Question

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

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

Step-by-step solution

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.

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: $$100 = 0.35 \times 2.03 \times 10^{-4} \times 5.67 \times 10^{-8} \times T^4$$

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) \(2640 \mathrm{~K}\).

Key concepts

Incandescent Light Bulb

An incandescent light bulb is a source of artificial light that works by heating a filament until it glows. This is done by passing an electric current through the filament, which is usually made of tungsten because of its high melting point. As electricity flows through the filament, it heats up and emits light. However, most of the energy consumed by the bulb is converted into heat rather than visible light. For a traditional incandescent bulb, over 90% of the energy becomes heat. This level of inefficiency is why newer, more energy-efficient lighting technologies have become popular. Nevertheless, incandescent bulbs are often favored for their warm light and continuous spectrum.

These bulbs are easy to use and have been the mainstay of household lighting for many years. However, because they waste a lot of electricity as heat, they are gradually being replaced by LED and CFL bulbs, which offer better energy efficiency and longer lifespan.

Tungsten Filament

The filament in an incandescent light bulb is made of tungsten, a material known for its remarkable properties. Tungsten is chosen primarily because of its high melting point, which is around 3422°C. This allows it to withstand the intense heat without melting as it emits light in a bulb. A tungsten filament efficiently emits light by resisting the electrical current passing through it, which heats the filament and causes it to glow brightly.

Tungsten's high melting point also means it can be heated to very high temperatures, making it ideal for use in incandescent bulbs. As it heats up, the tungsten emits thermal radiation, which glows as visible light. Over time, however, the filament in the bulb can become thin and eventually break, which ends the life of the bulb. This wear and tear are due to a process called sublimation, where atoms vaporize from the filament and get deposited inside the bulb's glass surface.

Emissivity

Emissivity is a key concept in understanding how materials emit heat. It is the measure of a material's ability to emit thermal radiation and is expressed as a ratio between 0 and 1. A perfect black body, which is an idealized physical body that absorbs all incident electromagnetic radiation, has an emissivity of 1. In contrast, an object that emits no thermal radiation would have an emissivity of 0.

For practical purposes, most materials have an emissivity value between these two extremes. Tungsten, the material used in incandescent light bulb filaments, has an emissivity of around 0.35 at high temperatures. This means that tungsten emits 35% as much thermal radiation as a perfect black body. In the context of an incandescent bulb, the lower emissivity means that much of the energy consumed is not effectively converted into light, but instead dissipates as heat.

Thermal Radiation

Thermal radiation is the emission of electromagnetic waves from all matter that has a temperature greater than absolute zero. These waves fall under the electromagnetic spectrum and can manifest as heat or light. The process of emitting thermal radiation is a natural consequence of a body's temperature.

All objects continuously emit thermal radiation, and the rate at which they do so depends on their temperature and surface characteristics. As objects get hotter, the emitted radiation becomes visible as light. This is precisely what happens in an incandescent light bulb. The filament gets so hot that it starts to glow, emitting visible light along with infrared radiation.

In the case of the tungsten filament inside a bulb, the Stefan-Boltzmann Law describes how much power is radiated due to temperature. The emitted power is determined by multiplying the emissivity, surface area, Stefan-Boltzmann constant, and the fourth power of the temperature in Kelvin. Due to the nature of tungsten and the high temperature of the filament, the bulb spreads most of its energy as heat rather than light.