D2.5 Solve power, energy, and time problems

Ontario Grade 11 Physics

Energy and Society

Solving Grade 11 problems with the relationship between transferred energy, power, and elapsed time

Energy describes the ability of a system to cause change or to be transferred. Power tells how quickly energy is transferred. Time tells how long the transfer lasts. These ideas can be connected with one simple relationship. Before calculating, choose the system you are describing. In this lesson, energy transferred into that system is positive. The problems use scalar quantities: they have size but no direction. Therefore, they do not need a vector positive direction. We will use positive values for the stated amounts of energy and power, and describe the transfer as entering or leaving the chosen system.

What you will learn

Explain how power, energy transfer, and time are related.
Use the relationship between power, energy, and time to find an unknown quantity.
Convert common units to SI units and report answers with suitable significant figures.
Check that a calculated answer has the correct units and is reasonable.

1. Prerequisite bridge: quantities and units

A physical quantity is something that can be measured or calculated. A scalar has a size but no direction. Energy, power, and time are scalars. This differs from a vector, which has both size and direction. You do not add a direction such as north or upward to a power or energy answer.

The SI unit of energy is the joule, symbol

\mathrm{J}

. The SI unit of power is the watt, symbol

\mathrm{W}

. One watt means one joule of energy transferred each second. The SI unit of time is the second, symbol

\mathrm{s}

. Use these units together when solving a problem.

Some questions give time in minutes or energy in kilojoules. Convert before substituting: one minute is

60\,\mathrm{s}

, and one kilojoule is

1000\,\mathrm{J}

. Keeping units beside the numbers helps reveal conversion errors.

Energy, power, and time are scalars, so their answers do not have vector directions.
Use joules, watts, and seconds for the standard calculation.
Choose the system and keep track of whether energy enters or leaves it.

2. The model: how energy, power, and time connect

Power is the rate of energy transfer. A larger power means more energy is transferred in a given time. For the same amount of energy, a larger power means the transfer takes less time.

Let

E

represent the energy transferred, in joules. Let

P

represent power, in watts. Let

t

represent elapsed time, in seconds. The relationship can be rearranged to solve for any one of the three quantities. First identify what is unknown, then choose the matching form.

This relationship applies to the energy transferred at a steady rate over the stated time. In a word problem, the system might be an appliance, a motor, or another object receiving or transferring energy. Use the power and energy values for that same system and time interval.

A positive numerical result gives the size of the energy transfer, power, or elapsed time. In this lesson's sign convention, energy entering the chosen system is positive. If a question describes energy leaving, state that in words; do not attach a vector direction to a scalar.

P=\frac{E}{t}=\frac{\Delta E}{\Delta t}

Power describes energy transferred per unit time.
Rearrange the relationship using algebra to find the unknown quantity.
A watt is equivalent to a joule per second.

3. A reliable calculation method

Start by naming the system, such as a kettle, and stating what interval the time covers. List the known quantities and the unknown. Convert all known values to SI units before substituting.

Write the governing relationship before inserting numbers. Rearrange it only as needed. Include units in the substitution so that the units in the result can be checked. Round the final value to a sensible number of significant figures, guided by the given data.

Finish by checking the units and the scale of the answer. For power, joules divided by seconds must give watts. For energy, watts multiplied by seconds must give joules. For time, joules divided by watts must give seconds. Ask whether the result makes sense: more energy at the same power should take more time, while greater power for the same energy should take less time.

Identify the system, known values, unknown, and time interval.
Convert to SI units before substituting.
Check both the units and the expected size of the answer.

4. Applying the relationship in context

In a question about an appliance, the stated power tells how quickly it transfers energy while operating. Multiplying by the operating time gives the energy transferred during that interval. The answer describes the amount of energy, not the direction of a force or motion.

Watch for wording such as 'each second' or 'in 4.0 minutes'. The first may already describe power; the second is an elapsed time that must be converted to seconds. Do not confuse a time interval with a clock reading. The calculation uses how long the transfer lasts.

The relationship does not by itself tell you what happens to every part of the energy. For this expectation, use only the stated power, energy, and time information to solve for the missing quantity.

Power is a rate, while energy is an amount transferred.
Use the time for the stated operating interval.
Do not infer extra information that the problem does not provide.

Worked example

Finding energy from power and time

A small heater transfers energy into the air in a room at a steady power of

750\,\mathrm{W}

for

4.0\,\mathrm{min}

. How much energy does it transfer?

Set the system and known values
Choose the heater and the air it warms as the system. Energy enters this system, so the transfer is positive by our convention. The known power is $750\,\mathrm{W}$ , and the time must be converted from minutes to seconds. Energy is the unknown.
Convert the time
Convert minutes to seconds because the SI relationship uses watts and seconds.
$4.0\,\mathrm{min}\times\frac{60\,\mathrm{s}}{1\,\mathrm{min}}=240\,\mathrm{s}$
Solve for energy
Use energy equals power multiplied by time. Substitute both values with their units.
$E=Pt=(750\,\mathrm{J/s})(240\,\mathrm{s})=180000\,\mathrm{J}$

Answer: The heater transfers

1.8\times10^{5}\,\mathrm{J}

, or

180\,\mathrm{kJ}

, into the system. The two significant figures reflect the given time.

Check: The units reduce from joules per second multiplied by seconds to joules. A heater operating for four minutes at hundreds of watts transfers much more than a few joules, so this result is reasonable. The energy transfer is positive into the chosen system.

Worked example

Finding power from energy and time

A device receives

36\,\mathrm{kJ}

of energy in

2.0\,\mathrm{min}

. What is its average power over that interval?

Define the system and unknown
Take the device as the system. The energy transfer is into the device and is positive. Power is the unknown. Convert the energy and time to joules and seconds before calculating.
Convert the given quantities
A kilojoule is one thousand joules, and a minute is sixty seconds.
$36\,\mathrm{kJ}=36000\,\mathrm{J},\qquad 2.0\,\mathrm{min}=120\,\mathrm{s}$
Calculate power
Divide the energy transferred by the elapsed time. The result is the average power during the stated interval.
$P=\frac{E}{t}=\frac{36000\,\mathrm{J}}{120\,\mathrm{s}}=300\,\mathrm{W}$

Answer: The device's average power is

3.0\times10^{2}\,\mathrm{W}

.

Check: Joules divided by seconds gives watts. Transferring

36\,\mathrm{kJ}

over two minutes corresponds to a few hundred joules each second, so

300\,\mathrm{W}

is reasonable. The energy direction is into the device; power itself is a scalar.

Worked example

Finding time from energy and power

A battery transfers

5400\,\mathrm{J}

to a sensor at a steady power of

18\,\mathrm{W}

. How long does the transfer take?

Identify the system and quantities
Choose the sensor as the system. Energy enters it, so the transfer is positive. The energy and power are known; elapsed time is unknown. They are already in SI units.
Rearrange the relationship
Since power is energy divided by time, multiply by time and then divide by power to isolate time.
$t=\frac{E}{P}$
Substitute and calculate
Divide the transferred energy by the power, keeping the units in the calculation.
$t=\frac{5400\,\mathrm{J}}{18\,\mathrm{J/s}}=300\,\mathrm{s}$

Answer: The transfer takes

3.0\times10^{2}\,\mathrm{s}

, or

5.0\,\mathrm{min}

.

Check: Joules divided by joules per second gives seconds. At

18\,\mathrm{J}

transferred each second,

300\,\mathrm{s}

gives

5400\,\mathrm{J}

, so the answer checks. The transfer is into the sensor.

Common mistakes and how to avoid them

Using minutes directly with power in watts.

Correction: Convert minutes to seconds before substituting, because a watt is a joule per second.

Treating power and energy as interchangeable.

Correction: Energy is an amount transferred; power is how quickly it is transferred. Use the correct rearrangement for the unknown.

Reporting an answer without a unit.

Correction: Attach joules, watts, or seconds as appropriate, then check that the units follow from the calculation.

Adding a direction such as upward to energy or power.

Correction: Energy, power, and time are scalars. Describe whether energy enters or leaves the chosen system in words.

Assuming that a larger power always means more energy in every situation.

Correction: Energy depends on both power and time. Compare powers only when the time intervals are the same.

Lesson summary

Power is the rate at which energy is transferred.
Use $P=E/t$ and rearrange it to find energy or time.
Use joules, watts, and seconds, converting other units first.
State the system and whether energy enters or leaves it; these quantities do not have vector directions.
Check units, significant figures, and whether the answer is physically reasonable.

Check your understanding

Question 1

A motor transfers

2400\,\mathrm{J}

in

30\,\mathrm{s}

. What is its power?

$80\,\mathrm{W}$
$72000\,\mathrm{W}$
$0.0125\,\mathrm{W}$
$80\,\mathrm{J}$

Show answer and explanation

80\,\mathrm{W}

Power is energy divided by time:

2400\,\mathrm{J}/30\,\mathrm{s}=80\,\mathrm{W}

. The units are joules per second.

Question 2

At a steady power of

50\,\mathrm{W}

, how much energy is transferred in

10\,\mathrm{s}

?

$5\,\mathrm{J}$
$500\,\mathrm{J}$
$0.20\,\mathrm{J}$
$500\,\mathrm{W}$

Show answer and explanation

500\,\mathrm{J}

Energy is power multiplied by time:

(50\,\mathrm{J/s})(10\,\mathrm{s})=500\,\mathrm{J}

.

Question 3

A device receives

900\,\mathrm{J}

at a steady power of

30\,\mathrm{W}

. How long does the transfer take?

$0.033\,\mathrm{s}$
$870\,\mathrm{s}$
$30\,\mathrm{s}$
$27000\,\mathrm{s}$

Show answer and explanation

30\,\mathrm{s}

Time is energy divided by power:

900\,\mathrm{J}/30\,\mathrm{J/s}=30\,\mathrm{s}

.

Key terms

Energy: A scalar quantity that can be transferred to or from a system. Its SI unit is the joule.
Power: The rate at which energy is transferred. Its SI unit is the watt.
Time interval: The elapsed duration of an event or energy transfer. Its SI unit is the second.
System: The object or group of objects chosen for the problem.
Scalar: A quantity with size but no direction.
Significant figures: Digits that show the precision supported by the given measurements or values.

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About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Physics (SPH3U), expectation D2.5. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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D2.5 · Solve power, energy, and time problems

Ontario Grade 11 Physics

Solving Grade 11 problems with the relationship between transferred energy, power, and elapsed time

What you will learn

1. Prerequisite bridge: quantities and units

2. The model: how energy, power, and time connect

3. A reliable calculation method

4. Applying the relationship in context

Finding energy from power and time

Finding power from energy and time

Finding time from energy and power

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through SPH3U

About this lesson and its review