IGCSE · 25 September 2026 · 6 min read
IGCSE Physics energy and efficiency questions: choose the useful output
An efficiency question is partly a calculation and partly a decision about what the device is meant to do. Warming water is useful for a kettle, but warming the casing is usually not counted as its useful output. A motor's purpose may be lifting a load, so the useful output is the load's gain in gravitational potential energy.
These original examples connect energy, work, power and efficiency. They also show why a percentage can look sensible while describing the wrong system. Write the intended output and time interval before calculating; that small step prevents many errors with electrical input and mechanical output.
Define the job and keep the accounting consistent
Write efficiency = useful output energy / total input energy. If both quantities refer to the same time interval, you can instead use useful output power / total input power. The result is a fraction; multiply by 100 to express it as a percentage. A 0.75 fraction and 75% describe the same efficiency, whereas 0.75% describes a much smaller fraction.
Choose a system boundary before identifying losses. For a motor lifting a load, electrical energy enters and the load's gravitational potential energy increases; energy may also warm the motor or surrounding air. The full input remains accounted for. Calling energy wasted means it is not useful for the chosen task, not that it has disappeared.
Worked example: electrical input and a lifted load
A motor lifts a 12 kg load vertically by 2.5 m in 8.0 s. Its electrical input power is 50 W. Using g = 10 N/kg for this example, the useful energy is ΔEₚ = mgh = 12 × 10 × 2.5 = 300 J. The total input energy is E = Pt = 50 × 8.0 = 400 J. The efficiency is 300 / 400 = 0.75 = 75%.
The useful output power is 300 / 8.0 = 37.5 W, giving the same efficiency through 37.5 / 50. The remaining 100 J is transferred in other ways during the lift, assuming the stated useful output is the only one being counted. The motor's 50 W input is not the same as its useful lifting power; treating both as 50 W would incorrectly imply 100% efficiency.
Worked example: heating water without confusing energy and temperature
A 1.5 kW kettle heats 0.50 kg of water from 20°C to 80°C in 100 s. Take the water's specific heat capacity as 4200 J/(kg °C). The useful energy gained by the water is E = mcΔT = 0.50 × 4200 × (80 − 20) = 126 000 J. Electrical input is 1500 × 100 = 150 000 J, so the efficiency for heating the water is 126 000 / 150 000 = 84%.
The temperature change is 60°C, not the final 80°C. The input power must be converted from 1.5 kW to 1500 W before multiplying by seconds. This calculation uses the stated temperature rise without a change of state. If water were boiling away, energy transferred during the phase change would need separate consideration rather than being represented by another temperature rise.
Use power to discuss speed, not total energy alone
Power is the rate of energy transfer. Two devices can transfer the same useful energy while taking different times. A more powerful heater can heat the same mass of water through the same temperature change more quickly if the useful heating power is greater. That does not automatically make it more efficient: efficiency also depends on its total input and other transfers.
For a lifting device, reducing the lifting time while keeping the load and height unchanged raises the required useful power but leaves mgh unchanged. If a question supplies power and asks for time, use t = E / P with the appropriate input or output energy. Matching useful energy with input power requires allowing for the efficiency.
Recognise impossible answers and misleading improvements
An efficiency above 100% for these simple devices signals inconsistent units, reversed division, mismatched intervals or an incomplete account of inputs. Recheck the useful-output numerator and total-input denominator. A figure below 100% can still be wrong, so also compare it with the energy balance and the device's stated purpose.
An improvement needs a physical reason. Insulation can reduce unwanted thermal transfer from hot water; lubrication can reduce energy transferred by friction in a mechanical system. Neither is a universal answer for every device. Explain which transfer changes and why that leaves a larger fraction of the same input available for the intended output.
Independent practice: work backwards from efficiency
A device transfers 3600 J usefully and is 60% efficient. Find the input energy and the energy not counted as useful. Convert 60% to 0.60. Since 0.60 = 3600 / input, the input is 3600 / 0.60 = 6000 J. The remaining energy is 6000 − 3600 = 2400 J. Multiplying 3600 by 0.60 instead would incorrectly make the input smaller than the useful output.
If the transfer lasts 30 s, the useful output power is 120 W and input power is 200 W. Their ratio confirms 60%. To revise, cover the solution, write a complete energy account and then compare both the calculation and its interpretation. Return later with a different useful output; the same relationship should work without copying the numbers.
Questions, explained
Choose a question for a direct answer, then explore the explanation and supporting resources. Each answer has its own link to save or share.
Can I calculate efficiency using power instead of energy?
Yes, use useful output power divided by total input power when both describe the same operating conditions. It gives the same ratio as energies transferred over a common interval. Do not divide useful energy by input power and call the result efficiency: joules divided by watts gives a time, showing that the quantities do not form the required ratio.
If energy is conserved, how can a device waste energy?
Wasted energy is energy transferred in ways that do not perform the intended useful task. For a motor lifting a load, warming the motor and surroundings is usually not counted as useful lifting output. The total energy is still conserved. Efficiency describes how the input is distributed, rather than how much energy survives the process.
How do I find input energy when efficiency and useful energy are given?
Rearrange efficiency = useful output / input to input = useful output / efficiency, using efficiency as a fraction. For 60% efficiency and 3600 J useful output, input = 3600 / 0.60 = 6000 J. Check that the input exceeds the useful output and that their ratio reproduces the stated efficiency.