Problem 19 - Physics

Question

The filament of a light bulb has a temperature of \(3.0 \times 10^{3}{\underline{\phantom{xx}}}^{\circ} \mathrm{C}\) and radiates sixty watts of power. The emissivity of the filament is \(0.36 .\) Find the surface area of the filament.

Step-by-Step Solution

Answer

The surface area of the filament is approximately \(6.49 \times 10^{-5}\) m².

Step 1: Convert Temperature to Kelvin

The given temperature is in Celsius, so we need to convert it to Kelvin. The temperature in Kelvin is calculated as follows: \[ T(K) = T(^{\circ}C) + 273.15 = 3.0 \times 10^3 + 273.15 = 3273.15 \text{ K} \]

Step 2: Use the Stefan-Boltzmann Law

The power radiated by a body can be found using the Stefan-Boltzmann law, which is given by: \[ P = \epsilon \sigma A T^4 \] where \(P\) is the power, \(\epsilon\) 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: Substitute Known Values

Substitute the known values into the Stefan-Boltzmann equation: \[ 60 = 0.36 \times 5.67 \times 10^{-8} \times A \times (3273.15)^4 \]

Step 4: Calculate Surface Area

Rearrange the equation to solve for \(A\): \[ A = \frac{60}{0.36 \times 5.67 \times 10^{-8} \times (3273.15)^4} \] Calculate the value: \[ A \approx 6.49 \times 10^{-5} \text{ m}^2 \]

Key Concepts

Emissivity

Emissivity is a measure of an object's ability to emit thermal radiation compared to a perfect black body. A black body has an emissivity value of 1, meaning it is a perfect emitter and absorber of radiation. In contrast, real objects have emissivity values less than 1.

In the context of the Stefan-Boltzmann Law, emissivity (\( \epsilon \)) is a vital factor that modifies the total power radiated by an object. The power radiated is proportional to the fourth power of the object's absolute temperature and its surface area but is also directly proportional to its emissivity.

• High emissivity means more radiation emitted and absorbed (closer to a black body)
• Low emissivity means less radiation efficiency
• For various materials, emissivity values can differ significantly

For example, in the given exercise, the filament has an emissivity of 0.36, which means it emits 36% of the radiation a black body would at the same temperature.

Surface Area Calculation

In physics, particularly when dealing with thermal radiation, calculating the surface area of an object is crucial for understanding how much energy it emits. According to the Stefan-Boltzmann law, a higher surface area leads to more radiation being emitted by the object.

To calculate surface area using the Stefan-Boltzmann law, you need the following known quantities:

• Power radiated (\( P \))
• Emissivity (\( \epsilon \))
• Temperature in Kelvin (\( T \))

The surface area (\( A \)) is calculated by rearranging the Stefan-Boltzmann equation to isolate \( A \):
\[ A = \frac{P}{\epsilon \times \sigma \times T^4} \]
In the exercise, using the known power of 60 watts, emissivity of 0.36, and temperature converted to Kelvin, the surface area was calculated to be approximately \( 6.49 \times 10^{-5} \text{ m}^2 \). This shows how effectively the filament emits energy into its surroundings.

Temperature Conversion

Temperature conversion is an essential skill in physics and engineering, necessary for accurate calculations, especially those involving thermal dynamics or energy transfer.

The problem begins with a temperature presented in Celsius, which must be converted to Kelvin for accurate use in the Stefan-Boltzmann law. This conversion is straightforward:

• Add 273.15 to the Celsius temperature to convert to Kelvin
• K is the unit used in absolute temperature scales, indispensable for scientific calculations

In the provided exercise, the original temperature of the filament is given as 3.0 \times 10^3degrees Celsius. By converting to Kelvin, we find:\[ T(K) = T(^{\circ}C) + 273.15 = 3273.15 \text{ K} \]
This conversion ensures that temperature-related equations remain dimensionally consistent and correct. The Kelvin scale is absolute, meaning 0 K represents the absence of thermal energy, unlike Celsius, which can handle negative temperatures, nonsensical for certain physics equations.