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(a) Use Stefan's law to calculate the total power radiated per unit area by a tungsten filament at a temperature of \(3000 \mathrm{~K}\). (Assume that the filament is an ideal radiator.) (b) If the tungsten filament of a lightbulb is rated at \(75 \mathrm{~W}\), what is the surface area of the filament? (Assume that the main energy loss is due to radiation.)

Short Answer

Expert verified
The total power radiated per unit area by the tungsten filament at a temperature of 3000 K is 459 W/m^2, and the surface area of the filament is 0.16 m^2.

Step by step solution

01

Calculate the total power radiated per unit area

To begin, use Stefan's law which is formulated as \( P = \sigma*T^4 \), where \( \sigma \) is the Stefan-Boltzmann constant. The constant \( \sigma \) equals \(5.67 * 10^{-8} W/m^2K^4\), and the temperature T is given as 3000 K. Therefore, the total power radiated per unit area \( P \) equals \( \sigma*T^4 = 5.67 * 10^{-8} * 3000^4 W/m^2 = 459 W/m^2.\)
02

Calculate the surface area of the filament

To find the surface area of the filament, use the formula \(W = P*A\), where W is the total power of the filament (75 W), P is the total power radiated per unit area (calculated in the previous step), and A is the surface area of the filament. Solving for A, you get \( A = W / P = 75W / 459W/m^2 = 0.16m^2\).

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Key Concepts

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

Tungsten Filament
Tungsten is a metal used mainly in electrical applications, like the filament of light bulbs. What makes tungsten special is its ability to withstand high temperatures, which is crucial for devices that emit light. Tungsten filaments are perfect for this purpose because they can be heated to glow brightly without melting. In a lighting context, a filament is the thin wire in a bulb that, when heated by an electric current, glows and produces light. Tungsten is chosen because of its high melting point (about 3422°C) and good electrical conductivity. It allows light bulbs to operate at lower energies while still being very bright. Therefore, tungsten filaments play a vital role in light-producing devices. They have the incredible ability to emit light for thousands of hours, due to their toughness and heat resistance, making them a reliable component in many household and industrial lighting solutions.
Thermal Radiation
Thermal radiation is the emission of energy as electromagnetic waves or particles when a material is at a certain temperature. It is how heat energy is transferred from an object. All objects emit thermal radiation as long as they have a temperature above absolute zero.Understanding thermal radiation involves considering three key components:
  • Temperature
  • Surface properties of the material
  • The environment surrounding the material
Thermal radiation is crucial in understanding how objects like lightbulb filaments operate. The Stefan-Boltzmann Law is a fundamental principle in this context, which relates temperature to the energy radiated. According to this law, the emitted power per unit area (\( P \)) is calculated using the formula \( P = \sigma T^4 \), where \( \sigma \) is the Stefan-Boltzmann constant, a known value. This demonstrates that the radiant energy depends strongly on the temperature, resulting in more energy being emitted at higher temperatures.
Surface Area Calculation
When calculating physical properties such as the surface area of an object, we often need to rely on known formulas that relate different aspects of the object's characteristics. In the context of a tungsten filament in a lightbulb, we consider the thermal radiation it emits and its total power rating. The relationship used here is: \( W = P \times A \), where \( W \) is the total power output, \( P \) is the power per unit area (found via the Stefan-Boltzmann law), and \( A \) is the area we want to solve for. By rearranging this equation, \( A = W / P \), the surface area can be calculated if the total power and power per unit area are known.Surface area calculations are significant because they affect how much radiation a filament can emit. A larger area can emit more energy, assuming other factors remain the same. For practical applications, understanding these calculations is essential for designing efficient lighting systems and ensuring proper heat management.

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Most popular questions from this chapter

The average power generated by the Sun has the value \(3.74 \times 10^{26} \mathrm{~W}\). Assuming the average wavelength of the Sun's radiation to be \(500 \mathrm{~nm}\), find the number of photons emitted by the Sun in \(1 \mathrm{~s}\).

In a Compton collision with an electron, a photon of violet light \((\lambda=4000 \AA)\) is backscattered through an angle of \(180^{\circ}\). (a) How much energy (eV) is transferred to the electron in this collision? (b) Compare your result with the energy this electron would acquire in a photoelectric process with the same photon. (c) Could violet light eject electrons from a metal by Compton collision? Explain.

Density of modes. The essentials of calculating the number of modes of vibration of waves confined to a cavity may be understood by considering a one-dimensional example. (a) Calculate the number of modes (standing waves of different wavelength) with wavelengths between \(2.0 \mathrm{~cm}\) and \(2.1 \mathrm{~cm}\) that can exist on a string with fixed ends that is \(2 \mathrm{~m}\) long. (Hint: use \(n(\lambda / 2)=L\), where \(n=1,2,3,4,5 \ldots \ldots\) Note that a specific value of \(n\) defines a specific mode or standing wave with different wavelength.) (b) Calculate, in analogy to our threedimensional calculation, the number of modes per unit wavelength per unit length, \(\frac{\Delta n}{L \Delta \lambda}\). (c) Show that in general the number of modes per unit wavelength per unit length for a string of length \(L\) is given by $$ \frac{1}{L}\left|\frac{d n}{d \lambda}\right|=\frac{2}{\lambda^{2}} $$ Does this expression yield the same numerical answer as found in (a)? (d) Under what conditions is it justified to replace \(\left|\left(\frac{\Delta n}{L \Delta \lambda}\right)\right|\) with \(\left|\left(\frac{d n}{L d \lambda}\right)\right| ?\) Is the expression \(n=2 L / \lambda\) a continuous function?

Photons of wavelength \(450 \mathrm{~nm}\) are incident on a metal. The most energetic electrons ejected from the metal are bent into a circular arc of radius \(20 \mathrm{~cm}\) by a magnetic field whose strength is equal to \(2.0 \times 10^{-5} \mathrm{~T}\). What is the work function of the metal?

Two light sources are used in a photoelectric experiment to determine the work function for a particular metal surface. When green light from a mercury lamp \((\lambda=546.1 \mathrm{~nm})\) is used, a retarding potential of \(1.70 \mathrm{~V}\) reduces the photocurrent to zero. (a) Based on this measurement, what is the work function for this metal? (b) What stopping potential would be observed when using the yellow light from a helium discharge tube \((\lambda=587.5 \mathrm{~nm}) ?\)

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