Physics

Problem

A tungsten filament inside an incandescent light bulb can be approximated as a uniform cylinder

4.0\,\text{cm}

long with a diameter of

0.020\,\text{cm}

. When the bulb is operating at its nominal voltage, the filament’s steady-state temperature is

2800\,\text{K}

and its surface emissivity is 0.32. The electrical power delivered to the filament is

60\,\text{W}

. Ignore any heat conduction through the filament’s leads and any convection within the bulb; assume radiation is the only way the filament loses energy. (a) Calculate the total external surface area of the filament. (Neglect the area of the filament’s circular ends.) (b) Determine the rate at which the filament radiates energy at

2800\,\text{K}

. (c) Compute the ratio of the radiated power found in part (b) to the electrical power supplied to the filament, and interpret this ratio in terms of the filament’s efficiency as a light source. (d) Briefly discuss one physical reason, other than radiation, that could cause the actual filament temperature to differ from the temperature calculated using an ideal radiation-only model.

Intuitive explanation

The filament reaches a steady-state temperature when the electrical power flowing in equals the thermal power radiated away. You'll use the Stefan–Boltzmann law, $P = \epsilon \sigma A T^4$ , which tells you how much energy a hot surface radiates based on its temperature, emissivity, and surface area. The key is recognizing that at equilibrium, the $60\,\text{W}$ input must equal the radiated power, so you can either verify the given temperature or solve for an unknown—but first you need to calculate the filament's surface area from its cylindrical geometry.

Intuitive explanation

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