Question 570: Thermodynamics & KTG

Question

A wire of length 10 cm and diameter 0.5 mm is used in a bulb. The temperature of the wire is $1727^{\circ} \mathrm{C}$ and power radiated by the wire is 94.2 W . Its emissivity is $\frac{x}{8}$ where $x=$ _________.
(Given $\sigma=6.0 	imes 10^{-8} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~K}^{-4}, \pi=3.14$ and assume that the emissivity of wire material is same at all wavelength.)

Accepted Answer

To determine the emissivity ($\varepsilon$) of the wire used in the bulb, we start with the given parameters and the formula for power radiated by a black body:
Length of the wire ($L$): 10 cm = 0.1 m
Diameter of the wire ($d$): 0.5 mm = 0.0005 m
Temperature of the wire ($T$): $1727^{\circ} \mathrm{C} = 2000 \mathrm{K}$
Power radiated ($P$): 94.2 W
The power radiated by the wire can be expressed as:
$ P = \varepsilon \sigma A T^4 $
Where:
$\sigma$ is the Stefan-Boltzmann constant ($6.0 	imes 10^{-8} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~K}^{-4}$)
$A$ is the surface area of the wire, calculated using the formula for the surface area of a cylinder: $A = \pi d L$
Substituting the given values into the formula:
$ 94.2 = \varepsilon 	imes (6 	imes 10^{-8}) 	imes (3.14 	imes 0.0005 	imes 0.1) 	imes (2000)^4 $
Simplifying the expression inside the parentheses:
$ 94.2 = \varepsilon 	imes (6 	imes 10^{-8}) 	imes 3.14 	imes 0.0005 	imes 0.1 	imes (2000)^4 $
On solving for $\varepsilon$:
$ \varepsilon = \frac{94.2}{[(6 	imes 10^{-8}) 	imes 3.14 	imes 0.0005 	imes 0.1 	imes (2000)^4]} $
Given that $\varepsilon$ is $\frac{x}{8}$, evaluating this expression provides the value of $x = 5$. Thus, the emissivity $\varepsilon$ is $\frac{5}{8}$.

Thermodynamics & KTG

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