Calculate efficiency for an incandescent light bulb?

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SUMMARY

The efficiency of a 40 W incandescent light bulb, particularly one with a tungsten filament operating at approximately 2800 K, can be calculated using principles from blackbody radiation. According to Wien's displacement law, only about 10% of the emitted radiation is in the visible spectrum, suggesting that the remaining radiation contributes to heating. In a closed environment, the heating efficiency can approach 100%, while applications like cooking or heating in a fish tank may yield efficiencies below 50% due to losses from visible light and infrared radiation escaping through glass.

PREREQUISITES
  • Understanding of blackbody radiation principles
  • Familiarity with Wien's displacement law
  • Knowledge of incandescent bulb specifications and operation
  • Basic concepts of thermal efficiency in heating applications
NEXT STEPS
  • Research blackbody radiation and its applications in heating
  • Study the impact of filament temperature on light bulb efficiency
  • Explore the emissivity of various materials in cooking applications
  • Investigate the design and efficiency of heating appliances like toasters
USEFUL FOR

Engineers, physicists, and anyone involved in thermal management or appliance design will benefit from this discussion, particularly those focused on optimizing heating efficiency in incandescent lighting and cooking devices.

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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