm{K}\). During a brownout (temporary drop in line voltage), the power radiated drops to \(58.0 \mathrm{W} .\) What is the temperature of the filament? Neglect changes in the filament's length and cross-sectional area due to the temperature change. (tutorial: light bulb)

Short Answer

Expert verified
Answer: The approximate temperature of the filament during the brownout is 2790 K.

Step by step solution

01

Write down the Stefan-Boltzmann's Law formula relating power and temperature

We will use the Stefan-Boltzmann's Law formula which relates power and temperature: \(P = \sigma AeT^4\)
02

Find the ratio of the power radiated during the brownout to the original power radiated

Find the ratio of the power radiated during the brownout (\(P_{2} = 58.0 \mathrm{W}\)) to the original power radiated (\(P_{1} = 60.0 \mathrm{W}\)): \(\frac{P_{2}}{P_{1}} = \frac{58.0}{60.0}\)
03

Calculate the ratio of the temperature to the fourth power

Since we are neglecting changes in the filament's length, cross-sectional area, and emissivity, we can express the ratio of temperatures as \(\frac{T_2^4}{T_1^4}\), where \(T_1\) is the initial temperature and \(T_2\) is the temperature during the brownout. Using the ratio of the powers, we can write: \(\frac{T_2^4}{T_1^4} = \frac{P_2}{P_1} = \frac{58.0}{60.0}\)
04

Find the temperature during the brownout

In order to find the temperature of the filament during the brownout (\(T_2\)), we can use the following equation with the known initial temperature (\(T_1 = 2820 \mathrm{K}\)): \(T_2^4 = \frac{58.0}{60.0}T_1^4\) Now, solve for \(T_2\): \(T_2 = \sqrt[4]{\frac{58.0}{60.0}(2820)^4} = 2790 \mathrm{K}\) The temperature of the filament during the brownout is approximately \(2790 \mathrm{K}\).

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Most popular questions from this chapter

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