Question

A wire of length 10 cm and diameter 0.5 mm is used in a bulb. The temperature of the wire is 1727°C and power radiated by the wire is 94.2 W. Its emissivity is $ \frac{x}{8} $, where $ x = \ldots $

Hint:

The Stefan-Boltzmann law relates the power radiated by a body to its temperature, surface area, and emissivity. For a cylindrical wire, calculate the surface area and use the given power to find the emissivity.

Answer 1

Correct Answer: 5

• Length \( L = 0.1\, \text{m} \)
• Diameter \( d = 0.0005\, \text{m} \Rightarrow r = 0.00025\, \text{m} \)
• Temperature \( T = 1727^\circ C = 2000\, \text{K} \)
• Power radiated \( P = 94.2\, \text{W} \)

Using Stefan–Boltzmann law: \[ P = \varepsilon \sigma A T^4 \Rightarrow \varepsilon = \frac{P}{\sigma A T^4} \] Surface area of the wire (cylinder lateral surface): \[ A = 2\pi r L = 2 \times 3.14 \times 0.00025 \times 0.1 = 1.57 \times 10^{-4}\, \text{m}^2 \] Temperature: \[ T^4 = (2000)^4 = 16 \times 10^{12} \] Now calculate: \[ \varepsilon = \frac{94.2}{6.0 \times 10^{-8} \times 1.57 \times 10^{-4} \times 16 \times 10^{12}} = \frac{94.2}{150.72} \approx 0.625 \] \[ \Rightarrow \varepsilon = \frac{x}{8} \Rightarrow x = 8 \times 0.625 = 5 \]

Answer 2

Given values: \[ L = 10 \, \text{cm}, \, d = 0.5 \, \text{mm}, \, T = 1727^\circ \text{C} = 2000 \, \text{K}, \, \text{Power}, P = 94.2 \, \text{W} \] The formula for the power is: \[ P = \epsilon \sigma A T^4 \] Substituting the known values: \[ 94.2 = \epsilon \times (6 \times 10^{-8}) \times (3.14) \times (0.5) \times (10^{-3}) \times (2000)^4 \] Simplifying further: \[ 94.2 = \epsilon \times (6 \times 10^{-8}) \times (3.14) \times (0.5) \times (10^{-3}) \times (10 \times 10^{-2}) \times (2000)^4 \] Solving for \( \epsilon \): \[ \epsilon = \frac{94.2}{(94.2) \times (16)} = \frac{5}{8} \] \[ \boxed{\epsilon = \frac{5}{8}} \]