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D2.8 · Determine device efficiency and identify energy transformations

Learn to determine device efficiency and identify energy transformations through clear examples and targeted practice.

Ontario Grade 9 Science

Physics: Principles and Applications of Electricity

Ontario Grade 9 Science — D2.8

A lamp changes electrical energy into light, but it also warms up. A speaker changes electrical energy into sound, and some energy warms its parts. These everyday observations lead to two science questions: What energy transformations happen in a device? How much of its input energy becomes the output we want? In this lesson, you will describe energy changes and calculate efficiency using supplied or clearly labelled hypothetical information.

What you will learn

  • Identify the input energy and useful energy output of a device.
  • Describe energy transformations in a device.
  • Calculate device efficiency using supplied energy values.
  • Explain what an efficiency result does and does not show.

1. Follow energy through a device

Energy is the ability to cause change. You have met forms of energy such as electrical, light, sound, thermal, chemical, and motion energy. A device takes in energy and changes it. An energy transformation is a change from one form of energy to another.
Consider a battery-powered flashlight. Chemical energy stored in the battery changes into electrical energy in the circuit. The lamp changes electrical energy into light and thermal energy. Light is usually the desired output for a flashlight. The warmth is another output, even if it is not the one wanted for that task.
A useful energy output is the output form that does the device’s intended job. The intended job matters. Thermal energy in a room is useful output from a room heater. Light is usually useful output from a flashlight. So, decide what task is being considered before naming the useful output.
A simple energy-flow model can show the main transformations. For example, a flashlight model might show chemical energy changing to electrical energy, then to light and thermal energy. Such a model helps organize ideas. It may not show every small output or detail.
  • A device changes input energy into one or more output forms.
  • Useful output depends on the task being considered.
  • An energy-flow model is a simplified way to show transformations.

2. Determine efficiency

Efficiency compares the useful energy output with the total energy input. It tells us what share of the input became the chosen useful output. A higher efficiency means a greater share became that output for the stated task.
To calculate efficiency, divide the useful output energy by the total input energy. Then multiply by 100 to express the result as a percentage. Use the same energy unit for both values, such as joules (J). The units then cancel in the division, so the percentage has no energy unit.
For example, if a device takes in 100 J and provides 30 J of useful energy, its efficiency is 30%. The remaining input is not destroyed. It has gone into other outputs or the surroundings. In a simple calculation, input minus useful output can represent the energy not counted as useful, when the values refer to the same device and time.
For this efficiency calculation, useful output should not be greater than total input. A result above 100% is a reason to check the supplied numbers, units, or choice of useful output. It is not a reasonable efficiency result for the stated comparison.
efficiency=useful energy outputtotal energy input×100%\text{efficiency}=\frac{\text{useful energy output}}{\text{total energy input}}\times100\%
  • Divide useful output energy by total input energy, then multiply by 100%.
  • Use matching energy units for input and output.
  • Efficiency describes a specific output for a specific task.

3. Use evidence carefully

To determine efficiency, you need values for total input energy and useful output energy. A device label, a supplied data set, or a carefully planned investigation might provide them. The practice examples in this lesson use hypothetical values. They are not experimental measurements.
In a real investigation, the task and useful output would need to be clearly defined. The measurements would also need to be reliable. A simplified energy-flow model can show the main transformations, but it may leave out small outputs or details. The model is not a complete record of every energy movement.
Efficiency is a scientific measure for a stated task and set of values. It does not by itself decide whether a device is safe, affordable, environmentally responsible, or right for a particular person. Those questions need other information and may involve judgments beyond the efficiency calculation.
Do not test mains-powered devices or open batteries at home. The practice calculations below need only the stated numbers. If a real investigation is assigned, follow the teacher’s instructions and use only approved equipment.
  • The result depends on the quality of the values and the definition of useful output.
  • A model can show important transformations without showing every detail.
  • Efficiency alone does not make a social or ethical decision.

Energy pathways for different tasks

DeviceInput energyUseful output for the stated taskPossible other output
FlashlightChemical energy in battery, then electrical energyLightThermal energy
MotorElectrical energyMotion energyThermal energy or sound
Room heaterElectrical energyThermal energy in the roomOther outputs may occur

Worked example

Flashlight efficiency

A hypothetical flashlight takes in 80 J of electrical energy. For this task, 24 J leaves as useful light energy. Determine its efficiency.
Flashlight energy changes
Flashlight energy changesSimplified model for the hypothetical example. It shows main energy forms, not measured amounts or every detail.transformationtransformationother outputChemical energy in batteryElectrical energyUseful lightThermal energyenergy

Simplified model for the hypothetical example. It shows main energy forms, not measured amounts or every detail.

  1. Choose the useful output
    The stated task is to provide light, so use 24 J as the useful output. The total input is 80 J.
  2. Calculate the percentage
    Divide useful output by total input and multiply by 100. Both values are in joules, so the units match.
    24 J80 J×100%=30%\frac{24\,\mathrm{J}}{80\,\mathrm{J}}\times100\%=30\%
  3. Interpret the result
    Using these hypothetical values, 30% of the input is counted as useful light output. The other energy was not counted as useful light; it was not destroyed.
Answer: The flashlight efficiency is 30%.
Check: The useful output, 24 J, is less than the 80 J input, and the calculated result is below 100%.

Worked example

A small motor

A hypothetical motor receives 150 J of electrical energy. It produces 90 J of useful motion energy for a task. Determine its efficiency and describe one other energy output that could occur.
Motor energy changes
Motor energy changesSimplified model for the hypothetical example. It is not a measurement of the motor or a complete list of its outputs.transformationother outputElectrical energyUseful motion energyThermal energy or soundenergy

Simplified model for the hypothetical example. It is not a measurement of the motor or a complete list of its outputs.

  1. Identify input and useful output
    The motor’s total input is 150 J of electrical energy. Its stated useful output is 90 J of motion energy.
  2. Calculate efficiency
    Divide the useful output by the total input, then express the fraction as a percentage.
    90 J150 J×100%=60%\frac{90\,\mathrm{J}}{150\,\mathrm{J}}\times100\%=60\%
  3. Describe another output
    The motor could also produce thermal energy as its parts warm, or sound. These are possible outputs even though the example counts motion as useful.
Answer: The motor efficiency is 60%. It may also produce thermal energy or sound.
Check: 90 J is less than 150 J, and 60% is below 100%.

Worked example

Compare two devices

Two hypothetical devices each take in 200 J for the same stated task. Device A provides 50 J of useful output. Device B provides 110 J of useful output. Calculate each efficiency and identify which has greater efficiency for the stated task.
  1. Calculate Device A
    Use 50 J as the useful output and 200 J as the total input.
    50 J200 J×100%=25%\frac{50\,\mathrm{J}}{200\,\mathrm{J}}\times100\%=25\%
  2. Calculate Device B
    Apply the same method because the input and stated task are the same for both devices.
    110 J200 J×100%=55%\frac{110\,\mathrm{J}}{200\,\mathrm{J}}\times100\%=55\%
  3. Compare the results
    Device B has greater efficiency for this task: 55% compared with 25%. This comparison does not tell us other things, such as cost or suitability.
Answer: Device A is 25% efficient and Device B is 55% efficient. Device B has greater efficiency for the stated task.
Check: Both results are below 100%, and Device B has the larger useful-output share.

Common mistakes and how to avoid them

Calling every output wasted energy.
Correction: Name the intended task first. An output that is not useful for one task may be useful for another.
Saying energy disappears when efficiency is below 100%.
Correction: Energy changes form or is transferred. The calculation counts only the output chosen as useful.
Dividing input by useful output.
Correction: Divide useful output by total input, then multiply by 100%.
Treating high efficiency as proof that a device is best in every way.
Correction: Efficiency answers a specific energy question. Other choices need other information.

Lesson summary

  • Devices transform input energy into one or more output forms.
  • Efficiency compares useful energy output with total energy input.
  • Use the same energy unit for both values and express the result as a percentage.
  • A calculation depends on the stated task and the quality of the information supplied.

Check your understanding

Question 1

A hypothetical device takes in 120 J and provides 36 J of useful output. What is its efficiency?
  1. 30%
  2. 36%
  3. 70%
  4. 300%
Show answer and explanation
30%
Divide 36 J by 120 J and multiply by 100%. The useful output is 30% of the input.

Question 2

A lamp changes electrical energy mainly into which two output forms?
  1. Light and thermal energy
  2. Chemical energy and motion energy
  3. Sound energy and chemical energy only
  4. Electrical energy and stored fuel
Show answer and explanation
Light and thermal energy
A lamp produces light and also warms. The exact amounts depend on the lamp.

Question 3

What must you decide before choosing the useful output in an efficiency calculation?
  1. The task the device is intended to do
  2. The device’s colour
  3. Whether the device is popular
  4. The size of its packaging
Show answer and explanation
The task the device is intended to do
Useful output depends on the task. Thermal energy is useful for heating, while light is usually useful for a flashlight.

Key terms

Energy transformation
A change from one form of energy to another.
Input energy
The total energy supplied to a device.
Useful energy output
The output energy form that does the intended task.
Efficiency
The percentage of input energy that becomes the chosen useful output.
Model
A simplified representation that helps explain a system but may leave out details.

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About this lesson and its review

Published by DoAssignment. This reviewed lesson follows Ontario Grade 9 Science (SNC1W), expectation D2.8. It is a study resource, not an official curriculum publication.

Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.

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