The cylindrical filament in a light bulb has a diameter of 0.050 mm, an emissivity of 1.0 , and a temperature of $3000 ^ { \circ } \mathrm { C } \text {. }$ How long should the filament be in order to radiate 60 W of power? $( \sigma =$ $\left. 5.67 \times 10 ^ { - 8 } \mathrm {~W} / \mathrm { m } ^ { 2 } \cdot \mathrm { K } ^ { 4 } \right)$

Question

The cylindrical filament in a light bulb has a diameter of  0.050 mm, an emissivity of  1.0 , and a temperature of $3000 ^ { \circ } \mathrm { C } \text {. }$ How long should the filament be in order to radiate  60 W of power? $( \sigma =$ $\left. 5.67 \times 10 ^ { - 8 } \mathrm {~W} / \mathrm { m } ^ { 2 } \cdot \mathrm { K } ^ { 4 } \right)$

Quizplus · Accepted Answer

The power radiated by the filament is given by:$$P=\epsilon \sigma A T^4$$where:* $P$ is the power radiated (in watts)* $\epsilon$ is the emissivity of the filament (a dimensionless number between 0 and 1)* $\sigma$ is the Stefan-Boltzmann constant ($5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^2 \cdot \mathrm{K}^4$)* $A$ is the surface area of the filament (in square meters)* $T$ is the temperature of the filament (in kelvins)The surface area of a cylinder is given by:$$A=2\pi r L$$where:* $r$ is the radius of the cylinder (in meters)* $L$ is the length of the cylinder (in meters)Substituting the given values into the equations above, we get:$$60\mathrm{~W} = (1.0)(5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^2 \cdot \mathrm{K}^4)(2\pi (2.5 \times 10^{-5} \mathrm{~m})^2 L)(3000 \mathrm{~K})^4$$Solving for $L$, we get:$$L = 5.9 \mathrm{~cm}$$Therefore, the filament should be 5.9 cm long in order to radiate 60 W of power.

The Cylindrical Filament in a Light Bulb Has a Diameter 3000∘C. 3000 ^ { \circ } \mathrm { C } \text {. }3000∘C.

The Cylindrical Filament in a Light Bulb Has a Diameter $3000 ^ { \circ } \mathrm { C } \text {. }$