Calculate efficiency for an incandescent light bulb?

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To calculate the efficiency of a 40 W incandescent light bulb with a tungsten filament at approximately 2800 K, it is noted that about 10% of its radiation is emitted in the visible spectrum, suggesting that the remaining energy contributes to heating. When used as a heater in a closed room, the heating efficiency can be considered nearly 100%. However, in applications like fish tanks or cooking appliances, efficiency varies due to the loss of visible light and infrared radiation through glass and other materials. For cooking, while the bulb may provide 40W, it is insufficient for effective food preparation, and the actual efficiency in a toaster setup is estimated to be less than 50%. The overall efficiency depends on the specific geometry and emissivity of the food being heated.
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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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