Calculate efficiency for an incandescent light bulb?

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Discussion Overview

The discussion revolves around calculating the efficiency of an incandescent light bulb, specifically a 40 W bulb with a tungsten filament operating at approximately 2800 K. Participants explore the implications of using the bulb for heating purposes, considering factors such as blackbody radiation and the visible spectrum of emitted light.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Experimental/applied

Main Points Raised

  • One participant questions how to calculate the efficiency of the bulb, noting that only about 10% of the radiation is emitted in the visible spectrum, suggesting that the rest contributes to heating.
  • Another participant argues that if the bulb is used as a heater in a closed room, its heating efficiency could be considered 100%.
  • A different participant raises the scenario of using the bulb in a cooking appliance, implying that the efficiency would depend on the specific application.
  • One participant mentions that the power output of the bulb might vary slightly due to engineering tolerances, estimating it could be around 40 W ± 10%.
  • Another participant suggests that for cooking, the efficiency could be less than 50% due to the geometry of the setup and the emissivity of the food, which changes as it cooks.

Areas of Agreement / Disagreement

Participants express differing views on the efficiency of the bulb when used for heating or cooking, with no consensus reached on a specific efficiency value. The discussion remains unresolved regarding the exact efficiency in various applications.

Contextual Notes

Participants highlight limitations related to assumptions about blackbody radiation, the impact of geometry on efficiency, and the variability in power output of the bulb.

Who May Find This Useful

Individuals interested in the practical applications of incandescent bulbs for heating or cooking, as well as those exploring the physics of light and thermal radiation.

germblaster
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How does one calculate efficiency for an incandescent light bulb? Assume the bulb is 40 W, has a tungsten filament at temp. ~ 2800 K. The bulb would be used for heating in a device.
If we assume blackbody:
From Wien's displacement law, the maximum wavelength is 9.99E-7 m. Fromt the blackbody curve for this temp., it appears that only ~10% of the radiation is emitted in the visible spectrum. Does this mean that the rest goes into heating?

Is if safe to assume an efficiency of greater than 90%?
 
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If you are using the bulb as a heater, then in a closed room it's heating efficiency is 100%.

If you are using it in something like a fish tank you would subtract the visible part that escapes - it's a bit less than that because quite a bit of infrared will go through the glass as well.
 


What about a cooking appliance?
 


You're pretty much going to get 40W out of it - the power changes a little as the filament heats up, and I don't know good the engineering tolerances are on light bulbs, so it might be 40W +- 10%
 


germblaster said:
What about a cooking appliance?

40W is not enough to cook food.

But assuming you were using enough power to cook, any radiation (visible or otherwise) that is absorbed by the food will heat it. The answer will come down to the particular geometry of your setup, plus the emissivity of the food which will change as it darkens while cooking.

For a simple toaster, I'd put the efficiency at less than 50%, since at least half the radiation is directed away from the toast. Factor in that the toast, while it is still white, will be reflecting a significant portion of the radiation.
 

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