Problem 69 - University Physics with Modern Physics

Question

The operating temperature of a tungsten filament in an incandescent light bulb is 2450 K, and its emissivity is 0.350. Find the surface area of the filament of a 150-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 \(6.96 \times 10^{-5} \, \text{m}^2\).

Step 1: Identify Known Values

We know the following values: the power of the bulb is 150 W, the temperature of the filament is 2450 K, and the emissivity of the filament is 0.350.

Step 2: Use the Stefan-Boltzmann Law

The Stefan-Boltzmann Law relates the power radiated by a black body to its temperature: \[ P = e\sigma A T^4 \]where \( P \) is the power, \( e \) is the emissivity, \( \sigma \) is the Stefan-Boltzmann constant \( (5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4) \), \( A \) is the surface area, and \( T \) is the temperature in Kelvin.

Step 3: Rearrange the Equation

To find the surface area \( A \), rearrange the equation as follows:\[ A = \frac{P}{e\sigma T^4} \]

Step 4: Substitute Known Values

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

Step 5: Calculate the Surface Area

Perform the calculation:\[ A = \frac{150}{0.350 \times 5.67 \times 10^{-8} \times 3.628 \times 10^{13}} \approx 6.96 \times 10^{-5} \, \text{m}^2 \]

Key Concepts

Emissivity

Emissivity is a measure of how effectively a surface emits thermal radiation. It is represented by the symbol \( e \) and ranges from 0 to 1. A perfect black body, which emits all possible radiation, has an emissivity of 1. Real-world objects, however, have emissivity values less than 1, indicating that they emit less radiation than a black body at the same temperature.

An emissivity of 0.350 for tungsten means the filament emits only 35% of the radiation a perfect black body would at its operating temperature. The concept of emissivity is crucial because it influences the total power radiated by the filament. When using the Stefan-Boltzmann Law, emissivity directly impacts the calculation of emitted power. Hence, accounting for emissivity allows for more accurate predictions of thermal radiation in engineering applications such as heating elements and incandescent bulbs.

Black Body Radiation

Black body radiation refers to the theoretical emission of electromagnetic waves by an idealized object that absorbs all incident radiation, known as a black body. This concept is pivotal in understanding the transfer of energy as electromagnetic radiation, especially at high temperatures.

• The Stefan-Boltzmann Law models the power radiated by a black body based on its temperature, formulated as \( P = \sigma A T^4 \).
• The law is modified for real objects using the emissivity factor \( e \), resulting in the equation \( P = e\sigma A T^4 \).

Here, \( \sigma \) is the Stefan-Boltzmann constant, with a value of \( 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \). This modified law helps calculate the radiation from incandescent bulbs, enabling the determination of how much electrical energy transforms into thermal radiation. Black body radiation principles are thus cornerstone concepts in thermodynamics and physics, informing design and optimization of thermal devices.

Surface Area Calculation

Calculating the surface area of a filament in a light bulb is essential to determine the total radiation emitted. The Stefan-Boltzmann Law can be rearranged to solve for surface area \( A \). The equation used is:

\[ A = \frac{P}{e\sigma T^4} \]

• \( P \) is the power output, which is 150 W.
• \( e \) is the emissivity of the tungsten filament, 0.350 in this scenario.
• \( T \) indicates the filament's temperature of 2450 K.

Substituting these values, the surface area \( A \) is calculated to be approximately \( 6.96 \times 10^{-5} \, \text{m}^2 \). This result means that the filament emits this power over its entire surface area. Understanding and calculating surface area in such applications reflects how effectively the energy converts to light and heat in a bulb, therefore linking theoretical concepts to practical outcomes in electrical engineering.