Efficiency Formula

The efficiency of a system is a measure of how successfully 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 energy is transferred from the thermal store of a kettle's heating element to the thermal store of the water, this is useful energy
- When energy is transferred to the plastic or metal casing of the kettle and to the surrounding air, this energy is wasted

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

$η = \frac{E (o u t p u t)}{E (i n p u t)} = \frac{P (o u t p u t)}{P (i n p u t)}$

Where:
- η = efficiency (the greek letter "eta")
- E = energy (J)
- P = power (W)
To turn this equation into a percentage, just × 100 %
It can also be written in words as:

$η = \frac{useful work out}{total work in} = \frac{useful power out}{total power in}$

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.

Answer:

Step 1: Write down the efficiency equation (as a percentage)

$η = \frac{u s e f u l p o w e r o u t}{u s e f u l p o w e r i n} \times 100 %$

Step 2: Rearrange equation for the useful power in

$u s e f u l p o w e r i n = \frac{u s e r f u l p o w e r o u t \times 100 %}{η}$

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 = $\frac{353.16}{3}$ = 117.72 W

Step 4: Substitute values into power input equation

$u s e f u l p o w e r i n = \frac{117.72 \times 100}{35} = 336 W$

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 transfers energy from the kinetic store to the gravitational potential store of the water
Since its efficiency is 20%, the kinetic energy can be multiplied by 0.2 since only 20% of the kinetic energy will be transferred (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. It is a ratio with both the numerator and denominator with the same units.

Efficiency Formula (HL IB Physics)

Revision Note