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Size of a Light-Bulb Filament. The operating temperature of a tungsten filament in an incandescent light bulb is 2450 \(\mathrm{K}\) , and its emissivity is 0.350 . Find the surface area of the filament of a \(150-\mathrm{W}\) bulb if all the electrical energy consumed by the bulb is radiated by the filament as electromagnetic waves. (Only a fraction of the radiation appears as visible light)

Short Answer

Expert verified
The surface area of the tungsten filament is approximately 0.039 m².

Step by step solution

01

Understand the Problem

The problem asks us to find the surface area of a tungsten filament given its operating temperature, emissivity, and power of the bulb. The filament radiates all its energy as electromagnetic waves due to its temperature.
02

Use the Stefan-Boltzmann Law

The Stefan-Boltzmann Law relates the power radiated by an object to its temperature and surface area. It is given by the formula:\[ P = \epsilon \sigma A T^4 \]where \(P\) is power, \(\epsilon\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant \(5.67 \times 10^{-8} \ \mathrm{W/m^2K^4}\), \(A\) is the surface area, and \(T\) is the temperature.
03

Set Up the Equation

Substitute the given values into the Stefan-Boltzmann Law:\[ 150 \ \mathrm{W} = 0.350 \times 5.67 \times 10^{-8} \ \mathrm{W/m^2K^4} \times A \times (2450 \ \mathrm{K})^4 \]
04

Simplify and Solve for A

Solve for \(A\) by isolating it on one side of the equation:\[ A = \frac{150}{0.350 \times 5.67 \times 10^{-8} \times (2450)^4} \]
05

Calculate the Value

Calculate the expression to get the surface area:\[ A \approx \frac{150}{0.350 \times 5.67 \times 10^{-8} \times 3.61 \times 10^{13}} \]\[ A \approx 0.039 \ \mathrm{m^2} \]

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Key Concepts

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

Tungsten Filament
A tungsten filament is a key component in incandescent light bulbs. It is a thin wire made of tungsten, a metal known for its high melting point and exceptional durability. Tungsten's melting point is around 3422 °C (6192 °F), making it suitable for high-temperature applications, such as in light bulb filaments where it's heated to high temperatures to produce light.
This high resistance to heat allows the tungsten filament to glow brightly without melting or breaking. As electric current passes through the tungsten filament, it heats up and emits light. The filament's effectiveness in producing light depends significantly on its temperature.
At approximately 2450 K, as indicated in our problem, the filament emits visible and other electromagnetic radiation. The efficiency and quality of light produced by a tungsten filament is part of why this material is preferred in traditional light bulbs.
Emissivity
Emissivity is a measure of an object's ability to emit thermal radiation. It is a dimensionless number ranging from 0 to 1. A higher emissivity means the object is more effective at radiating energy.
In our given problem, the tungsten filament has an emissivity of 0.350. This means that the filament emits 35% of the energy it theoretically could if it were a perfect emitter. For materials like tungsten used in filaments, the emissivity impacts how much energy is lost as heat.
Understanding emissivity is crucial when calculating thermal radiation from objects, such as when applying the Stefan-Boltzmann Law to determine the surface area of a filament. It helps us account for the actual amount of energy the filament is capable of radiating.
Electromagnetic Radiation
Electromagnetic radiation encompasses a broad spectrum of wavelengths and frequencies, including radio waves, microwaves, infrared, visible light, ultraviolet, and X-rays.
For a tungsten filament, as it heats up, it emits electromagnetic radiation across a range of wavelengths. Most commonly in light bulbs, this includes visible light, which enables illumination, and infrareds, contributing to heat. - Electromagnetic radiation is a wave phenomenon that carries energy through space. - The particular wavelength of light emitted depends on the temperature of the filament. In the context of light bulbs, only a fraction of the emitted radiation falls within the visible spectrum. The rest is often in the form of non-visible light, such as infrared, which doesn't contribute to lighting but still involves energy loss.
Thermal Radiation
Thermal radiation is the process by which energy is emitted by a body's surface due to its temperature. This form of radiation is part of the electromagnetic spectrum and is produced by the thermal motion of charged particles in matter.
As the tungsten filament in a light bulb heats to approximately 2450 K, it radiates energy. This energy is initially in the form of infrared radiation and eventually extends into the visible light range as the temperature increases. - All objects, regardless of temperature, emit some thermal radiation. - The intensity and type of radiation depend heavily on the object's temperature and emissivity. Thermal radiation is an essential factor to consider when designing devices like light bulbs, as it directly relates to the efficiency and brightness of the emitted light. By applying concepts such as the Stefan-Boltzmann Law, one can precisely determine how much energy is released and in what form.

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Most popular questions from this chapter

A tube leads from a \(0.150-\mathrm{kg}\) calorimeter to a flask in which water is boiling under atmospheric pressure. The calorimeter has specific heat capacity 420 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) , and it originally contains 0.340 \(\mathrm{kg}\) of water at \(15.0^{\circ} \mathrm{C}\) . Steam is allowed to condense in the calorimeter at atmospheric pressure until the temperature of the calorimeter and contents reaches \(71.0^{\circ} \mathrm{C}\) , at which point the total mass of the calorimeter and its contents is found to be 0.525 \(\mathrm{kg}\) . Compute the heat of vaporization of water from these data.

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