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

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Discussion Overview

The discussion revolves around the apparent discrepancy in energy output from a standard incandescent light bulb, specifically addressing the extra 267 J of energy that seems to be radiated compared to the electrical energy input. Participants explore the physics of the filament, including temperature, surface area, and emissivity, as well as the implications of these factors on power calculations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant calculates the radiant power of a filament at 2500 K using the Stefan-Boltzmann law, arriving at 367 watts, and questions the source of the additional 267 J of energy.
  • Another participant asserts that the filament does not burn and therefore does not store energy to be released.
  • Some participants suggest that the assumptions regarding the filament's dimensions and emissivity may be flawed, leading to discrepancies in calculations.
  • There are discussions about the surface area of the filament, with varying estimates based on different assumptions about its shape and dimensions.
  • One participant notes that tungsten evaporates, which could imply that the filament is consumable, while another counters that this evaporation occurs at hot points and is not part of the filament's design.
  • Several participants express uncertainty about the effective surface area of a coiled filament and its implications for radiation absorption and emission.
  • One participant proposes that the interior of a coiled filament may absorb much of its own radiation, affecting the overall power output.
  • Another participant discusses the effects of coiling on the filament's brightness and the physics behind it, noting the complexity of the interactions involved.

Areas of Agreement / Disagreement

Participants express multiple competing views regarding the calculations and assumptions about the filament's properties, and the discussion remains unresolved with no consensus on the source of the extra energy or the accuracy of the models presented.

Contextual Notes

Participants acknowledge limitations in their assumptions about filament dimensions, emissivity, and the effects of coiling, which may impact the accuracy of their calculations and conclusions.

  • #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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