Problem 63 - College Physics

Question

\(\cdot\) Size of a lightbulb filament. The operating temperature of a tungsten filament in an incandescent lightbulb is \(2450 \mathrm{K},\) and its emissivity is \(0.35 .\) 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 light. (In reality, only a small fraction of the radiation appears as visible light.)

Step-by-Step Solution

Answer

The surface area of the tungsten filament is approximately 20.9 m².

Step 1: Identify Relevant Formula

To determine the surface area of the filament, we use the Stefan-Boltzmann Law, which is given by the formula: \[ P = e imes \sigma imes A imes T^4 \]where \( P \) is the power (150 W), \( e \) is emissivity (0.35), \( \sigma \) is the Stefan-Boltzmann constant (approximately \( 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \)), \( A \) is the surface area of the filament, and \( T \) is the temperature (2450 K).

Step 2: Rearrange the Formula

The formula needs to be rearranged to solve for the surface area \( A \). Therefore, we have:\[ A = \frac{P}{e \times \sigma \times T^4} \]

Step 3: Insert Known Values

Substitute the known values into the rearranged formula:\[ A = \frac{150}{0.35 \times 5.67 \times 10^{-8} \times (2450)^4} \]

Step 4: Calculate Temperature to the Fourth Power

Calculate \( 2450^4 \):\[ 2450^4 = 3.61025 \times 10^{13} \]

Step 5: Complete the Calculation

Substitute \( 2450^4 \) back in and compute the surface area:\[ A = \frac{150}{0.35 \times 5.67 \times 10^{-8} \times 3.61025 \times 10^{13}} \]Perform the arithmetic to find the surface area in square meters:

Step 6: Perform Arithmetic Operations

Calculate the denominator:\[ 0.35 \times 5.67 \times 10^{-8} \times 3.61025 \times 10^{13} \approx 7.161 \]Calculate the surface area:\[ A = \frac{150}{7.161} \approx 20.95 \text{ m}^2 \]

Step 7: Round and Conclude

Since results need to be practical, round the surface area to three significant figures: \[ A \approx 20.9 \text{ m}^2 \].

Key Concepts

Surface Area Calculation

When you're dealing with objects like a lightbulb filament, one key aspect is determining the surface area. Why? Because this helps understand how much area is available for emitting thermal radiation. In this exercise, we use a formula derived from the Stefan-Boltzmann Law to find this area.

• The formula is \[ A = \frac{P}{e \times \sigma \times T^4} \]where:
  • \( P \) is the power in watts.
  • \( e \) is the emissivity.
  • \( \sigma \) is the Stefan-Boltzmann constant.
  • \( T \) is the temperature in Kelvin.

To apply this, rearrange the formula to solve for \( A \), which is dependent on knowing the emissivity, the temperature raised to the fourth power, and the power consumption of the bulb. Dropping these values in helps compute a specific figure for the surface area — crucial for understanding heat transfer in this tiny, hot object. Knowing the exact surface area allows engineers to determine how effectively an object can radiate heat, particularly in devices engineered for thermal efficiency. This concept is widely applicable across various technologies involving heat exchange. Understanding surface area calculations ensures optimal thermal management in systems relying on design precision.

Thermal Radiation

Thermal radiation is a fascinating natural law where any object with temperature radiates energy. This process is governed by the Stefan-Boltzmann Law. In simpler terms, it's about how objects emit energy as electromagnetic waves.

• This emitted energy is proportional to:
  • Surface area \( A \)
  • Emissivity \( e \)
  • Temperature \( T \), raised to the fourth power.

This scenario with a lightbulb filament provides a real-world application. The bulb uses electric power, turning it into thermal energy that gets radiated out. At \( 2450 \text{K} \), significant thermal radiation occurs, influencing not just light but heat felt from the bulb.Thermal radiation is integral in heat exchange for various industries, from climate systems to manufacturing processes, even extending to astrophysics. Understanding it helps in leveraging natural heat for energy efficiency and optimal performance of devices designed to radiate energy in controlled manners.

Emissivity of Materials

Emissivity is a key concept in the realm of heat transfer and thermal radiation. It represents how efficiently a material emits thermal radiation compared to an ideal black body. Think of it as a material's thermal signature.

• Measured between 0 and 1:
  • An emissivity of 1 means it behaves like a perfect black body, which is optimal in theory for emitting radiation.
  • An emissivity of 0 means it emits no thermal radiation.

In this context, our tungsten filament has an emissivity of 0.35, indicating it's less efficient than a black body but still adequate for emitting substantial radiation. This property varies significantly across materials and conditions (like temperature or surface finish). Engineers and scientists gauge emissivity to improve the thermal management of machines and buildings, making systems energy-efficient and more sustainable. Through detailed knowledge of emissivity, one can choose the right materials for heat-emitting or heat-resistant applications, optimizing for different environmental needs.