/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 \(\mathrm{A} 100 \mathrm{~W}\) i... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\(\mathrm{A} 100 \mathrm{~W}\) incandescent light bulb has a cylindrical tungsten filament \(30.0 \mathrm{~cm}\) long. \(0.40 \mathrm{~mm}\) in diameter, and with an emissivity of \(0.26 .\) (a) What is the temperature of the filament? (b) For what wavelength does the spectral emittance of the bulb peak? (c) Incandescent light bulbs are not very efficient sources of visible light. Explain why this is so.

Short Answer

Expert verified
In the exercise, it is calculated that the temperature of the filament is approximately 2939 K. The spectral emittance of the bulb peaks in the infrared region at a wavelength of approximately \(9.88 × 10^{-7}m\). Incandescent light bulbs are not efficient as sources of visible light because most of the energy is emitted as infrared radiation (heat), rather than visible light.

Step by step solution

01

Calculate the filament surface area

First, the surface area (\(A\)) of the filament has to be calculated. Since it has a cylindrical shape, use the formula for the surface area of a cylinder: \(A = \pi d l\), where \(d\) is the diameter and \(l\) is the length. Convert the diameter from mm to m: \(d = 0.40 mm = 0.40 × 10^{-3} m\). Then substitute values and calculate: \(A = \pi × 0.40 × 10^{-3} m × 30.0 cm = 0.0377 m^2 \).
02

Solve for temperature using the Stefan-Boltzmann Law

Then, use the Stefan-Boltzmann Law: \(P = e σ A T^4\), where \(P\) is power, \(e\) is emissivity, \(σ\) is Stefan’s constant (\(5.67 × 10^{-8} W/m^{2}K^{4}\)) and \(T\) is temperature. Rewrite to solve for \(T\): \(T = \left[\frac{P}{e \sigma A}\right]^{(\frac{1}{4})}\). Substitute the values into the equation: \(T = \left[\frac{100 W}{0.26 × 5.67 × 10^{-8} W/m^{2}K^{4} × 0.0377 m^{2}}\right]^{(\frac{1}{4})}\) and calculate \(T ≈ 2939 K\).
03

Use Wien’s Law to determine spectral emittance peak

For part (b), use Wien’s Law (\(λ_{max} = \frac{b}{T}\)), where \(λ_{max}\) is the peak wavelength, \(b\) is Wien’s constant (\(2.9 × 10^{-3} mK\)) and \(T\) is temperature. Substitute the values into the equation: \(λ_{max} = \frac{2.9 × 10^{-3}mK}{2939K} ≈ 9.88 × 10^{-7} m\). This is in the infrared region, showing that the bulb emits mostly infrared light.
04

Explain the efficiency of incandescent light bulbs

For part (c), explain: Incandescent light bulbs are not efficient as sources of visible light because most of the energy is emitted as infrared radiation (heat), rather than visible light. The prior calculations showed that the peak emittance is in the infrared region. This means that a lot of power is wasted as heat instead of light.

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stefan-Boltzmann Law
The Stefan-Boltzmann Law is fundamental in understanding how incandescent light bulbs work, particularly in determining the temperature of the filament. This law states that the power radiated by a black body is proportional to the fourth power of its absolute temperature. Mathematically, it's expressed as \(P = e \sigma A T^4\).
This equation can tell us much about the filament's temperature based on the power output, emissivity of the material, and the surface area of the filament. In the case of an incandescent bulb, the law was used to calculate the filament temperature as approximately 2939 K.
  • \(P\) represents the power emitted.
  • \(e\) is the emissivity, a measure of how efficiently a surface emits thermal radiation compared to a perfect emitter or black body.
  • \(\sigma\) is Stefan's constant, \(5.67 \times 10^{-8} \text{W/m}^2\text{K}^4\).
  • \(A\) is the surface area, and \(T\) is the absolute temperature.
By understanding this law, students can comprehend the immense power loss as heat rather than visible light in incandescent bulbs.
Wien's Law
Wien's Law is crucial for determining the peak spectral emittance of an incandescent bulb. It illustrates the relationship between the temperature of a black body and the wavelength at which it emits the most radiation. The law is expressed in the formula: \(\lambda_{max} = \frac{b}{T}\).
Here, \(\lambda_{max}\) is the peak wavelength, where the bulb emits the most intensity. The constant \(b\) is Wien's displacement constant, \(2.9 \times 10^{-3} \text{mK}\).
Through Wien's Law, we find out that as the temperature increases, the wavelength decreases, implying a shift towards the blue spectrum. However, for the incandescent bulb's filament at about 2939 K, the wavelength is around \(9.88 \times 10^{-7} \text{m}\), peaking in the infrared range. This indicates that much of the energy emitted is not visible light but infrared radiation (heat).
Energy Efficiency
Energy efficiency is a significant consideration in evaluating incandescent light bulbs. Despite their classic design and wide usage, incandescent bulbs emit more heat than light, making them less efficient compared to modern options.
  • Incandescent bulbs transform less than 10% of their energy consumption into visible light.
  • The majority of the energy is wasted as infrared radiation, which is not visible to the human eye.
  • This inefficiency contributes to higher energy costs and more substantial environmental impacts.
Understanding this energy dynamic helps in appreciating why more efficient lighting solutions, such as LEDs and CFLs, are preferred in today’s environmental and cost-conscious society.
Infrared Radiation
Infrared radiation refers to the type of electromagnetic radiation emitted by incandescent bulbs in large quantities. The filament in these bulbs emits light when heated, but much of this energy falls outside the visible spectrum into the infrared region.
Infrared light comprises longer wavelengths than visible light, making it invisible to the human eye but felt as heat.
  • This radiation is why incandescent bulbs feel hot to touch when on.
  • It contributes to the inefficiency of these bulbs since a large portion of electrical energy is converted into heat.
  • This property is also leveraged in applications where heat is needed, such as in heat lamps or incubators.
Knowing about infrared radiation enhances understanding of the inefficiencies and alternate uses of incandescent bulbs.
Tungsten Filament
The tungsten filament is a crucial component of an incandescent bulb. Tungsten is chosen due to its high melting point and ability to withstand high temperatures, making it ideal for producing light in incandescents.
  • The filament heats up to temperatures around 2500-3000 K.
  • Tungsten's high melting point means it's durable and can glow brightly without melting or breaking.
  • Despite these qualities, tungsten filaments are still inefficient since much energy is lost as heat.
This inefficiency dramatically contrasts the preferred lighting technologies today, which utilize materials and methods that offer greater energy conversion into visible light. Understanding the role of tungsten filaments helps to clarify the limitations of incandescent light bulbs in energy efficiency.

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

BIO Removing Birthmarks. Pulsed dye lasers emit light of wavelength \(585 \mathrm{nm}\) in \(0.45 \mathrm{~ms}\) pulses to remove skin blemishes such as birthmarks. The beam is usually focused onto a circular spot \(5.0 \mathrm{~mm}\) in diameter. Suppose that the output of one such laser is \(20.0 \mathrm{~W}\). (a) What is the energy of each photon, in eV? (b) How many photons per square millimeter are delivered to the blemish during each pulse?

Imagine another universe in which the value of Planck's constant is \(0.0663 \mathrm{~J} \cdot \mathrm{s},\) but in which the physical laws and all other physical constants are the same as in our universe. In this universe, two physics students are playing catch. They are \(12 \mathrm{~m}\) apart, and one throws a \(0.25 \mathrm{~kg}\) ball directly toward the other with a speed of \(6.0 \mathrm{~m} / \mathrm{s}\). (a) What is the uncertainty in the ball's horizontal momentum, in a direction perpendicular to that in which it is being thrown, if the student throwing the ball knows that it is located within a cube with volume \(125 \mathrm{~cm}^{3}\) at the time she throws it? (b) By what horizontal distance could the ball miss the second student?

(a) In an electron microscope, what accelerating voltage is needed to produce electrons with wavelength \(0.0600 \mathrm{nm} ?\) (b) If protons are used instead of electrons, what accelerating voltage is needed to produce protons with wavelength \(0.0600 \mathrm{nm} ?\) (Hint: In each case the initial kinetic energy is negligible.)

Sirius \(\mathrm{B}\). The brightest star in the sky is Sirius, the Dog Star. It is actually a binary system of two stars, the smaller one (Sirius B) being a white dwarf. Spectral analysis of Sirius B indicates that its surface temperature is \(24,000 \mathrm{~K}\) and that it radiates energy at a total rate of \(1.0 \times 10^{25} \mathrm{~W}\). Assume that it behaves like an ideal blackbody. (a) What is the total radiated intensity of Sirius \(\mathrm{B}\) ? (b) What is the peak-intensity wavelength? Is this wavelength visible to humans? (c) What is the radius of Sirius B? Express your answer in kilometers and as a fraction of our sun's radius. (d) Which star radiates more total energy per second, the hot Sirius \(\mathrm{B}\) or the (relatively) cool sun with a surface temperature of \(5800 \mathrm{~K}\) ? To find out, calculate the ratio of the total power radiated by our sun to the power radiated by Sirius B.

High-speed electrons are used to probe the interior structure of the atomic nucleus. For such electrons the expression \(\lambda=h / p\) still holds, but we must use the relativistic expression for momentum, \(p=m v / \sqrt{1-v^{2} / c^{2}}\). (a) Show that the speed of an electron that has de Broglie wavelength \(\lambda\) is \(v=\frac{c}{\sqrt{1+(m c \lambda / h)^{2}}}\) (b) The quantity \(h / m c\) equals \(2.426 \times 10^{-12} \mathrm{~m}\). (As we saw in Section 38.3 , this same quantity appears in Eq. \((38.7),\) the expression for Compton scattering of photons by electrons.) If \(\lambda\) is small compared to \(h / m c,\) the denominator in the expression found in part (a) is close to unity and the speed \(v\) is very close to \(c .\) In this case it is convenient to write \(v=(1-\Delta) c\) and express the speed of the electron in terms of \(\Delta\) rather than \(v .\) Find an expression for \(\Delta\) valid when \(\lambda \ll h / m c .[\)Hint: Use the binomial expansion \((1+z)^{n}=1+n z+\left[n(n-1) z^{2} / 2\right]+\cdots,\) valid for the case \(|z|<1 .]\) (c) How fast must an electron move for its de Broglie wavelength to be \(1.00 \times 10^{-15} \mathrm{~m}\), comparable to the size of a proton? Express your answer in the form \(v=(1-\Delta) c\), and state the value of \(\Delta\).

See all solutions

Recommended explanations on Physics Textbooks

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.