Question

A light bulb is constructed from 2cm of tungsten wire of diameter $$50\mu \text{m}$$ and is enclosed in an evacuated glass tube. What temperature does the filament reach when it is operated at power of 1 watt? (Assume the emissivity of the tungsten surface is 0.35, Stefan’s constant=$$5.67\times {10}^{-4}\text{c}{\text{m}}^{-2}\cdot {\text{K}}^{-4}$$)
A.$$2000\text{K}$$
B.$$4000\text{K}$$
C.$$2500\text{K}$$
D.$$1200\text{K}$$

Answer 1

Hint: Use the formula for the area of the cylinder and determine the area of the cylindrical tungsten filament. Use the expression for Stefan’s law to determine the surface temperature of the filament. Stefan's law gives the relation between the power radiated by the filament, Stefan’s constant, surface area of the tungsten filament, the emissivity of the tungsten surface and surface temperature of the filament.

Formulae used:
The surface area $$A$$ of the cylinder is given by
$$A=2\pi rh$$ …… (1)
Here, $$r$$ is the radius of the cylinder and $$h$$ is the height of the cylinder.
The expression for Stefan’s law is given by
$$P=\sigma Ae{T}^4$$ …… (2)
Here, $$P$$ is the power radiated, $$\sigma$$ is the Stefan’s constant, $$A$$ is the surface area of the body, $$e$$ is the emissivity and $$T$$ is the surface temperature of the body.

Complete step by step solution:
We have given that the length of the tungsten wire is $$2\text{cm}$$ and diameter is $$50\mu \text{m}$$.
$$l=2\text{cm}$$
$$d=50\mu \text{m}$$
The operating power of the light bulb is $$1\text{W}$$.
$$P=1\text{W}$$
Convert the unit of the length of the wire in the SI system of units.
$$l=\left(2\text{cm}\right)\left({\displaystyle \frac{{10}^{-2}\text{m}}{1\text{cm}}}\right)$$
$$\Rightarrow l=0.02\text{m}$$
Convert the unit of diameter of the wire in the SI system of units.
$$d=\left(50\mu \text{m}\right)\left({\displaystyle \frac{{10}^{-6}\text{m}}{1\mu \text{m}}}\right)$$
$$\Rightarrow d=50\times {10}^{-6}\text{m}$$
Calculate the surface area of the tungsten wire using equation (1).
Rewrite equation (1) for surface area of the tungsten wire.
$$A=2\pi {\displaystyle \frac{d}{2}}l$$
$$\Rightarrow A=\pi dl$$
Substitute $$3.14$$ for $$\pi$$, $$50\times {10}^{-6}\text{m}$$ for $$d$$ and $$0.02\text{m}$$ for $$l$$ in the above equation.
$$A=\left(3.14\right)\left(50\times {10}^{-6}\text{m}\right)\left(0.02\text{m}\right)$$
$$\Rightarrow A=3.14\times {10}^{-6}{\text{m}}^2$$
Hence, the surface area of the cylinder is $$3.14\times {10}^{-6}{\text{m}}^2$$.
We can determine the surface temperature of the filament using equation (2).
Rearrange equation (2) for the surface temperature of the filament.
$$T={\left({\displaystyle \frac{P}{\sigma Ae}}\right)}^{{\displaystyle \frac{1}{4}}}$$
Substitute $$1\text{W}$$ for $$P$$, $$5.67\times {10}^{-8}\text{c}{\text{m}}^{-2}\cdot {\text{K}}^{-4}$$ for $$\sigma$$, $$3.14\times {10}^{-6}{\text{m}}^2$$ for $$A$$ and $$0.35$$ for $$e$$ in the above equation.
$$T={\left({\displaystyle \frac{1\text{W}}{\left(5.67\times {10}^{-8}\text{c}{\text{m}}^{-2}\cdot {\text{K}}^{-4}\right)\left(3.14\times {10}^{-6}{\text{m}}^2\right)\left(0.35\right)}}\right)}^{{\displaystyle \frac{1}{4}}}$$
$$\Rightarrow T=2001.5\text{K}$$
$$\Rightarrow T=2000\text{K}$$
Therefore, the filament will reach the temperature of $$2000\text{K}$$.

So, the correct answer is “Option A”.

Note:
The students should not forget to convert the units of length of the tungsten filament and diameter of the tungsten filament in the SI system of units. The surface temperature of the filament obtained by calculation is in degree kelvin and not in degree Celsius. So there is no need for unit conversion.