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Light bulb 1 operates with a higher filament temperature than light bulb 2 , but both filaments have the same emissivity. (a) How is the power \(P\) expressed in terms of the energy \(Q\) radiated by a bulb and the time \(t\) during which the energy is radiated? (b) Does a higher filament temperature generate more radiated power or less radiated power? (c) Does a smaller area for radiation promote more radiated power or less radiated power? (d) Suppose that both bulbs radiate the same power. Is the filament area of bulb 1 greater than, less than, or the same as the filament area of bulb \(2 ?\) Give your reasoning.

Short Answer

Expert verified
(a) \( P = \frac{Q}{t} \) (b) More radiated power (c) Less radiated power (d) Filament area of bulb 1 is less.

Step by step solution

01

Understand Power in Terms of Energy and Time

The power \( P \) is defined as the energy \( Q \) radiated by a bulb per unit time \( t \). Therefore, the power can be expressed as \( P = \frac{Q}{t} \), where \( Q \) is the total energy radiated and \( t \) is the time during which the energy is radiated.
02

Effect of Filament Temperature on Radiated Power

A higher filament temperature leads to more radiated power. According to the Stefan-Boltzmann law, the power radiated from a black body (such as a filament) is proportional to the fourth power of its temperature: \( P \propto T^4 \). Thus, when the temperature \( T \) increases, the radiated power \( P \) increases.
03

Effect of Radiating Area on Power

The radiated power is also proportional to the area \( A \) of the emitting surface. From the Stefan-Boltzmann law, \( P = \epsilon \sigma A T^4 \), where \( \epsilon \) is the emissivity and \( \sigma \) the Stefan-Boltzmann constant. Therefore, a smaller area results in less radiated power, given a constant emissivity and temperature.
04

Comparing Filament Areas with Equal Radiated Power

If both bulbs radiate the same power but bulb 1 operates at a higher temperature, the area of filament 1 must be smaller compared to the filament 2. Since \( P = \epsilon \sigma A T^4 \), for the power to remain constant as \( T \) increases, the area \( A \) must decrease.

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Key Concepts

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

Thermal Radiation
Thermal radiation is a type of electromagnetic radiation emitted by all objects with a temperature above absolute zero. It is the process where heat energy is transformed into light waves that can radiate through space. Every object emits thermal radiation depending on its temperature. - The higher the temperature, the more thermal radiation is emitted. - This radiation can occur in a variety of forms, from infrared waves to visible light, depending on the temperature of the object. When you hold your hand near a hot bulb, the warmth you feel is due to thermal radiation. This is a fundamental property of materials and is essential for understanding how objects exchange energy through radiation.
Emissivity
Emissivity is a measure of how efficiently a surface emits thermal radiation compared to a perfect emitter known as a black body. It is a dimensionless value that ranges between 0 and 1. - A perfect black body has an emissivity of 1, meaning it emits the maximum possible radiation. - Real-world objects have emissivities less than 1, depending on their material and surface characteristics. Understanding emissivity is crucial for predicting how different materials and objects radiate energy. For example, a light bulb filament, regardless of the specific bulb, emits radiation based on its emissivity and temperature.
Black Body Radiation
Black body radiation refers to the theoretical thermal radiation that would be emitted by a "perfect" emitter. A black body absorbs all incident radiation, reflecting none, and radiates energy at the maximum possible intensity for a given temperature. - The concept helps scientists predict the radiation profiles of real-world objects. - Black body radiation serves as a benchmark for understanding actual emissive behavior. The Stefan-Boltzmann Law, which describes the power radiated by a black body related to its temperature, illustrates the fourth power temperature relationship, affirming that even small increases in temperature can dramatically increase radiative power.
Temperature and Power Relationship
The temperature and power relationship is fundamental in understanding radiative heat transfer. According to the Stefan-Boltzmann Law, the power (\(P\) radiated by an object is proportional to the fourth power of its absolute temperature (\(T\): \(P = \epsilon \sigma A T^4\), where \(\epsilon\) is emissivity, \(\sigma\) is the Stefan-Boltzmann constant, and \(A\) is the surface area.- This relationship indicates that a small increase in temperature results in a significant increase in power output.- For example, if the temperature doubles, the power increases by a factor of 16.In practical terms, this means that devices operating at higher temperatures, like a light bulb filament, emit more power. Even with the same emissivity, a higher temperature ensures more efficient energy radiation, affecting power generation and heating capabilities.

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Most popular questions from this chapter

A solid sphere has a temperature of \(773 \mathrm{~K}\). The sphere is melted down and recast into a cube that has the same emissivity and emits the same radiant power as the sphere. What is the cube's temperature?

Multiple-Concept Example 3 discusses an approach to problems such as this. The ends of a thin bar are maintained at different temperatures. The temperature of the cooler end is \(11^{\circ} \mathrm{C},\) while the temperature at a point \(0.13 \mathrm{~m}\) from the cooler end is \(23{ }^{\circ} \mathrm{C}\) and the temperature of the warmer end is \(48^{\circ} \mathrm{C}\). Assuming that heat flows only along the length of the bar (the sides are insulated), find the length of the bar.

Multiple-Concept Example 8 discusses the ideas on which this problem depends. Interactive LearningWare \(13.2\) at reviews the concepts that are involved in this problem. Suppose the skin temperature of a naked person is \(34{ }^{\circ} \mathrm{C}\) when the person is standing inside a room whose temperature is \(25^{\circ} \mathrm{C}\). The skin area of the individual is \(1.5\) \(\mathrm{m}^{2}\). (a) Assuming the emissivity is \(0.80\), find the net loss of radiant power from the body. (b) Determine the number of food Calories of energy (1 food Calorie \(=4186 \mathrm{~J}\) ) that are lost in one hour due to the net loss rate obtained in part (a). Metabolic conversion of food into energy replaces this loss.

The amount of radiant power produced by the sun is approximately \(3.9 \times 10^{26} \mathrm{~W}\). Assuming the sun to be a perfect blackbody sphere with a radius of \(6.96 \times 10^{8} \mathrm{~m},\) find its surface temperature (in kelvins).

A person is standing outdoors in the shade where the temperature is \(28{ }^{\circ} \mathrm{C}\). (a) What is the radiant energy absorbed per second by his head when it is covered with hair? The surface area of the hair (assumed to be flat) is \(160 \mathrm{~cm}^{2}\) and its emissivity is \(0.85 .\) (b) What would be the radiant energy absorbed per second by the same person if he were bald and the emissivity of his head were \(0.65 ?\)

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