Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 1: Problem 158

Over 90 percent of the energy dissipated by an incandescent light bulb is in the form of heat, not light. What is the temperature of a vacuum-enclosed tungsten filament with an exposed surface area of \(2.03 \mathrm{~cm}^{2}\) in a \(100 \mathrm{~W}\) incandescent light bulb? The emissivity of tungsten at the anticipated high temperatures is about \(0.35\). Note that the light bulb consumes \(100 \mathrm{~W}\) of electrical energy, and dissipates all of it by radiation. (a) \(1870 \mathrm{~K}\) (b) \(2230 \mathrm{~K}\) (c) \(2640 \mathrm{~K}\) (d) \(3120 \mathrm{~K}\) (e) \(2980 \mathrm{~K}\)

Short Answer

Expert verified
Answer: The temperature of the tungsten filament is approximately 2640 K.

Step by step solution

01

Write down the Stefan-Boltzmann Law formula

The formula for Stefan-Boltzmann Law is given by $$P = e \times A \times \sigma \times T^4$$ where \(P\) is the power radiated by the object, \(e\) is the emissivity, \(A\) is the surface area of the object, \(\sigma\) is the Stefan-Boltzmann constant, and \(T\) is the temperature in Kelvin.
02

Substitute the given values into the formula

We are given the following values: - \(P = 100 \mathrm{~W}\) (power consumed by the light bulb and dissipated by radiation) - \(e = 0.35\) (emissivity of tungsten) - \(A = 2.03 \mathrm{~cm}^2 = 2.03 \times 10^{-4} \mathrm{~m}^2\) (exposed surface area of the filament) - \(\sigma = 5.67 \times 10^{-8} \mathrm{W m^{-2} K^{-4}}\) (Stefan-Boltzmann constant) Now we can substitute these values into the formula: $$100 = 0.35 \times 2.03 \times 10^{-4} \times 5.67 \times 10^{-8} \times T^4$$
03

Solve for the temperature T

To solve for the temperature T, we can first simplify the equation as follows: $$100 = 1.260245 \times 10^{-11} \times T^4$$ Now, divide by \(1.260245 \times 10^{-11}\) to isolate \(T^4\): $$T^4 = \frac{100}{1.260245 \times 10^{-11}}$$ Evaluate the expression on the right side: $$T^4 = 7.935 \times 10^{11}$$ Now, take the fourth root of the result to find the temperature T: $$T = (7.935 \times 10^{11})^{\frac{1}{4}}$$ Calculate the temperature T: $$T \approx 2640 \mathrm{~K}$$ Hence, the temperature of the tungsten filament is around \(2640 \mathrm{~K}\). The correct answer is (c) \(2640 \mathrm{~K}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Engine valves \(\left(c_{p}=440 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.\) and \(\left.\rho=7840 \mathrm{~kg} / \mathrm{m}^{3}\right)\) are to be heated from \(40^{\circ} \mathrm{C}\) to \(800^{\circ} \mathrm{C}\) in \(5 \mathrm{~min}\) in the heat treatment section of a valve manufacturing facility. The valves have a cylindrical stem with a diameter of \(8 \mathrm{~mm}\) and a length of \(10 \mathrm{~cm}\). The valve head and the stem may be assumed to be of equal surface area, with a total mass of \(0.0788 \mathrm{~kg}\). For a single valve, determine ( \(a\) ) the amount of heat transfer, \((b)\) the average rate of heat transfer, \((c)\) the average heat flux, and \((d)\) the number of valves that can be heat treated per day if the heating section can hold 25 valves and it is used 10 h per day.

A \(3-\mathrm{m}^{2}\) black surface at \(140^{\circ} \mathrm{C}\) is losing heat to the surrounding air at \(35^{\circ} \mathrm{C}\) by convection with a heat transfer coefficient of \(16 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and by radiation to the surrounding surfaces at \(15^{\circ} \mathrm{C}\). The total rate of heat loss from the surface is (a) \(5105 \mathrm{~W}\) (b) \(2940 \mathrm{~W}\) (c) \(3779 \mathrm{~W}\) (d) \(8819 \mathrm{~W}\) (e) \(5040 \mathrm{~W}\)

A 2-in-diameter spherical ball whose surface is maintained at a temperature of \(170^{\circ} \mathrm{F}\) is suspended in the middle of a room at \(70^{\circ} \mathrm{F}\). If the convection heat transfer coefficient is \(15 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\) and the emissivity of the surface is \(0.8\), determine the total rate of heat transfer from the ball.

A flat-plate solar collector is used to heat water by having water flow through tubes attached at the back of the thin solar absorber plate. The absorber plate has a surface area of \(2 \mathrm{~m}^{2}\) with emissivity and absorptivity of \(0.9\). The surface temperature of the absorber is \(35^{\circ} \mathrm{C}\), and solar radiation is incident on the absorber at \(500 \mathrm{~W} / \mathrm{m}^{2}\) with a surrounding temperature of \(0^{\circ} \mathrm{C}\). Convection heat transfer coefficient at the absorber surface is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the ambient temperature is \(25^{\circ} \mathrm{C}\). Net heat rate absorbed by the solar collector heats the water from an inlet temperature \(\left(T_{\text {in }}\right)\) to an outlet temperature \(\left(T_{\text {out }}\right)\). If the water flow rate is \(5 \mathrm{~g} / \mathrm{s}\) with a specific heat of \(4.2 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\), determine the temperature rise of the water.

\(80^{\circ} \mathrm{C}\). Also, determine the convection heat transfer coefficients at the beginning and at the end of the heating process. 1-133 It is well known that wind makes the cold air feel much colder as a result of the wind chill effect that is due to the increase in the convection heat transfer coefficient with increasing air velocity. The wind chill effect is usually expressed in terms of the wind chill temperature (WCT), which is the apparent temperature felt by exposed skin. For outdoor air temperature of \(0^{\circ} \mathrm{C}\), for example, the wind chill temperature is \(-5^{\circ} \mathrm{C}\) at \(20 \mathrm{~km} / \mathrm{h}\) winds and \(-9^{\circ} \mathrm{C}\) at \(60 \mathrm{~km} / \mathrm{h}\) winds. That is, a person exposed to \(0^{\circ} \mathrm{C}\) windy air at \(20 \mathrm{~km} / \mathrm{h}\) will feel as cold as a person exposed to \(-5^{\circ} \mathrm{C}\) calm air (air motion under \(5 \mathrm{~km} / \mathrm{h}\) ). For heat transfer purposes, a standing man can be modeled as a 30 -cm- diameter, 170-cm-long vertical cylinder with both the top and bottom surfaces insulated and with the side surface at an average temperature of \(34^{\circ} \mathrm{C}\). For a convection heat transfer coefficient of \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat loss from this man by convection in still air at \(20^{\circ} \mathrm{C}\). What would your answer be if the convection heat transfer coefficient is increased to \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) as a result of winds? What is the wind chill temperature in this case?
See all solutions

Recommended explanations on Physics Textbooks

Work Energy and Power

Read Explanation

Thermodynamics

Read Explanation

Radiation

Read Explanation

Modern Physics

Read Explanation

Waves Physics

Read Explanation

Translational Dynamics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free