Problem 73 - University Physics with Modern Physics

Question

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.)

Step-by-Step Solution

Answer

The surface area of the filament is approximately \(4.91 \times 10^{-4} \ \mathrm{m}^2\).

Step 1: Understand the Problem

We need to find the surface area of a tungsten filament in a 150-W bulb, with a known emissivity and temperature, under the assumption that all electrical energy is emitted as heat.

Step 2: Use Stefan-Boltzmann Law

The power radiated by a black body is given by the Stefan-Boltzmann law: \[ P = \epsilon \sigma A T^4 \] where \(P\) is the power emitted (150 W), \(\epsilon\) is the emissivity (0.350), \(\sigma\) is the Stefan-Boltzmann constant \(5.67 \times 10^{-8} \mathrm{W/m}^2 \mathrm{K}^{4}\), \(A\) is the surface area of the filament, and \(T\) is the temperature (2450 K).

Step 3: Rearrange the Formula to Solve for Surface Area

Rearrange the Stefan-Boltzmann law to solve for the surface area \(A\): \[ A = \frac{P}{\epsilon \sigma T^4} \]

Step 4: Substitute Values into the Formula

Substitute the known values into the rearranged formula: \[ A = \frac{150}{0.350 \times 5.67 \times 10^{-8} \times (2450)^4} \]

Step 5: Calculate the Surface Area

Perform the calculation: \[ A \approx \frac{150}{0.350 \times 5.67 \times 10^{-8} \times 3.60 \times 10^{12}} \] After completing this computation, we find \( A \approx 4.91 \times 10^{-4} \ \mathrm{m}^2\).

Key Concepts

Tungsten Filament

Tungsten is a commonly used material in the filaments of incandescent light bulbs. Its high melting point makes it especially suitable for this purpose since it can withstand high temperatures without melting. This material is capable of emitting electromagnetic radiation, which is a mix of visible light and heat, when heated to high temperatures.

The filament's primary role is to convert electrical energy into light and heat. When electrical current passes through the tungsten filament, it heats up and starts to glow, emitting light. The choice of tungsten is critical because its properties allow it to operate effectively at the high temperatures where incandescent lighting is most efficient.

Emissivity

Emissivity is a measure of how effectively a surface emits energy as thermal radiation. It is represented by a value between 0 and 1, with 1 being a perfect emitter (also known as a black body).

A tungsten filament with an emissivity value of 0.350 means that it emits only 35% of the energy compared to a perfect black body. This is an important factor when calculating the energy emitted from the filament in the form of radiation.

Understanding emissivity is crucial in applying the Stefan-Boltzmann Law, allowing us to find the radiant heat energy leaving the filament and, consequently, assists in determining other properties such as the surface area.

Electromagnetic Radiation

Electromagnetic radiation is energy that spreads out as it travels through space. This energy is emitted by the tungsten filament in the light bulb as it heats up. It includes a variety of waves such as visible light, infrared, ultraviolet, and others, depending on the temperature of the filament.

In the case of a tungsten filament, when electrical energy is applied, it converts into thermal energy, which then emits as electromagnetic radiation. Not all of this radiation is visible light; much of it is actually heat.

• Visible light: The part of the spectrum we can see with our eyes.
• Infrared radiation: Less visible, primarily felt as heat.

Overall, understanding electromagnetic radiation allows us to grasp how energy is transferred in the filament and how efficiency can vary based on wavelength sensitivity.

Surface Area Calculation

Calculating the surface area of the tungsten filament involves using the Stefan-Boltzmann Law. This physical law helps us relate the power emitted by a heatsource to other factors such as emissivity, temperature, and the physical dimensions of the source.

Here, the law is set in the form \[ P = \epsilon \sigma A T^4 \] where:- \(P\) is the power (150 W)- \(\epsilon\) is the emissivity (0.350)- \(\sigma\) is the Stefan-Boltzmann constant - \(A\) is the surface area- \(T\) is the temperature (2450 K)
The formula is rearranged to find the surface area\[ A = \frac{P}{\epsilon \sigma T^4} \]By substituting the known values, we obtain\[ A \approx \frac{150}{0.350 \times 5.67 \times 10^{-8} \times (2450)^4} \]
Performing this calculation, the resulting surface area is approximately \(4.91 \times 10^{-4} \, \mathrm{m}^2\). This calculation informs us of the effective area from which the filament radiates energy, crucial for understanding its performance in emitting light.