Efficiency of a Light Bulb: Calculating Energy Transfer

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SUMMARY

The discussion centers on calculating the energy transfer of a light bulb with an efficiency rating of 85% when supplied with 60J of electrical energy. The correct calculation reveals that 51J of energy is converted to light energy, while 9J is lost primarily as heat. The term "efficient" in this context refers to the energy that is effectively utilized as light, confirming that the light bulb emits 51J of useful light energy.

PREREQUISITES
  • Understanding of energy efficiency concepts
  • Basic knowledge of energy transfer calculations
  • Familiarity with the principles of electrical energy
  • Concept of heat loss in electrical devices
NEXT STEPS
  • Research the principles of energy efficiency in electrical appliances
  • Learn about the calculations involved in energy transfer and conversion
  • Explore the impact of heat loss on the efficiency of light bulbs
  • Investigate different types of light bulbs and their efficiency ratings
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Students studying physics, electrical engineers, and anyone interested in understanding energy efficiency in lighting technology.

Molly1235
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Homework Statement

"A light bulb is 85% efficient. If it is supplied with 60J of electrical energy, how much of this energy is transferred to its surroundings as light energy?


The attempt at a solution

So I calculated that 85% of 60 is 51J, however I don't know if being "efficient" refers to energy that remains in the circuit, or energy that is released as light energy, seeing as this is the main purpose of a light bulb? I'm thinking it's the first, which would mean 9J is lost as light energy?
 
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I think it's safe to assume that the useful energy in this case is the light being emitted from the bulb. Which would mean that 51 J of light energy is emitted. The other energy doesn't "remain in the circuit" it is lost to heat.
 
Millacol88 said:
I think it's safe to assume that the useful energy in this case is the light being emitted from the bulb. Which would mean that 51 J of light energy is emitted. The other energy doesn't "remain in the circuit" it is lost to heat.

Ah ok! Thank you very much :)
 

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