Power as a rate of transfer

Definition

Power

Power is the rate at which energy is transferred, which is the same as the rate at which work is done.

Definition

Watt

The watt is the unit of power, equal to an energy transfer of one joule each second.

Power is the rate at which energy is transferred, and because work done and energy transferred are the same quantity it is also the rate of doing work.
Power answers how many joules each second, not how many joules altogether.
Two machines that transfer the same energy have different powers if they take different times, and the quicker one is the more powerful.
The unit of power is the watt, $W\text{W}$ , and $J/s1\ \text{W} = 1\ \text{J/s}$ .
A device rated at $W60\ \text{W}$ transfers $J60\ \text{J}$ every second it is switched on.
Larger powers are quoted in kilowatts and megawatts, where $W1\ \text{kW} = 1000\ \text{W}$ and $W1\ \text{MW} = 1\,000\,000\ \text{W}$ .
A large power does not guarantee a large total energy transfer, because a small power running for a long time can transfer more.

The power equation

Power is calculated from $\dfrac{E}{t}$ .
Here $P$ is the power in watts, $E$ is the work done or energy transferred in joules and $t$ is the time taken in seconds.
Rearranging gives $\times t$ for the energy transferred and $\dfrac{E}{P}$ for the time taken.
Convert every time into seconds first, so $s2.5\ \text{minutes} = 150\ \text{s}$ and $s1\ \text{hour} = 3600\ \text{s}$ .
Convert kilowatts into watts before substituting, so $W2.4\ \text{kW} = 2400\ \text{W}$ .
For a lift or a hoist, work out $\times g \times \Delta h$ first and then divide by the time.
For a machine pulling against a steady force, work out $\times d$ first and then divide by the time.

Example

Power of an electric motor

A motor does $J3600\ \text{J}$ of work in $s30\ \text{s}$ .
The equation is $\dfrac{E}{t}$ .
Substituting gives $\dfrac{3600}{30}$ .
The power is $120\ \text{W}$ , which means the motor transfers $J120\ \text{J}$ every second.

Example

Power of a lift raising a load

A lift raises a load of mass $kg500\ \text{kg}$ through a vertical height of $m12\ \text{m}$ in $s20\ \text{s}$ , with $10\ \text{N/kg}$ .
The useful energy transferred is $\times g \times \Delta h = 500 \times 10 \times 12$ .
This gives $60\,000\ \text{J}$ .
The power is $\dfrac{E}{t} = \dfrac{60\,000}{20}$ .
The useful power output of the lift is $3000\ \text{W}$ , which is $kW3.0\ \text{kW}$ .

Example

Energy transferred by a rated appliance

A $kW2.4\ \text{kW}$ shower runs for $minutes8.0\ \text{minutes}$ .
Converting the values gives $2400\ \text{W}$ and $\times 60 = 480\ \text{s}$ .
Rearranging gives $\times t = 2400 \times 480$ .
The energy transferred is $1\,152\,000\ \text{J}$ , which is $kJ1152\ \text{kJ}$ .

Practical

Measuring your own power output

Aim: to measure the useful power output of a student climbing a flight of stairs.
Apparatus: bathroom scales or a balance, metre rule, stopwatch reading to $s0.01\ \text{s}$ , a staircase with a countable number of steps, and a calculator.
Variables: the time taken is the measured variable, the power output is the calculated variable, and the student, the staircase and the vertical height climbed are kept the same for every run.
Method, set-up:
- Measure the student's mass in kilograms on the scales and record it.
- Measure the vertical height of one step with the metre rule, then multiply by the number of steps to get the vertical height of the whole flight.
- Mark a clear start line at the bottom step and a finish line at the top step so every run covers the same vertical height.
Method, measurements:
- Start the stopwatch as the student's foot leaves the start line and stop it as the foot lands on the finish line.
- Climb at a brisk but safe walking pace, keeping one hand free for the handrail.
- Record the time in seconds.
- Repeat three times with a rest between runs, then calculate the mean time.
- Repeat with a second student so two power outputs can be compared.
Results: the student who climbs the same flight in the shorter time has the greater power output, and a heavier student climbing in the same time has a greater power output still.
Maths: calculate the useful energy transferred from $\times g \times \Delta h$ with $10\ \text{N/kg}$ , then calculate the power from $\dfrac{E}{t}$ ; a student of mass $kg60\ \text{kg}$ climbing $m3.0\ \text{m}$ in $s5.0\ \text{s}$ transfers $J1800\ \text{J}$ and has a power output of $W360\ \text{W}$ .
Watch out: reaction time in starting and stopping the stopwatch is the largest source of uncertainty, so use a longer flight and repeat every run; the height must be the vertical rise of the flight rather than the length of the staircase; the figure found is the useful power only, because the body also dissipates energy by heating.
Safety: keep the staircase clear, do not run, use the handrail, and let anyone who feels unwell sit the activity out.

Comparing power in real machines

Two cranes raising identical loads to the same height do the same work, so the crane that finishes sooner has the greater power.
A $kW2.0\ \text{kW}$ kettle and a $W10\ \text{W}$ phone charger differ by a factor of $200$ in the joules they transfer each second.
The charger left plugged in overnight can still transfer more energy in total than the kettle used for two minutes.
The power rating printed on an appliance gives the energy it transfers each second while it is working normally.
A car engine quoted in kilowatts is being described by how quickly it can do work, not by how far the car can travel.
Comparing powers fairly means comparing the energy transferred over the same time, or the time taken to transfer the same energy.

Exam technique

Answering power questions

Write $\dfrac{E}{t}$ before substituting, and show any conversion of minutes into seconds on its own line.
Work out the energy first when the question gives a mass and a height, or a force and a distance, then divide by the time.
Give the unit as $W\text{W}$ , or as $kW\text{kW}$ only when the question asks for kilowatts.
When comparing two devices, say which quantity is the same and which differs, such as the same energy is transferred in a shorter time, so the power is greater.
Use the word rate in the definition, because an answer that says only that power is energy does not gain the mark.

Common Mistake

Do not give a power in joules, because joules measure the energy transferred while watts measure the rate of transfer.
Do not substitute a time in minutes or hours, because the equation needs seconds.
Do not assume the more powerful device always transfers more energy, because that holds only for equal times.
Do not confuse $kW\text{kW}$ with $kJ\text{kJ}$ , because one is a power and the other is an energy.

Self review

Define power and give its unit.
Write the equation linking power, energy transferred and time.
Complete the statement $J/s1\ \text{W} = \ldots\ \text{J/s}$ .
Describe how to measure your own power output on a flight of stairs.
Explain how a $W10\ \text{W}$ charger can transfer more energy than a $kW2\ \text{kW}$ kettle.

Power as a rate of transfer

Definition

Power

Power is the rate at which energy is transferred, which is the same as the rate at which work is done.

Definition

Watt

The watt is the unit of power, equal to an energy transfer of one joule each second.

Power is the rate at which energy is transferred, and because work done and energy transferred are the same quantity it is also the rate of doing work.
Power answers how many joules each second, not how many joules altogether.
Two machines that transfer the same energy have different powers if they take different times, and the quicker one is the more powerful.
The unit of power is the watt, $W\text{W}$ , and $J/s1\ \text{W} = 1\ \text{J/s}$ .
A device rated at $W60\ \text{W}$ transfers $J60\ \text{J}$ every second it is switched on.
Larger powers are quoted in kilowatts and megawatts, where $W1\ \text{kW} = 1000\ \text{W}$ and $W1\ \text{MW} = 1\,000\,000\ \text{W}$ .
A large power does not guarantee a large total energy transfer, because a small power running for a long time can transfer more.

The power equation

Power is calculated from $\dfrac{E}{t}$ .
Here $P$ is the power in watts, $E$ is the work done or energy transferred in joules and $t$ is the time taken in seconds.
Rearranging gives $\times t$ for the energy transferred and $\dfrac{E}{P}$ for the time taken.
Convert every time into seconds first, so $s2.5\ \text{minutes} = 150\ \text{s}$ and $s1\ \text{hour} = 3600\ \text{s}$ .
Convert kilowatts into watts before substituting, so $W2.4\ \text{kW} = 2400\ \text{W}$ .
For a lift or a hoist, work out $\times g \times \Delta h$ first and then divide by the time.
For a machine pulling against a steady force, work out $\times d$ first and then divide by the time.

Example

Power of an electric motor

A motor does $J3600\ \text{J}$ of work in $s30\ \text{s}$ .
The equation is $\dfrac{E}{t}$ .
Substituting gives $\dfrac{3600}{30}$ .
The power is $120\ \text{W}$ , which means the motor transfers $J120\ \text{J}$ every second.

Example

Power of a lift raising a load

A lift raises a load of mass $kg500\ \text{kg}$ through a vertical height of $m12\ \text{m}$ in $s20\ \text{s}$ , with $10\ \text{N/kg}$ .
The useful energy transferred is $\times g \times \Delta h = 500 \times 10 \times 12$ .
This gives $60\,000\ \text{J}$ .
The power is $\dfrac{E}{t} = \dfrac{60\,000}{20}$ .
The useful power output of the lift is $3000\ \text{W}$ , which is $kW3.0\ \text{kW}$ .

Example

Energy transferred by a rated appliance

A $kW2.4\ \text{kW}$ shower runs for $minutes8.0\ \text{minutes}$ .
Converting the values gives $2400\ \text{W}$ and $\times 60 = 480\ \text{s}$ .
Rearranging gives $\times t = 2400 \times 480$ .
The energy transferred is $1\,152\,000\ \text{J}$ , which is $kJ1152\ \text{kJ}$ .

Practical

Measuring your own power output

Aim: to measure the useful power output of a student climbing a flight of stairs.
Apparatus: bathroom scales or a balance, metre rule, stopwatch reading to $s0.01\ \text{s}$ , a staircase with a countable number of steps, and a calculator.
Variables: the time taken is the measured variable, the power output is the calculated variable, and the student, the staircase and the vertical height climbed are kept the same for every run.
Method, set-up:
- Measure the student's mass in kilograms on the scales and record it.
- Measure the vertical height of one step with the metre rule, then multiply by the number of steps to get the vertical height of the whole flight.
- Mark a clear start line at the bottom step and a finish line at the top step so every run covers the same vertical height.
Method, measurements:
- Start the stopwatch as the student's foot leaves the start line and stop it as the foot lands on the finish line.
- Climb at a brisk but safe walking pace, keeping one hand free for the handrail.
- Record the time in seconds.
- Repeat three times with a rest between runs, then calculate the mean time.
- Repeat with a second student so two power outputs can be compared.
Results: the student who climbs the same flight in the shorter time has the greater power output, and a heavier student climbing in the same time has a greater power output still.
Maths: calculate the useful energy transferred from $\times g \times \Delta h$ with $10\ \text{N/kg}$ , then calculate the power from $\dfrac{E}{t}$ ; a student of mass $kg60\ \text{kg}$ climbing $m3.0\ \text{m}$ in $s5.0\ \text{s}$ transfers $J1800\ \text{J}$ and has a power output of $W360\ \text{W}$ .
Watch out: reaction time in starting and stopping the stopwatch is the largest source of uncertainty, so use a longer flight and repeat every run; the height must be the vertical rise of the flight rather than the length of the staircase; the figure found is the useful power only, because the body also dissipates energy by heating.
Safety: keep the staircase clear, do not run, use the handrail, and let anyone who feels unwell sit the activity out.

Comparing power in real machines

Two cranes raising identical loads to the same height do the same work, so the crane that finishes sooner has the greater power.
A $kW2.0\ \text{kW}$ kettle and a $W10\ \text{W}$ phone charger differ by a factor of $200$ in the joules they transfer each second.
The charger left plugged in overnight can still transfer more energy in total than the kettle used for two minutes.
The power rating printed on an appliance gives the energy it transfers each second while it is working normally.
A car engine quoted in kilowatts is being described by how quickly it can do work, not by how far the car can travel.
Comparing powers fairly means comparing the energy transferred over the same time, or the time taken to transfer the same energy.

Exam technique

Answering power questions

Write $\dfrac{E}{t}$ before substituting, and show any conversion of minutes into seconds on its own line.
Work out the energy first when the question gives a mass and a height, or a force and a distance, then divide by the time.
Give the unit as $W\text{W}$ , or as $kW\text{kW}$ only when the question asks for kilowatts.
When comparing two devices, say which quantity is the same and which differs, such as the same energy is transferred in a shorter time, so the power is greater.
Use the word rate in the definition, because an answer that says only that power is energy does not gain the mark.

Common Mistake

Do not give a power in joules, because joules measure the energy transferred while watts measure the rate of transfer.
Do not substitute a time in minutes or hours, because the equation needs seconds.
Do not assume the more powerful device always transfers more energy, because that holds only for equal times.
Do not confuse $kW\text{kW}$ with $kJ\text{kJ}$ , because one is a power and the other is an energy.

Self review

Define power and give its unit.
Write the equation linking power, energy transferred and time.
Complete the statement $J/s1\ \text{W} = \ldots\ \text{J/s}$ .
Describe how to measure your own power output on a flight of stairs.
Explain how a $W10\ \text{W}$ charger can transfer more energy than a $kW2\ \text{kW}$ kettle.

8.3.1 Power

Power as a rate of transfer

The power equation

Power of an electric motor

Power of a lift raising a load

Energy transferred by a rated appliance

Measuring your own power output

Comparing power in real machines

Answering power questions

Review 8.3.1 Power by teaching Genie

8.3.1 Power

Power as a rate of transfer

The power equation

Power of an electric motor

Power of a lift raising a load

Energy transferred by a rated appliance

Measuring your own power output

Comparing power in real machines

Answering power questions

Review 8.3.1 Power by teaching Genie

8.3.1 Power

Power as a rate of transfer

The power equation

Power of an electric motor

Power of a lift raising a load

Energy transferred by a rated appliance

Measuring your own power output

Comparing power in real machines

Answering power questions

Review 8.3.1 Power by teaching Genie

8.3.1 Power Revision Guide

8.3.1 Power

Power as a rate of transfer

The power equation

Power of an electric motor

Power of a lift raising a load

Energy transferred by a rated appliance

Measuring your own power output

Comparing power in real machines

Answering power questions

Review 8.3.1 Power by teaching Genie

8.3.1 Power Revision Guide