Grade 11 · lesson 3 of 4

Distinguish Useful Efficiency from Conservation

Compute an output fraction and complete an energy ledger.

Download the grade pack

Supplied materials

  • Use the supplied fictional observations or model. Paper and pencil are optional; spoken answers are welcome. No equipment or real experiment is needed.

Learn the idea

Efficiency describes the fraction of input energy serving a stated purpose; energy conservation accounts for all destinations. For a lighting task, useful efficiency here is emitted-light energy / electrical input energy × 100%. Thermal transfers count in conservation even when they are not the desired light output. Specify the device boundary and interval before adding amounts.

During a fictional interval, a lamp receives 300 J electrically and emits 75 J as light and 210 J as thermal transfers to its surroundings. Its stored internal energy increases by 15 J. These are the only listed transfers. The ledger balances: 75 + 210 + 15 = 300 J. Lighting efficiency is 75/300 = 25%, not 100%. This simple energy fraction does not measure human-perceived brightness, which also depends on wavelength.

  1. Write electrical input and the three accounted destinations.
  2. Add destinations to check 300 J.
  3. Calculate the lighting energy fraction as a percentage.
  4. Explain why 25% useful efficiency can coexist with a complete 100% energy ledger.
Worked example

See the method in action

Lighting efficiency is 25%. Thermal transfer plus storage account for the other 225 J, so the low useful fraction does not imply energy destruction.

Your turn

Check your reasoning

Your answer is temporary and is not saved to an account. Download your current answer before leaving or reloading. Written work on paper or in another app stays in your own file.

A no-storage-change lighting model receives 500 J and emits 150 J light; all remaining energy is thermal output. Which pair is correct?

Feedback appears here after you check. This is practice, not a graded school assessment.

See the answer and explanation

30% useful and 350 J thermal

150/500 × 100 = 30%, and 500 − 150 = 350 J thermal under the supplied complete-output model.

Take it further

Change the useful job to warming an enclosure and give 400 J input, 320 J thermal transfer into that enclosure, and 80 J to the outside. Useful efficiency for that stated job is 320/400 = 80%; usefulness depends on the goal.

Nothing here is saved to an account or sent to a teacher. The answer download includes your current choice and any checked feedback.

Check the subject sources

These LifeStarter passages, scenarios and exercises are authored for this Academy with AI assistance and review. External resources keep their publishers’ terms. A grade label suggests a starting level; it does not establish alignment with your school’s complete requirements.