Calculating Wire Emissivity Using Radiation Power

This solution explains how to find the emissivity of a wire in a bulb based on its physical characteristics and the power it radiates, using the Stefan-Boltzmann law.

Given Parameters

• Wire Length, $L = 10 \text{ cm} = 0.1 \text{ m}$
• Wire Diameter, $d = 0.5 \text{ mm} = 0.5 \times 10^{-3} \text{ m}$
• Wire Radius, $r = \frac{d}{2} = 0.25 \times 10^{-3} \text{ m}$
• Temperature, $T_{\circ C} = 1727^{\circ}\text{C}$
• Power Radiated, $P = 94.2 \text{ W}$
• Stefan-Boltzmann constant, $\sigma = 6.0 \times 10^{-8} \text{ W m}^{-2} \text{ K}^{-4}$
• $\pi = 3.14$
• Emissivity, $\epsilon = \frac{x}{8}$

Step 1: Convert Temperature to Kelvin

The Stefan-Boltzmann law requires temperature in Kelvin.

$ T(\text{K}) = T(^{\circ}\text{C}) + 273 $

$ T = 1727 + 273 = 2000 \text{ K} $

Step 2: Calculate the Surface Area of the Wire

The wire is treated as a cylinder. The surface area ($A$) is calculated using the formula $A = 2 \pi r L$.

$ A = 2 \times \pi \times r \times L $

$ A = 2 \times 3.14 \times (0.25 \times 10^{-3} \text{ m}) \times (0.1 \text{ m}) $

$ A = 6.28 \times 0.025 \times 10^{-3} \text{ m}^2 $

$ A = 0.157 \times 10^{-3} \text{ m}^2 $

Step 3: Apply the Stefan-Boltzmann Law

The power ($P$) radiated by a surface is given by $P = \epsilon \sigma A T^4$. We need to find the emissivity ($\epsilon$).

$ \epsilon = \frac{P}{\sigma A T^4} $

Step 4: Calculate Emissivity ($\epsilon$)

Substitute the known values into the formula:

$ \epsilon = \frac{94.2 \text{ W}}{(6.0 \times 10^{-8} \text{ W m}^{-2} \text{ K}^{-4}) \times (0.157 \times 10^{-3} \text{ m}^2) \times (2000 \text{ K})^4} $

Calculate $T^4$:

$ T^4 = (2000)^4 = (2 \times 10^3)^4 = 16 \times 10^{12} \text{ K}^4 $

Now substitute $T^4$ back into the emissivity equation:

$ \epsilon = \frac{94.2}{(6.0 \times 10^{-8}) \times (0.157 \times 10^{-3}) \times (16 \times 10^{12})} $

$ \epsilon = \frac{94.2}{(6.0 \times 0.157 \times 16) \times 10^{-8 - 3 + 12}} $

$ \epsilon = \frac{94.2}{150.72 \times 10^{1}} $

$ \epsilon = \frac{94.2}{1507.2} \approx 0.625 $

Step 5: Determine the value of x

The problem states that the emissivity is $\epsilon = \frac{x}{8}$. We found $\epsilon \approx 0.625$.

$ \frac{x}{8} = 0.625 $

$ x = 0.625 \times 8 $

$ x = 5 $

The value of $x$ is 5, which lies between 5 and 5.