Why Efficiency Cannot Exceed 100% in an Energy Budget

A motor receives \(2.0\ \text{kJ}\) of electrical energy and delivers \(2.3\ \text{kJ}\) of useful mechanical energy. Divide the output by the input and you get 115% efficiency.

Has the motor broken the law of conservation of energy?

No. Something is missing from the energy budget. Perhaps the motor has released energy stored in a flywheel, spring, or battery. Perhaps an additional energy input has not been measured. Or perhaps the input and output have been measured over different time intervals.

Efficiency cannot exceed 100% because useful output energy cannot exceed the total energy supplied to a system when every energy source is properly accounted for. This follows directly from conservation of energy.

The important skill is recognising what counts as an input, what counts as useful output, and where the remaining energy goes.

01Start with the energy budget

Imagine an electric motor powering a machine. Electrical energy enters through a cable, and the motor transfers mechanical energy through a rotating shaft.

The mechanical energy is useful because it performs the motor’s intended job.

But is all the electrical energy converted into mechanical energy?

No. Some energy is also transferred to the surroundings through heating, sound, and vibration.

Suppose the motor receives 100 J of electrical energy. Its energy budget might look like this:

Notice that all 100 J are accounted for.

The motor hasn’t destroyed 30 J of energy. That energy has been transferred into forms that aren’t useful for the motor’s intended task.

This gives us our first important distinction:

• Total energy output includes every form of energy transferred out of the system.
• Useful energy output includes only the energy that performs the intended task.

For this motor, the useful output is 70 J, not 100 J.

02Why efficiency has a maximum of 100%

Efficiency measures the proportion of supplied energy that becomes useful output energy.

Its standard definition is:

\[ \boxed{\eta=\frac{E_{\text{useful output}}} {E_{\text{input}}}} \]

Here:

• \(\eta\) is efficiency, expressed as a decimal or percentage.
• \(E_{\text{useful output}}\) is the energy transferred usefully, measured in joules (J).
• \(E_{\text{input}}\) is the total energy supplied, measured in joules (J).

To express efficiency as a percentage, multiply by 100:

\[ \boxed{\eta(\%)= \frac{E_{\text{useful output}}} {E_{\text{input}}}\times100} \]

For our motor:

\[ \eta=\frac{70}{100}\times100 \]

\[ \boxed{\eta=70\%} \]

Why can’t this answer exceed 100%?

Because useful output is only one part of the energy leaving the system. The remaining energy may be transferred as heat, sound, or other unwanted forms.

Conservation of energy requires:

\[ E_{\text{input}}= E_{\text{useful output}}+ E_{\text{other output}} \]

This equation assumes that the system has no net change in stored energy during the measurement.

Since the other output cannot be negative:

\[ E_{\text{useful output}}\leq E_{\text{input}} \]

Therefore:

\[ \boxed{0\leq\eta\leq1} \]

Or, when expressed as a percentage:

\[ \boxed{0\%\leq\eta\leq100\%} \]

What would 100% efficiency actually mean?

Imagine an ideal motor receiving 100 J of electrical energy.

If all 100 J became useful mechanical energy, with no energy transferred into unwanted forms, its efficiency would be:

\[ \eta=\frac{100}{100}\times100 \]

\[ \boxed{\eta=100\%} \]

This is the theoretical upper limit for the standard energy-conversion efficiency.

Real devices generally have efficiencies below 100% because some energy is transferred into forms that don’t contribute to their intended task.

However, don’t confuse this with conservation of energy.

A motor operating at 40% efficiency still obeys conservation of energy perfectly. It simply transfers 40% of its input into useful mechanical output and the remaining 60% into other forms.

03Energy and power: the amount-versus-rate distinction

A common reason students get confused about efficiency is that they mix up energy and power.

They are related, but they measure different things.

Power is the rate of energy transfer:

\[ P_{\text{avg}}=\frac{E}{t} \]

where:

• \(P_{\text{avg}}\) is average power in watts (W).
• \(E\) is energy transferred in joules (J).
• \(t\) is the time taken in seconds (s).

Remember:

\[ 1\ \text{W}=1\ \text{J s}^{-1} \]

A device transferring 400 J in one second has an average power of 400 W.

A device transferring the same 400 J over four seconds has an average power of 100 W.

Both transfer the same amount of energy. They simply transfer it at different rates.

Can efficiency be calculated using power instead?

Yes.

Since power is energy transferred per unit time, efficiency can also be calculated using:

\[ \boxed{\eta= \frac{P_{\text{useful output}}} {P_{\text{input}}}} \]

This is valid when both power values are average powers measured over the same time interval, with no unaccounted change in stored energy.

For example, suppose a motor receives an average electrical input power of 500 W and delivers an average useful mechanical output power of 350 W.

Its efficiency is:

\[ \eta=\frac{350}{500}\times100 \]

\[ \boxed{\eta=70\%} \]

The energy calculation would give exactly the same result if both powers were measured over the same interval.

For a more detailed explanation of the distinction between energy and its rate of transfer, see Can an Object Have Kinetic Energy While Net Power Is Zero?.

A useful warning about power

Suppose a machine receives 200 W of electrical power but delivers 260 W of useful mechanical power for a short period.

Does this prove its efficiency is 130%?

Not necessarily.

The machine might be releasing energy stored in a rotating flywheel or compressed spring. Its output power can temporarily exceed its measured electrical input power because another energy source is contributing.

Power tells you how quickly energy is transferred. It doesn’t tell you whether all the energy being transferred came from the input you’re measuring.

We’ll examine this situation in more detail shortly.

04What if an efficiency calculation gives more than 100%?

If your calculation produces an efficiency above 100%, don’t immediately change your answer to 100%.

Investigate the energy budget.

There are several possible explanations.

Explanation 1: An energy store is being discharged

A battery, spring, raised weight, or rotating flywheel can store energy.

When that stored energy decreases, it supplies energy to the system.

Imagine a spring-powered winch. During a test, it receives 300 J of electrical energy and releases 100 J of stored elastic potential energy.

The winch now has 400 J of energy available.

It could transfer 360 J into useful lifting work and 40 J into unwanted thermal energy.

The energy budget is:

\[ 300+100=360+40 \]

Everything balances.

However, if you divide the useful output by only the electrical input, you obtain:

\[ \frac{360}{300}\times100=120\% \]

That is not the winch’s efficiency. You’ve ignored the 100 J supplied by the spring.

The correct calculation is:

\[ \eta=\frac{360}{300+100}\times100 \]

\[ \boxed{\eta=90\%} \]

The apparent efficiency above 100% came from leaving an energy source out of the denominator.

A useful way to picture this is a bank account. You might spend $130 when your income for the week is only $100 because you withdraw another $30 from savings.

Your spending exceeds your weekly income, but it doesn’t exceed your total available funds.

Similarly, a machine can produce more energy than its measured external input if it also draws on an existing energy store. The stored energy must be included in the budget.

Explanation 2: The system boundary is incorrect

A system boundary determines which objects and energy stores are included in your calculation.

Consider a motor connected to a flywheel.

If you measure electrical energy entering the motor but measure mechanical energy leaving the combined motor-and-flywheel arrangement, you may be comparing different systems.

The flywheel might be supplying some of the output energy.

You must either:

• Measure the energy transfers for the motor alone, or
• Include the flywheel and account for its change in stored kinetic energy.

Mixing the two boundaries can produce misleading efficiency values.

Explanation 3: The measurement intervals don’t match

Energy calculations must refer to the same time interval.

Suppose you measure the electrical input during a 10-second period but measure the mechanical output during a period that includes energy stored before those 10 seconds.

The output may include energy supplied earlier.

This is another way to obtain an apparent efficiency above 100%.

For a fair comparison, measure both input and output over the same interval and account for changes in stored energy.

A more complete energy-budget equation

When the system can store or release energy, the basic energy equation needs one extra term:

\[ \boxed{ E_{\text{external input}}= E_{\text{useful output}}+ E_{\text{other output}}+ \Delta E_{\text{stored}} } \]

Here:

\[ \Delta E_{\text{stored}}= E_{\text{final}}-E_{\text{initial}} \]

A positive value means the system has gained stored energy.

A negative value means the system has released stored energy.

This equation gives us three possibilities:

The key point is that a decrease in stored energy is an additional energy source.

When calculating efficiency, you must account for that released energy as part of the total energy supplied.

If there is no net change in stored energy, the standard HSC energy-budget equation is sufficient.

05An important contrast: How can a heat pump have a COP above 100%?

You might encounter a heat pump described as having an efficiency greater than 100%.

This sounds like a contradiction, but the terminology matters.

A heat pump transfers thermal energy from outside a building into the building using electrical energy.

It doesn’t generate all the delivered heat from electricity. It also transfers existing thermal energy from the surroundings.

Its energy balance can be written as:

\[ Q_H=Q_C+W \]

where:

• \(Q_H\) is the thermal energy delivered indoors.
• \(Q_C\) is the thermal energy extracted from the outside environment.
• \(W\) is the electrical work supplied.

A heat pump is often assessed using its coefficient of performance (COP):

\[ COP=\frac{Q_H}{W} \]

Notice that the denominator includes only the electrical work supplied, not the thermal energy extracted from outside.

Consequently, the COP can be greater than 1.

For example, a COP of 4 means the heat pump delivers four units of thermal energy for every unit of electrical work supplied.

The other three units come from the environment.

This does not violate conservation of energy because the electrical work is not the only energy input.

Importantly, COP is a performance ratio, not the standard energy-conversion efficiency defined earlier. The two quantities should not be confused.

For ordinary HSC efficiency calculations, stick to the useful energy output divided by the total energy input.

06A reliable method for efficiency questions

When approaching an HSC Physics question, use this sequence.

Step 1: Identify the intended task.

Decide which energy transfer is useful.

For a motor, it might be mechanical energy. For an electric heater, it might be thermal energy.

Step 2: Identify the system boundary.

Determine which objects and energy stores are included.

Step 3: Identify every energy input.

Don’t automatically assume that the electrical input is the only source. Check whether a battery, spring, flywheel, or other store contributes energy.

Step 4: Check the time interval.

Make sure the input and output energy measurements refer to the same period.

Step 5: Calculate efficiency.

Use:

\[ \eta=\frac{E_{\text{useful output}}} {E_{\text{total input}}}\times100 \]

Or, when appropriate:

\[ \eta=\frac{P_{\text{useful output}}} {P_{\text{input}}}\times100 \]

Step 6: Check the energy budget.

Ask whether the remaining energy has been accounted for through heating, sound, energy storage, or another transfer.

If your result exceeds 100%, revisit the assumptions and measurements rather than accepting the number.

07Worked examples

08Questions and solutions

The following questions progress from a straightforward efficiency calculation to more demanding energy-accounting problems.

Question 1

A desk fan receives 2.40 kJ of electrical energy during a test. Of this energy, 1.68 kJ is transferred usefully into the kinetic energy of the moving air.

Calculate:

a) The efficiency of the fan.

b) The amount of energy transferred into other forms.

Solution 1

The fan’s efficiency is 70%, and 0.72 kJ is transferred into other forms.

Part a: Calculate efficiency.

Use:

\[ \eta=\frac{E_{\text{useful output}}} {E_{\text{input}}}\times100 \]

Substitute:

\[ \eta=\frac{1.68}{2.40}\times100 \]

\[ \boxed{\eta=70\%} \]

Part b: Calculate the other energy transfers.

Using conservation of energy:

\[ E_{\text{other output}}= E_{\text{input}}-E_{\text{useful output}} \]

\[ =2.40-1.68 \]

\[ \boxed{E_{\text{other output}}=0.72\ \text{kJ}} \]

The remaining 30% of the input energy is transferred into other forms, including heating and sound.

Question 2

An electric water pump raises 100 kg of water through a vertical height of 2.4 m in 25 seconds.

The pump draws electrical power at an average rate of 120 W.

Assume the water begins and finishes at rest, and there is no net change in other stored energy.

Calculate:

a) The useful gravitational potential energy gained by the water.

b) The average useful output power of the pump.

c) The efficiency of the pump.

Use \(g=9.8\ \text{m s}^{-2}\).

Solution 2

The pump has an efficiency of 78.4%, and its average useful output power is approximately 94.1 W.

Part a: Calculate the gravitational potential energy gained.

The useful output is the increase in gravitational potential energy:

\[ E_{\text{useful output}}=mgh \]

Substitute:

\[ E_{\text{useful output}} =100\times9.8\times2.4 \]

\[ \boxed{E_{\text{useful output}}=2352\ \text{J}} \]

Part b: Calculate the average useful output power.

Use:

\[ P_{\text{useful output}}= \frac{E_{\text{useful output}}}{t} \]

\[ P_{\text{useful output}}= \frac{2352}{25} \]

\[ \boxed{P_{\text{useful output}}=94.08\ \text{W}} \]

Part c: Calculate the efficiency.

First, calculate the total electrical energy supplied:

\[ E_{\text{input}}=P_{\text{input}}t \]

\[ E_{\text{input}}=120\times25 \]

\[ E_{\text{input}}=3000\ \text{J} \]

Now calculate efficiency:

\[ \eta=\frac{2352}{3000}\times100 \]

\[ \boxed{\eta=78.4\%} \]

The pump converts 78.4% of its electrical input into useful gravitational potential energy.

The remaining energy is:

\[ E_{\text{other output}}=3000-2352 \]

\[ \boxed{E_{\text{other output}}=648\ \text{J}} \]

This energy is transferred into other forms, including heating and sound.

Question 3

A student tests an electric motor and records the following illustrative measurements:

The motor’s intended task is to drive a mechanical load.

Two students interpret the results differently:

• Student A claims the motor is 70% efficient because only the mechanical output is useful.
• Student B claims the motor is 100% efficient because the total energy output equals the electrical input.

Determine which student is correct. Explain the difference between their calculations and why conservation of energy does not require the motor’s efficiency to be 100%.

Solution 3

Student A is correct: the motor’s efficiency is 70%. Student B has confused total energy conservation with useful energy conversion.

First, calculate the useful output efficiency:

\[ \eta=\frac{E_{\text{useful output}}} {E_{\text{input}}}\times100 \]

\[ \eta=\frac{1.05}{1.50}\times100 \]

\[ \boxed{\eta=70\%} \]

The motor is designed to transfer mechanical energy to a load. Therefore, the 1.05 kJ mechanical output is the useful energy transfer.

Now consider Student B’s calculation.

The total energy output is:

\[ E_{\text{total output}}= 1.05+0.30+0.15 \]

\[ E_{\text{total output}}=1.50\ \text{kJ} \]

Therefore:

\[ \frac{E_{\text{total output}}} {E_{\text{input}}}\times100=100\% \]

This calculation is correct, but it does not calculate useful efficiency.

It simply confirms that the total energy output equals the total energy input.

The remaining 30% of the electrical energy is still transferred away from the motor, but it doesn’t perform the intended mechanical task.

This is exactly why energy conservation and efficiency are different ideas:

• Conservation of energy accounts for all energy transfers.
• Efficiency measures the proportion of energy transferred into a useful form.

A device can conserve energy perfectly while having an efficiency well below 100%.

Question 4

A spring-assisted winch receives 300 J of electrical energy during a 5.0-second test.

The winch contains a preloaded spring. During the test, the spring releases 100 J of stored elastic potential energy.

The winch transfers 360 J of useful mechanical energy to a load and 40 J into thermal energy.

Assume these are all the energy transfers during the test.

A student calculates:

\[ \eta=\frac{360}{300}\times100=120\% \]

The student concludes that the winch is more than 100% efficient.

a) Explain why this conclusion is incorrect.

b) Calculate the correct efficiency of the winch, including the spring’s contribution.

c) Calculate the average electrical input power and average useful mechanical output power.

d) Explain why the useful output power can exceed the electrical input power without violating conservation of energy.

Solution 4

The winch is 90% efficient. The apparent 120% result occurs because the student ignores the energy released by the spring.

Part a: Explain the error.

The electrical supply is not the only energy source.

The spring releases 100 J of previously stored energy during the test.

Therefore, the total energy supplied is:

\[ E_{\text{total input}}=300+100 \]

\[ E_{\text{total input}}=400\ \text{J} \]

The student has used only 300 J as the denominator, leaving out an important energy source.

Part b: Calculate the correct efficiency.

Use:

\[ \eta=\frac{E_{\text{useful output}}} {E_{\text{total input}}}\times100 \]

Substitute:

\[ \eta=\frac{360}{400}\times100 \]

\[ \boxed{\eta=90\%} \]

Check the energy budget:

\[ E_{\text{input}}=300+100=400\ \text{J} \]

\[ E_{\text{output}}=360+40=400\ \text{J} \]

The energy budget balances exactly.

Part c: Calculate the average powers.

Electrical input power:

\[ P_{\text{input}}=\frac{E_{\text{electrical input}}}{t} \]

\[ P_{\text{input}}=\frac{300}{5.0} \]

\[ \boxed{P_{\text{input}}=60\ \text{W}} \]

Useful mechanical output power:

\[ P_{\text{useful output}}= \frac{E_{\text{useful output}}}{t} \]

\[ P_{\text{useful output}}=\frac{360}{5.0} \]

\[ \boxed{P_{\text{useful output}}=72\ \text{W}} \]

Part d: Explain the difference in power.

The useful output power is 72 W, while the measured electrical input power is only 60 W.

The difference is possible because the spring is releasing stored energy.

Its average energy release rate is:

\[ P_{\text{spring}}=\frac{100}{5.0} \]

\[ P_{\text{spring}}=20\ \text{W} \]

The winch therefore receives an average of:

\[ 60+20=80\ \text{W} \]

Its useful output is 72 W, and the remaining 8 W is transferred as thermal energy.

The energy accounting is consistent:

\[ 80=72+8 \]

The apparent power efficiency of 120% is not the true efficiency. It compares useful output power with only one of the energy sources.

Question 5

An electric drive is connected to a flywheel and a mechanical load. The flywheel is included inside the system boundary.

The drive operates through two consecutive 10-second stages.

During Stage A, the flywheel slows down and releases stored kinetic energy. During Stage B, it speeds up and stores that energy again.

The following illustrative measurements are recorded:

Use the sign convention:

• A positive change in flywheel energy means the flywheel gains stored energy.
• A negative change means the flywheel releases stored energy.

Assume there are no other energy stores changing and all energy transfers have been measured accurately.

Two students make different claims:

• Student A says Stage A has an efficiency of 130%, so the drive has exceeded the 100% limit.
• Student B says the complete 20-second cycle must have an efficiency of 100% because energy is conserved.

Evaluate both claims.

In your answer:

a) Calculate the apparent efficiency in Stage A using only the electrical input.

b) Determine the average electrical input power and useful mechanical output power during each stage.

c) Calculate the efficiency of the complete 20-second cycle.

d) Explain why neither student’s conclusion is correct. In particular, explain why the flywheel’s change in stored energy matters when comparing the two stages.

Solution 5

The complete drive system has an efficiency of 88% over the full 20-second cycle. Student A has ignored the flywheel’s energy release, while Student B has confused conservation of energy with useful energy efficiency.

This question requires you to track the system through both stages rather than treating each measurement as an isolated result.

Part a: Calculate the apparent Stage A efficiency.

Using only the electrical input gives:

\[ \eta_{\text{apparent}}= \frac{E_{\text{useful output}}} {E_{\text{electrical input}}}\times100 \]

\[ \eta_{\text{apparent}}= \frac{2.6}{2.0}\times100 \]

\[ \boxed{\eta_{\text{apparent}}=130\%} \]

This is the ratio Student A obtains.

However, the flywheel releases 0.8 kJ during Stage A.

The total energy supplied during this stage is:

\[ E_{\text{total input}}=2.0+0.8 \]

\[ E_{\text{total input}}=2.8\ \text{kJ} \]

The actual Stage A efficiency is therefore:

\[ \eta_A=\frac{2.6}{2.8}\times100 \]

\[ \boxed{\eta_A\approx92.9\%} \]

Stage A does not violate the efficiency limit. The apparent 130% ratio simply ignores the flywheel’s contribution.

Part b: Calculate the average powers during each stage.

For Stage A, the electrical input power is:

\[ P_{\text{input,A}}=\frac{2000}{10} \]

\[ \boxed{P_{\text{input,A}}=200\ \text{W}} \]

The useful mechanical output power is:

\[ P_{\text{useful,A}}=\frac{2600}{10} \]

\[ \boxed{P_{\text{useful,A}}=260\ \text{W}} \]

The useful output power exceeds the electrical input power by 60 W.

This is possible because the flywheel is releasing energy.

Its average energy release rate is:

\[ P_{\text{flywheel}}=\frac{800}{10} \]

\[ P_{\text{flywheel}}=80\ \text{W} \]

Of this additional 80 W, 20 W is transferred into thermal energy, leaving 60 W of additional useful mechanical output power.

For Stage B, the electrical input power is:

\[ P_{\text{input,B}}=\frac{3000}{10} \]

\[ \boxed{P_{\text{input,B}}=300\ \text{W}} \]

The useful mechanical output power is:

\[ P_{\text{useful,B}}=\frac{1800}{10} \]

\[ \boxed{P_{\text{useful,B}}=180\ \text{W}} \]

During Stage B, the flywheel gains 0.8 kJ of kinetic energy.

Its average energy storage rate is:

\[ P_{\text{stored}}=\frac{800}{10} \]

\[ P_{\text{stored}}=80\ \text{W} \]

The 300 W input is accounted for by:

• 180 W of useful mechanical output.
• 40 W of thermal energy.
• 80 W stored in the flywheel.

Therefore:

\[ 300=180+40+80 \]

Both stages obey conservation of energy.

Part c: Calculate the efficiency over the complete cycle.

Now consider the entire 20-second operation.

First, calculate the total external electrical input:

\[ E_{\text{input}}=2.0+3.0 \]

\[ E_{\text{input}}=5.0\ \text{kJ} \]

Next, calculate the total useful mechanical output:

\[ E_{\text{useful output}}=2.6+1.8 \]

\[ E_{\text{useful output}}=4.4\ \text{kJ} \]

The flywheel loses 0.8 kJ in Stage A and gains 0.8 kJ in Stage B.

Therefore, its net change in stored energy is:

\[ \Delta E_{\text{flywheel}}=-0.8+0.8 \]

\[ \boxed{\Delta E_{\text{flywheel}}=0\ \text{kJ}} \]

The flywheel finishes with the same stored energy it had at the beginning.

We can now calculate the complete-cycle efficiency using the external electrical input:

\[ \eta_{\text{cycle}}= \frac{E_{\text{useful output}}} {E_{\text{input}}}\times100 \]

\[ \eta_{\text{cycle}}= \frac{4.4}{5.0}\times100 \]

\[ \boxed{\eta_{\text{cycle}}=88\%} \]

We can also check the energy budget.

Total thermal energy:

\[ E_{\text{thermal}}=0.2+0.4 \]

\[ E_{\text{thermal}}=0.6\ \text{kJ} \]

Total energy output:

\[ E_{\text{total output}}=4.4+0.6 \]

\[ E_{\text{total output}}=5.0\ \text{kJ} \]

This equals the total electrical input.

The full-cycle budget balances, with 88% of the supplied energy transferred usefully and 12% transferred into thermal energy.

Part d: Evaluate both students’ claims.

Student A is incorrect because the Stage A calculation excludes energy released by the flywheel.

The 130% figure is an apparent output-to-electrical-input ratio, not the efficiency of the complete system.

Student B is also incorrect.

Conservation of energy requires the total energy output to equal the total energy input when there is no net change in stored energy.

It does not require all the input energy to become useful output.

In this example, 0.6 kJ is transferred into thermal energy. That energy is conserved, but it doesn’t contribute to the intended mechanical task.

There is another subtle point here.

You cannot calculate the complete-cycle efficiency by simply averaging the two stage efficiencies.

The flywheel transfers energy between the stages. Its 0.8 kJ release in Stage A is balanced by its 0.8 kJ gain in Stage B.

That internal transfer must not be counted as an additional external energy source for the complete cycle.

The correct approach is to add the external input energies and useful output energies over the entire interval, then calculate their ratio.

The result is 88%, not 130%, 100%, or the simple average of two stage percentages.

08Where to go next

Understanding why efficiency cannot exceed 100% gives you the foundation for more complicated energy calculations.

The next useful step is learning to rearrange the efficiency equation to find an unknown output power or input power.

For example, if a motor has an input power of 800 W and an efficiency of 75%, its useful mechanical output power is:

\[ P_{\text{useful output}}=0.75\times800 \]

\[ \boxed{P_{\text{useful output}}=600\ \text{W}} \]

Work through How to Find Useful Output Power from Efficiency and Input Power to practise rearranging the relationship.

You can also extend the same energy-accounting method to multiple devices connected in sequence, where each device receives only the useful output transferred by the previous one. See How to Calculate Total Efficiency for Two Devices in Sequence.

The central habit to carry into your HSC Physics questions is simple: before calculating efficiency, make sure you know exactly where the energy came from, where it went, and whether any of it was stored or released during the measurement.