K323 / 2027
Energy overview

Lesson 5 of 6

Efficiency

  • I can calculate efficiency as useful energy output divided by total energy input.Syllabus

    Syllabus K323, 6(f). Calculate efficiency as useful energy output divided by total energy input.

Make a guess. You are not marked.

An electric motor is 40% efficient at lifting loads. What happens to the other 60% of the energy supplied to it?

  1. It stays in the motor for the next lift.
  2. It heats the motor and the surroundings.
  3. It is destroyed inside the motor.
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It heats the motor and the surroundings.

Energy is conserved. The 60% that is not useful is transferred to internal stores of the motor, bearings and air by heating and sound, where it is no longer useful for lifting.

Efficiency is the fraction of the total input that becomes the useful output for a stated purpose.

Conservation accounts for all the energy. Efficiency identifies which part serves the intended task. Use the same process and time interval for input and output.

Efficiency = useful energy output / total energy inputThe energy units cancel, giving a fraction with no unit. Multiply the fraction by 100% to express it as a percentage.

For a lifting motor, the intended increase in gravitational energy is useful. Energy increasing the motor's or surroundings' internal stores does not perform that lifting task. For a heater, an increase in the intended object's internal energy can be the useful output instead.

Useful lifting accounts for 300 J of the 400 J input

The intended result is an increase in the gravitational energy store of the load and Earth. The remaining 100 J increases internal stores of the motor and surroundings.

An energy account for a lifting system with 75 percent efficiencyThe electrical energy input is 400 J. Of this, 300 J is the useful increase in the gravitational energy of the load and Earth, and 100 J increases internal stores of the motor and surroundings. The input bar and the combined store-increase bar have equal lengths. Both use the same energy scale. The useful and internal segments are in an exact three-to-one ratio. The useful fraction is 300 divided by 400, or 75 percent. The remaining energy is accounted for.Transferred electrically400 J inputResulting store increases300 J100 J

300 J useful: gravitational store increase

100 J: internal store increases

Efficiency = 300 / 400 = 0.75 = 75%.

The motor receives 400 J electrically. Of that input, 300 J increases the intended gravitational store and 100 J increases internal stores. The full 400 J remains accounted for.

Worked example

Useful lifting output from a 400 J input

  1. State the useful output: the intended gravitational increase is 300 J.
  2. Divide by the total input: efficiency = 300 / 400 = 0.75.
  3. Convert to a percentage: 0.75 x 100% = 75%.
  4. Account for the remainder: 400 - 300 = 100 J increases internal stores, corresponding to 25% of the input.

The 100 J has not been destroyed. It has a destination, but it is not useful for the stated lifting purpose.

Try it step by step

Work it out in steps: Useful output mgh in J, then Energy not used for lifting in J, then Efficiency in %.

  1. Problem 1. A motor receives 600 J of energy and raises a 15 kg load through 3.0 m. Take g = 10 N/kg. Find the efficiency.
  2. Problem 2. A motor receives 2500 J and raises a 50 kg load through 4.0 m. Take g = 10 N/kg. Find the efficiency.
  3. Problem 3. A motor receives 1500 J and raises a 30 kg load through 3.0 m. Take g = 10 N/kg. Find the efficiency.
Show the answer

Problem 1.

Useful output mgh in J: Useful output = mgh = 15 x 10 x 3.0 = 450 J

Energy not used for lifting in J: Not useful = 600 - 450 = 150 J (to internal stores)

Efficiency in %: Efficiency = 450 / 600 x 100% = 75% (2 sf)

Problem 2.

Useful output mgh in J: Useful output = 50 x 10 x 4.0 = 2000 J

Energy not used for lifting in J: Not useful = 2500 - 2000 = 500 J

Efficiency in %: Efficiency = 2000 / 2500 x 100% = 80% (2 sf)

Problem 3.

Useful output mgh in J: Useful output = 30 x 10 x 3.0 = 900 J

Energy not used for lifting in J: Not useful = 1500 - 900 = 600 J

Efficiency in %: Efficiency = 900 / 1500 x 100% = 60% (2 sf)

Use the fraction when finding an unknown

Useful output = efficiency x total input, with efficiency written as a fraction. A device that is 0.75 efficient and receives 600 J gives 0.75 x 600 = 450 J of useful output.

Alternatively, total input = useful output / efficiency. Supplying 300 J of useful output at 0.75 efficiency requires 300 / 0.75 = 400 J. Substituting 75 instead of 0.75 would confuse a percentage with a fraction.

For a complete account with no uncounted initial store supplying the process, useful output cannot exceed total input, so the efficiency lies between 0 and 1, or 0% and 100%. An apparent result above 100% indicates a reversed ratio or an incomplete or inconsistent account.

Efficiency and power answer different questions

A powerful motor transfers energy rapidly. An efficient motor directs a large fraction of its input to the intended output. A motor could have high power but low efficiency, so one property does not establish the other.

Reducing friction in a lifting mechanism can reduce the transfer to internal stores, leaving a larger useful fraction for the same total input. The appropriate improvement depends on where the unwanted transfers occur.

Define useful before calculating. Compare energies over the same interval and use the full input. "Useful" depends on the task; it is not the name of a special energy store.

Watch out for this

A student says: "This motor is 70% efficient, so it destroys 30% of the energy it is given." What is wrong with this?

Show the answer

Energy is never destroyed. The other 30% is still there, transferred to less useful stores, mainly the internal energy of the motor and its surroundings.

Check your understanding

A lifting device receives 500 J electrically during a lift. Its intended gravitational energy increase is 200 J; the remaining 300 J increases internal stores. What is its efficiency for this purpose?

  1. 40%
  2. 60%
  3. 250%
Show the answer

40%

Efficiency = useful output / total input = 200 / 500 = 0.40 = 40%. The remaining 60% is still accounted for in internal stores.

Practise exam questions

Write your answer first, then tick the mark points it makes.

Exam-style question

A 60 W filament lamp gives out 3.0 W of light. An LED lamp gives out the same 3.0 W of light using only 5.0 W of electrical power. Calculate the efficiency of each lamp. Explain why the LED lamp is cheaper to run. [3]

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Exam-style question

Data-based question: Energy used by a lift [10]

A lift in a block of flats has no counterweight. A student reads the electricity meter for single trips up from the ground floor. Each time the lift carries the same passengers, and the total moving mass of the lift car and passengers is 800 kg. Take g = 10 N/kg.

Table 1. Electrical energy used for one trip up from the ground floor
floor reachedheight gained / melectrical energy used / kJ
39.0130
618.0220
927.0310
1236.0400
  1. (a) Calculate the gain in gravitational potential energy of the lift and passengers on the trip to floor 12. [2]
  2. (b) Calculate the efficiency of the lift for this trip. [2]
  3. (c) State one assumption you made in (a) or (b). [1]
  4. (d) The trip to floor 12 takes 24 s. Calculate the average useful power output of the lift. [2]
  5. (e) A resident says: "The lift wastes the same fraction of its energy whichever floor you go to." Use the data to evaluate this claim. [3]

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