Compare input, useful output, and efficiency of energy systems

Ontario Grade 11 Physics

Energy and Society

Input energy, useful output, and efficiency

Energy systems receive energy and transfer it into different forms. A lamp, for example, receives electrical energy and produces light, as well as heating. To compare systems fairly, first decide what each system is meant to do. Then identify its input, its useful output, and its efficiency. This lesson uses energy values in joules and a simple percentage relationship.

What you will learn

Identify the energy input and useful energy output of a system.
Explain why some input energy is not useful for the system’s purpose.
Calculate and compare the efficiency of energy systems.
Use energy values and units to judge what an efficiency comparison means.

1. Set the boundary and identify the energy

A system is the object or group of objects being studied. Its boundary is the chosen edge between the system and its surroundings. Energy that enters across the boundary is the input. Energy that leaves is an output. Naming the system matters: for a lamp, the system might be the lamp itself, not the entire room.

Energy is a scalar. A scalar has a size but no direction, unlike a vector, which has both size and direction. Energy values therefore do not need a positive or negative direction. In this lesson, input and output describe transfers across the system boundary, not vector directions.

The SI unit of energy is the joule, abbreviated J. A kilojoule, abbreviated kJ, is one thousand joules. Convert values to the same unit before comparing or calculating.

Useful output is the part of the output that serves the system’s intended purpose. For a lamp, light is usually the useful output. Heating may be an output too, but it is not usually the purpose of the lamp. For a room heater, heating the room is useful output. The same kind of energy can be useful in one system and not useful in another.

Define the system and its purpose before deciding which output is useful.
Input crosses into the system; output crosses out.
Energy is measured in joules and is a scalar.

2. Compare input, useful output, and other output

Energy systems often produce more than one output. A motor may produce motion that serves its purpose, while also heating and sound. Those other outputs still carry energy; they are simply not useful for the motor’s stated purpose. Calling an output non-useful does not mean the energy has disappeared.

For a simple comparison, treat the input energy as being divided among useful output and other output. In a real system, the boundary and measurements must be chosen carefully. The system’s purpose must also be stated, or two people may classify the same output differently.

Efficiency compares useful output energy with input energy. It tells what fraction of the input became the desired output. Multiply that fraction by one hundred to write it as a percentage. Efficiency has no unit because the energy units cancel.

For the comparison to be meaningful, use the same energy units for input and useful output, and use the same purpose when judging what counts as useful. Under this model, useful output cannot be greater than input, so efficiency cannot exceed one hundred percent.

\text{efficiency} = \frac{E_{\text{useful output}}}{E_{\text{input}}} \times 100\%

Other outputs still carry energy, even when they do not serve the chosen purpose.
Efficiency is a ratio of useful output energy to input energy.
A larger efficiency means a larger fraction of the input becomes useful output.

3. Read comparisons with care

Efficiency is not the same as the amount of useful energy produced. Two systems can have the same efficiency but different input energies. The one receiving more input can produce more useful output while keeping the same fraction. A comparison should say whether it concerns efficiency, input, or useful output.

A system with lower efficiency can still produce more useful output if it receives much more input. For example, knowing only that one device is more efficient does not tell you which device produced more useful energy. Check the given energy amounts as well as the percentages.

The purpose of a comparison may affect the result. If the task is to warm a room, heating is useful output. If the task is to provide light, light is useful output. State the purpose before using the equation, and keep that choice consistent throughout the calculation.

Efficiency compares fractions; useful output compares amounts.
A higher efficiency alone does not prove that a system produces more useful energy.
Make the intended purpose explicit.

4. A reliable calculation routine

First name the system and its purpose. Next list the input energy and the useful output energy. Convert them to the same unit if needed. Then substitute into the efficiency relationship and calculate. Round the result to a sensible number of significant figures based on the supplied values.

Finally, check the result. Efficiency should be between zero and one hundred percent for this comparison. The units should cancel, leaving a percentage with no energy unit. Ask whether the result makes sense: for example, an efficiency of fifty percent means half the input energy became the chosen useful output.

No direction is assigned to energy in these calculations. The words input and output show which way energy crosses the chosen system boundary. They do not turn energy into a vector.

E_{\text{input}} = E_{\text{useful output}} + E_{\text{other output}}

Keep energy units consistent before substituting.
Report efficiency as a percentage and check that it is physically reasonable.
Check both the result and what the comparison actually tells you.

Worked example

Efficiency of a lamp

A lamp receives 1,200 J of electrical energy. Its intended purpose is to produce light, and it gives 180 J of light energy. Find its efficiency.

Define the system
The system is the lamp. Its purpose is producing light, so the useful output is 180 J. The input is 1,200 J. Energy is a scalar, so no positive direction is needed.
Use the efficiency relationship
Divide useful output energy by input energy, then multiply by one hundred to express the fraction as a percentage.
$\text{efficiency} = \frac{180\ \mathrm{J}}{1200\ \mathrm{J}} \times 100\%$
Calculate and check
The joule units cancel. The inputs have two significant figures, so report two significant figures. The result is below one hundred percent, as expected.
$\text{efficiency} = 15\%$

Answer: The lamp’s efficiency for producing light is 15%.

Check: The units cancel, and 15% means 15% of the input became light. The remaining input energy is associated with other outputs, not missing energy.

Worked example

Compare energy transfers in a motor

A motor receives 2.4 kJ of energy and produces 1.8 kJ of useful motion energy. Find its efficiency and the amount of other output energy in this simple energy comparison.

Identify the purpose and values
The system is the motor, and its purpose here is to produce motion. The useful output is 1.8 kJ and the input is 2.4 kJ. The values are already in the same unit.
Calculate efficiency
Use the useful output divided by the input. The kilojoule units cancel, leaving a percentage.
$\text{efficiency} = \frac{1.8\ \mathrm{kJ}}{2.4\ \mathrm{kJ}} \times 100\% = 75\%$
Find other output energy
For this simple comparison, subtract the useful output from the input. This gives energy in outputs that are not the motor’s intended motion output.
$E_{\text{other output}} = 2.4\ \mathrm{kJ} - 1.8\ \mathrm{kJ} = 0.6\ \mathrm{kJ}$

Answer: The motor is 75% efficient for producing motion, with 0.6 kJ in other output energy.

Check: The useful and other outputs add to the 2.4 kJ input. The efficiency is below 100%, and both energy amounts retain the unit kJ.

Worked example

Same efficiency, different useful output

System A receives 500 J and produces 200 J of useful output. System B receives 1,500 J and produces 600 J of useful output. Compare their efficiencies and useful output amounts.

Calculate System A’s efficiency
For each system, divide useful output by input. Both values are in joules, so the units cancel.
$\text{efficiency}_{A} = \frac{200\ \mathrm{J}}{500\ \mathrm{J}} \times 100\% = 40\%$
Calculate System B’s efficiency
Use the same relationship for System B. Keeping the method the same makes the comparison fair.
$\text{efficiency}_{B} = \frac{600\ \mathrm{J}}{1500\ \mathrm{J}} \times 100\% = 40\%$
Compare the results
The efficiencies are equal, but the useful output amounts are not. System B produces more useful energy because it receives more input energy.
$600\ \mathrm{J} > 200\ \mathrm{J}$

Answer: Both systems are 40% efficient. System B produces 600 J of useful output, compared with 200 J from System A.

Check: The calculations use consistent SI energy units and give efficiencies within the allowed range. Equal efficiency does not mean equal useful output.

Common mistakes and how to avoid them

Calling every output useful.

Correction: Decide what the system is meant to do. Count only the output serving that purpose as useful output.

Treating other output energy as energy that has vanished.

Correction: Other output still carries energy. It is not useful for the stated purpose, but it is still an output.

Assuming the system with higher efficiency always produces more useful energy.

Correction: Compare the useful output amounts too. Systems with different inputs can have equal efficiency but different useful outputs.

Leaving different units in the numerator and denominator.

Correction: Convert energy values to a common unit before calculating efficiency.

Lesson summary

Define the system boundary and intended purpose before sorting energy transfers.
Input energy enters the system; output energy leaves it.
Useful output is the output that serves the system’s stated purpose.
Efficiency is useful output energy divided by input energy, expressed as a percentage.
Compare both efficiency and energy amounts when deciding what a system produces.

Check your understanding

Question 1

A device receives 800 J and gives 240 J of useful output. What is its efficiency?

30%
70%
300%
1,040%

Show answer and explanation

30%

Divide 240 J by 800 J and multiply by 100%. The joule units cancel, giving 30%.

Question 2

Two systems are both 50% efficient. System X receives 400 J, while System Y receives 1,000 J. Which statement is correct?

They must produce the same useful output.
System X produces 500 J of useful output.
System Y produces more useful output than System X.
System Y is more efficient because it receives more energy.

Show answer and explanation

System Y produces more useful output than System X.

At 50% efficiency, X produces 200 J and Y produces 500 J of useful output. Their efficiencies are equal, but Y’s useful output amount is larger.

Key terms

System: The object or group of objects being studied.
Boundary: The chosen edge separating the system from its surroundings.
Input energy: Energy transferred into the system across its boundary.
Useful output: Output energy that serves the system’s stated purpose.
Efficiency: The fraction of input energy that becomes useful output, often stated as a percentage.
Scalar: A quantity with a size but no direction, such as energy.

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About this lesson and its review

Published by DoAssignment. This reviewed lesson follows Ontario Grade 11 Physics (SPH3U), expectation D2.7. 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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