8.12-8.15 - Power and efficiency

Power is a rate

Imagine two motors lifting identical loads through the same height. Each motor does 1200 J of work, so each transfers the same amount of energy to the gravitational potential store of its load-Earth system. Motor A takes 3 s; motor B takes 6 s. Motor A is more powerful because it completes the same energy transfer in less time.

The SI unit of power is the watt, symbol W.

Motor A transfers $1200 \mathrm{J} / 3 \mathrm{s} = 400 \mathrm{J}/s$, so its power is 400 W. Motor B transfers $1200 \mathrm{J} / 6 \mathrm{s} = 200 \mathrm{J}/s$, so its power is 200 W. A power of 400 W means that 400 J of energy is transferred each second while the motor operates at that power.

Power and energy are different quantities. A joule measures an amount of energy transferred or work done. A watt measures how quickly that transfer happens. A low-power machine running for long enough can transfer more energy in total than a high-power machine running briefly.

Calculating power from energy and time

For a transfer over a measured time interval:

Work done is an energy transfer caused by a force acting through a distance, so the energy transferred has unit joule. The Edexcel equation sheet provides $P = E/t$; you still have to select the correct values and put them into joules and seconds before substitution.

Worked example

A lifting motor does 3.60 kJ of work in 15.0 s. Calculate its average power.

First convert kilojoules to joules:

$3.60 \mathrm{kJ} = 3.60 \times 1000 \mathrm{J} = 3600 \mathrm{J}$

Select the equation and substitute:

$P = E / t$

$P = 3600 \mathrm{J} / 15.0 \mathrm{s} = 240 \mathrm{W}$

The answer is 240 W, or $2.40 \times 10^{2} \mathrm{W}$ to three significant figures, matching the data. It means that the motor transfers an average of 240 J each second. The result is sensible because $240 \mathrm{J}/s \times 15.0 \mathrm{s} = 3600 \mathrm{J}$, recovering the stated work done.

The same relationship can be rearranged when a different quantity is unknown:

• $E = P \times t$ finds the work done or energy transferred;
• $t = E / P$ finds the time taken.

Keep the unit meaning visible when rearranging. For example, a 650 W motor operating for 20 s transfers $650 \mathrm{J}/s \times 20 \mathrm{s} = 13 000 \mathrm{J}$.

Efficiency needs a clear boundary

Draw an imaginary boundary around the device and choose one complete operating interval. Across that same boundary and interval, identify:

• the total energy supplied to the device;
• the useful energy transferred by the device for its intended purpose;
• energy transferred in unwanted, dissipated ways.