DP IB Physics: SL

Revision Notes

Syllabus Edition

First teaching 2014

Last exams 2024

|

2.3.7 Efficiency

Test Yourself

Efficiency

  • The efficiency of a system is a measure of how well energy is transferred in a system
  • Efficiency is defined as:

The ratio of the useful power or energy transfer output from a system to its total power or energy transfer input

  • If a system has high efficiency, this means most of the energy transferred is useful
  • If a system has low efficiency, this means most of the energy transferred is wasted
  • Determining which type of energy is useful or wasted depends on the system
    • When electrical energy is converted to light in a lightbulb, the light energy is useful and the heat energy produced is wasted
    • When electrical energy is converted to heat for a heater, the heat energy is useful and the sound energy produced is wasted

  • Efficiency is represented as a percentage, and can be calculated using the equation:

Efficiency equation, downloadable AS & A Level Physics revision notes

  • The energy can be of any form e.g. gravitational potential energy, kinetic energy
  • The efficiency equation can also be written in terms of power:

Efficiency equation 2, downloadable AS & A Level Physics revision notes
  • Where power is defined as the energy transferred per unit of time

Worked example

An electric motor has an efficiency of 35 %. It lifts a 7.2 kg load through a height of 5 m in 3 s. Calculate the power of the motor.

Step 1: Write down the efficiency equation 

Efficiency equation

Step 2: Rearrange for the power input  Worked Example Power Input Equation

Step 3: Calculate the power output

    • The power output is equal to energy ÷ time
    • The electric motor transferred electric energy into gravitational potential energy to lift the load

Gravitational potential energy = mgh = 7.2 × 9.81 × 5 = 353.16 J

Power = 353.16 ÷ 3 = 117.72 W

Step 4: Substitute values into power input equation Power Input Worked Example Answer

Worked example

Worked example

The diagram shows a pump called a hydraulic ram.

In one such pump, the long approach pipe holds 700 kg of water. A valve shuts when the speed of this water reaches 3.5 m s-1 and the kinetic energy of this water is used to lift a small quantity of water by height of 12m.

The efficiency of the pump is 20%.

Which mass of water could be lifted 12 m?

A. 6.2 kg               B. 4.6 kg               C. 7.3 kg               D. 0.24 kg

  • The pump is what converts the water’s kinetic energy into gravitational potential energy
  • Since its efficiency is 20%, the kinetic energy can be multiplied by 0.2 since only 20% of the kinetic energy will be converted (not 20% of the gravitational potential energy)

Exam Tip

Efficiency can be in a ratio (between 0 and 1) or percentage format (between 0 and 100%)If the question asks for efficiency as a ratio, give your answer as a fraction or decimal.If the answer is required as a percentage, remember to multiply the ratio by 100 to convert it: if the ratio = 0.25, percentage = 0.25 × 100 = 25 %Remember that efficiency has no units

You've read 0 of your 0 free revision notes

Get unlimited access

to absolutely everything:

  • Downloadable PDFs
  • Unlimited Revision Notes
  • Topic Questions
  • Past Papers
  • Model Answers
  • Videos (Maths and Science)

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Did this page help you?

Katie M

Author: Katie M

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.