Uncovering the Source of Extra 267 J in Standard Incandescent Light Bulb

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SUMMARY

The discussion centers on the discrepancy between the electrical power input and the radiant power output of standard incandescent light bulbs, specifically a 100W bulb. Calculations based on the Stefan-Boltzmann law indicate that the filament, at approximately 2500 K, radiates around 367 watts, leading to an excess of 267 J over the expected 100 J input. Participants clarify that the filament does not burn or store energy, and the calculations must consider factors such as filament geometry and emissivity to reconcile the differences in power measurements.

PREREQUISITES
  • Understanding of the Stefan-Boltzmann law and its application in thermal radiation.
  • Familiarity with electrical power calculations, specifically P = V^2/R.
  • Knowledge of filament materials, particularly tungsten, and their properties at high temperatures.
  • Basic concepts of emissivity and its impact on thermal radiation.
NEXT STEPS
  • Research the Stefan-Boltzmann law in detail and its implications for thermal radiation.
  • Explore the properties of tungsten as a filament material in incandescent bulbs.
  • Investigate the effects of filament geometry on light bulb efficiency and performance.
  • Learn about the relationship between electrical power input and thermal radiation output in incandescent light bulbs.
USEFUL FOR

Physicists, electrical engineers, lighting designers, and anyone interested in the principles of incandescent lighting and thermal radiation.

  • #31
I see. So then if you want to dissipate 100 watts you have to balance length and thickness to get the resistance at the right amount and also keep the temperature from either being too low (which will make a dim, inefficient bulb) or too high (reducing filament life).

Also, do you know if my assumption is post #13 about using less surface area in the calculation is accurate at all?
 
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  • #32
I said heat, not temperature. The thing that destroys bulbs is heat.

I am not sure how important the shadowing effect is. If you model a coil as a hollow tube, the light's dependence on the inner radius is only through the thickness of the tube and thus its resistivity.
 
  • #33
Vanadium 50 said:
I said heat, not temperature. The thing that destroys bulbs is heat.

I don't know what you mean by this. Could you elaborate?
 
  • #34
I don't have a reference but I recall coiling a filament reduces failure due to thermal shock. Time lapse movies of lights turning on show considerable vibration due to sudden temperature changes.
 
  • #35
Drakkith said:
I don't know what you mean by this. Could you elaborate?

Heat and temperature are two different quantities - I can stick my hand in a 350 degree oven without injury, but I can't stick it in a 212 degree pot of boiling water. I wrote heat because I meant heat. That's all I meant.
 

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