Efficiency
Why does an incandescent lamp get so hot that it can burn your finger, even though its job is to give light, and why does an LED lamp shine just as brightly with much less electricity?
Prior knowledge
- Energy is neither created nor destroyed. It changes form, and part of it often becomes heatchapter Forms of energy and conservation of energy
- The unit of energy is the joule, and when energy is transformed, part of it ends up as heatchapter Energy transformations and the joule
- Power tells how fast energy is transferred, the unit is the watt, and at constant power energy is power multiplied by timechapter Power P = ΔE/Δt
Core path: follow the steps in order. Tasks marked Optional give extra practice. Optional tasks: 4.
Think about it
Two lamps are lit in a room, an incandescent lamp and an LED lamp, and they shine equally brightly. The incandescent lamp takes a power of $60\,\text{W}$ from the electricity grid, the LED lamp only $10\,\text{W}$. When you touch the lamps after they have been on for a moment, the incandescent lamp is burning hot and the LED lamp is only lukewarm.
Where does the electrical energy taken by the incandescent lamp go, and how can you measure how good a device is?
Recap: power and energy
Predict
Make your predictions before investigating 2 questions
Prediction 1: efficiency of a petrol engine
Optional Prediction 2: running time and efficiency
Investigate
Choose a device and adjust the power taken. See how the incoming arrow divides into useful energy and losses.
Simulation
How to use
Choose a device from the menu. Adjust the power taken $P_\text{o}$ and the time $t$ with the sliders. The green arrow is the useful power $P_\text{a}$ and the orange arrow the loss power $P_\text{h}$. The readouts show the powers and the energies in the time $t$. The speed only changes how fast the stripes move in the arrows.
Assumptions of the model
In the model, each device has a constant efficiency. The values are typical rounded values, not measurements of any particular device: incandescent lamp $5\,\%$, LED lamp $30\,\%$, kettle $90\,\%$, electric motor $90\,\%$, petrol engine of a car $25\,\%$ and solar panel $20\,\%$. The efficiency of a real device depends on the load: strongly for a petrol engine, because at a small load it is clearly lower, and only a little for an electric motor when it runs close to the power it was designed for. So the model does not tell how efficiency depends on the power taken. The model assumes that the device runs at constant power. The losses are heat, and in a solar panel they also include reflected light. The readouts of the simulation are values of the model, not measurements.
How the calculation works
The useful power is $P_\text{a} = \eta \cdot P_\text{o}$ and the loss power $P_\text{h} = P_\text{o} - P_\text{a}$. In the time $t$ the energies are $E_\text{o} = P_\text{o} \cdot t$, $E_\text{a} = \eta \cdot E_\text{o}$ and $E_\text{h} = E_\text{o} - E_\text{a}$. In the picture the incoming arrow is always $100\,\text{px}$ thick, the useful arrow $\eta \cdot 100\,\text{px}$ and the loss arrow $(1 - \eta) \cdot 100\,\text{px}$.
Experiment 1: petrol engine
Test prediction 1. The device is a petrol engine of a car, and its input power is locked at $60\,\text{kW}$, because the efficiency of a real engine depends on the load. Change only the time $t$ and record at least three measurements. Look at the energies $E_\text{o}$, $E_\text{a}$ and $E_\text{h}$. The model efficiency $25\,\%$ is a typical value that the model receives as initial data: the simulation does not measure it from the engine but uses it to calculate the energies.
Initial state: device: car petrol engine, input power $P_o$ = $60\,000\,\text{W}$, time $t$ = $10\,\text{s}$. Kept constant: device, input power $P_o$. Varied: time $t$.
| time $t$ | $E_o$ $(\text{J})$ | $E_a$ $(\text{J})$ | $E_h$ $(\text{J})$ |
|---|
For each row of the table, calculate the quotient $E_\text{h} : E_\text{o}$.
What fraction of the energy taken by the engine becomes losses in the model? Answer as a percentage. You get the efficiency from this when you remember that the useful energy and the losses together make up the whole energy taken.
Hint 1: a guiding question
The share of losses is the share of the loss energy in the energy taken. Choose one row of the table and read $E_\text{h}$ and $E_\text{o}$ from it.
Hint 2: what is the connection
The share is $\dfrac{E_\text{h}}{E_\text{o}}$. Convert the ratio to a percentage by multiplying it by $100\,\%$.
Hint 3: an intermediate step
For example, in the row $t = 10\,\text{s}$ the energy taken is $600\,000\,\text{J}$ and the loss energy $450\,000\,\text{J}$, so $450\,000 : 600\,000 = 0.75$.
Show the solution
Given
a row of the table with time $t = 10\,\text{s}$, energy taken $E_\text{o} = 600\,000\,\text{J}$ and loss energy $E_\text{h} = 450\,000\,\text{J}$ (values of the model, power taken locked at $60\,\text{kW}$)
Find
the share of the losses in the energy taken, as a percentage
Principle
The share is calculated by dividing the loss energy by the energy taken. The share is the same on every row, because the engine runs at constant power: both energies are power multiplied by time, and the time cancels from the ratio.
Formula
Conversions
Not needed: both energies are in joules, so the units cancel. The ratio is converted to a percentage by multiplying it by $100\,\%$.
Substitution
Answer
$75\,\%$
Two significant figures, because the efficiency of the model is given to two figures. The efficiency is the rest: $100\,\% - 75\,\% = 25\,\%$.
Reasonableness
The useful energy is $E_\text{a} = 150\,000\,\text{J}$ and $150\,000\,\text{J} + 450\,000\,\text{J} = 600\,000\,\text{J}$. The share is below $100\,\%$ and it is the largest part, as for a petrol engine, whose exhaust gases and cooling carry away a large part of the energy.
Optional Experiment 2: energy and power
Test prediction 2. The device is an incandescent lamp and its input power is locked at $40\,\text{W}$. Change only the time $t$ and record at least three measurements. Look at the input energy $E_\text{o}$ and the useful energy $E_\text{a}$.
Initial state: device: incandescent lamp, input power $P_o$ = $40\,\text{W}$, time $t$ = $10\,\text{s}$. Kept constant: device, input power $P_o$. Varied: time $t$.
| time $t$ | $E_o$ $(\text{J})$ | $E_a$ $(\text{J})$ |
|---|
Conclusion: what efficiency depends on
Based on your table, write: how is efficiency calculated from energies and powers? Did it change when the running time changed, and why? What did the experiment tell you about the petrol engine?
Model answer
In experiment 1 the petrol engine took a constant power of $60\,\text{kW}$. The energies grew with time: when $t = 10\,\text{s}$, $E_\text{o} = 600\,\text{kJ}$ and $E_\text{a} = 150\,\text{kJ}$, and when $t = 20\,\text{s}$, $E_\text{o} = 1200\,\text{kJ}$ and $E_\text{a} = 300\,\text{kJ}$. The ratio $E_\text{a} : E_\text{o} = 25\,\%$ stayed the same, because at constant power $E = P \cdot t$ and the time cancels. The same ratio is obtained from the powers: $15\,\text{kW} : 60\,\text{kW} = 25\,\%$. The remaining $75\,\%$ was losses, that is, heat. So efficiency does not depend on running time when the device runs steadily. The values are values of the simulation model. In the model the efficiency is constant even when the power taken changes, but that is an assumption of the model: the efficiency of a real engine depends on the load.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Theory summary
Key idea
Energy does not disappear, but not all the energy taken by a device becomes the form of energy that you want from the device. The job of an incandescent lamp is to give light, so its useful energy is light. However, most of the electrical energy that the lamp takes becomes heat, which warms the lamp and its surroundings but does not give light. This part is called losses. Losses are not lost energy but energy that has changed into a useless form, usually heat.
The flow of energy through a device can be shown with an energy flow diagram, also called a Sankey diagram. In the diagram the thickness of an arrow is proportional to energy: the incoming arrow divides into a thin arrow of useful energy and a thick arrow of losses. When the lamp takes electrical energy of $100\,\text{J}$, only $5\,\text{J}$ becomes light and the remaining $95\,\text{J}$ is heat. The thicknesses of the outgoing arrows add up to the thickness of the incoming arrow, because energy is conserved.
Losses depend on the purpose of the device
In a kettle, heat is useful when it warms the water, and a loss when it escapes through the sides of the kettle into the room. The heat of an incandescent lamp, on the other hand, is counted as a loss, even if it warms a room in winter, because the benefit of the lamp has been defined as light. So what is useful depends on the purpose of the device, and the value of the efficiency depends on what is defined as useful and what is counted as part of the device.
Efficiency tells the share of the useful part
Efficiency is the share of useful energy in the energy taken by the device. Efficiency is at most $100\,\%$, because energy is conserved and you cannot get more useful energy out than you put in. In real devices it is often below $100\,\%$, because part of the energy taken does not end up in the defined useful purpose: for example, it becomes heat in friction or in conductors, and the heat spreads into the surroundings. When the purpose of a device is something other than heat, efficiency varies a lot: an electric motor reaches about $90\,\%$, but a lamp and a combustion engine stay clearly below $100\,\%$. In an electric heater the purpose is the heat itself: in practice all the electrical energy becomes heat that warms the room, so the efficiency is almost $100\,\%$. An ideal, lossless device would have an efficiency of $100\,\%$. The efficiencies of devices vary a lot:
| Device and its useful output | typical $\eta$ |
|---|---|
| incandescent lamp: light | $5\,\%$ |
| LED lamp: light | $30\,\%$ |
| kettle: warming of water | $90\,\%$ |
| electric motor: motion | $90\,\%$ |
| petrol engine: motion of a car | $25\,\%$ |
| solar panel: electricity | $20\,\%$ |
The values in the table are typical rounded values, not measurements of any particular device. The real efficiency depends on the model of the device and on the way it is used. The value for the petrol engine is typical for steady motorway driving; in city driving and at other small loads the efficiency is clearly lower. The efficiency of an electric motor changes little when it runs close to the power it was designed for.
Efficiency
Efficiency $\eta$ (eta) is the ratio of the useful energy to the energy taken by the device. When the device runs at constant power, the same ratio is also obtained from powers:
$$\eta = \frac{E_\text{a}}{E_\text{o}} = \frac{P_\text{a}}{P_\text{o}}$$
where $\eta$ is the efficiency (a ratio, often given as a percentage), $E_\text{a}$ the useful energy given by the device (J), $E_\text{o}$ the energy taken by the device (J), $P_\text{a}$ the useful power (W) and $P_\text{o}$ the power taken by the device (W). A ratio is converted to a percentage by multiplying it by $100\,\%$, for example $0.30 \cdot 100\,\% = 30\,\%$. The multiplication does not change the number, because $1\,\% = 0.01$ and so $100\,\% = 100 \cdot 0.01 = 1$.
The ratio of the powers is the same as the ratio of the energies, because at constant power energy is power multiplied by time (chapter Power P = ΔE/Δt). While the device runs for a time $t$, the energies are $E_\text{a} = P_\text{a} \cdot t$ and $E_\text{o} = P_\text{o} \cdot t$, and the time cancels:
$$\begin{aligned} \eta &= \frac{E_\text{a}}{E_\text{o}} \\ &= \frac{P_\text{a} \cdot t}{P_\text{o} \cdot t} \\ &= \frac{P_\text{a}}{P_\text{o}} \end{aligned}$$
So efficiency does not depend on whether energy or power is measured, or on the running time.
Conservation of energy and losses
Conservation of energy divides the energy taken by the device into useful energy and losses $E_\text{h}$:
$$E_\text{o} = E_\text{a} + E_\text{h}$$
The same holds for powers, $P_\text{o} = P_\text{a} + P_\text{h}$. When the efficiency is known, the useful energy is $E_\text{a} = \eta \cdot E_\text{o}$ and the energy taken is $E_\text{o} = \tfrac{E_\text{a}}{\eta}$.
Comparing two lamps
A lamp with a low efficiency needs a higher power to get the same useful output. For a light power of $3\,\text{W}$ an incandescent lamp needs $60\,\text{W}$ but an LED lamp only $10\,\text{W}$, because the efficiencies are $5\,\%$ and $30\,\%$. In the diagram the thicknesses of the arrows are on the same scale: the green arrows of the light power are equally thick, but the incoming arrow of the incandescent lamp is six times thicker.
In a diagram drawn to scale you can see the efficiency directly: it is the thickness of the useful arrow as a share of the thickness of the incoming arrow. If the incoming arrow is $5.0\,\text{cm}$ thick and the useful arrow $4.0\,\text{cm}$, the efficiency is $4.0 : 5.0 = 0.80$, that is, $80\,\%$.
Common misconception
Many people think that energy disappears in the losses. However, energy is conserved in every device. The energy leaving an incandescent lamp is equal to the energy entering it: $3\,\text{W}$ of light and $57\,\text{W}$ of heat make $60\,\text{W}$ in total. A loss only means that part of the energy ended up in a form that is of no use.
Another mistake is to think that a high-power device is efficient. Efficiency is a ratio, not an amount. A $100\,\text{W}$ incandescent lamp gives $5\,\text{W}$ of light and a $60\,\text{W}$ lamp $3\,\text{W}$, but both have an efficiency of $5\,\%$. A high-power device uses more energy, but power does not tell what share of the energy is useful.
Summary
- Efficiency is the share of useful energy in the energy taken by the device: $\eta = \tfrac{E_\text{a}}{E_\text{o}} = \tfrac{P_\text{a}}{P_\text{o}}$. It is often given as a percentage and is at most $100\,\%$. In real devices it is often below $100\,\%$, because part of the energy does not end up as the useful output; for an electric heater it is almost $100\,\%$.
- Energy is conserved: $E_\text{o} = E_\text{a} + E_\text{h}$. Losses are usually heat that spreads into the surroundings.
- In an energy flow diagram the thickness of an arrow is proportional to energy or power, and efficiency is the share of the useful arrow in the incoming arrow.
- The efficiency of a device that runs steadily does not depend on running time, because the time cancels from the ratio of the energies. The way the device is used can change it: for example, the efficiency of a petrol engine is clearly lower at a small load.
Optional Return to your predictions
Compare your predictions with the readouts of the simulation. Then return to the opening question: where does the electrical energy taken by the incandescent lamp go, and how can you measure how good a device is?
Would you change your prediction?
Apply
Worked example
Problem
A kettle has an efficiency of $90\,\%$ and takes a power of $2.0\,\text{kW}$ from the electricity grid. The kettle is on for $3.0\,\text{min}$. How much electrical energy becomes losses, that is, heat that does not warm the water?
Step 1/7
Known and unknown. Efficiency $\eta = 90\,\%$, power taken $P_\text{o} = 2.0\,\text{kW}$ and running time $t = 3.0\,\text{min}$. We want the loss energy $E_\text{h}$.
Step 2/7
Principle. The kettle runs at constant power, so the energy taken is power multiplied by time. Efficiency tells what share of the energy taken warms the water. Energy is conserved, so the rest is losses.
Step 3/7
Rearranged formula. The energy taken, the useful energy and the losses are found in order:
$$\begin{aligned} E_\text{o} &= P_\text{o} \cdot t \\ E_\text{a} &= \eta \cdot E_\text{o} \\ E_\text{h} &= E_\text{o} - E_\text{a} \end{aligned}$$
Step 4/7
Unit conversions. Power to watts:
Time to seconds:
Efficiency to a number: $90\,\% = 90 \cdot 0.01 = 0.90$.
Step 5/7
Substitution. $$\begin{aligned} E_\text{o} &= P_\text{o} \cdot t \\ &= 2000\,\text{W} \cdot 180\,\text{s} \\ &= 360\,000\,\text{J} \\[6pt] E_\text{a} &= \eta \cdot E_\text{o} \\ &= 0.90 \cdot 360\,000\,\text{J} \\ &= 324\,000\,\text{J} \\[6pt] E_\text{h} &= E_\text{o} - E_\text{a} \\ &= 360\,000\,\text{J} - 324\,000\,\text{J} \\ &= 36\,000\,\text{J} \end{aligned}$$
Step 6/7
Result. The energy that becomes losses is
The result is given with two significant figures, because the initial values $90\,\%$, $2.0\,\text{kW}$ and $3.0\,\text{min}$ are given to two figures.
Step 7/7
Reasonableness. The losses must be $100\,\% - 90\,\% = 10\,\%$ of the energy taken, and the energy taken is
so $0.10 \cdot 360\,\text{kJ} = 36\,\text{kJ}$. In addition, the useful energy and the losses add up to the energy taken: $324\,\text{kJ} + 36\,\text{kJ} = 360\,\text{kJ}$.
Find the error
Apply to a new situation
Radiation energy of $4.5\,\text{MJ}$ hits a solar panel during one hour. The efficiency of the panel is $20\,\%$.
What is the average electrical power of the panel? Give the answer in watts.
Hint 1: a guiding question
What do you want from the panel as useful output, and what is the energy taken here? Power tells how fast energy is transferred.
Hint 2: what is the connection
First calculate the useful energy $E_\text{a} = \eta \cdot E_\text{o}$ and then divide it by the time: $P_\text{a} = E_\text{a} : t$. Convert megajoules to joules and the hour to seconds.
Hint 3: an intermediate step
$t = 1 \cdot 3600\,\text{s}$, so $P_\text{a} = \dfrac{E_\text{a}}{t}$.
Show the solution
Given
energy hitting the panel $E_\text{o} = 4.5\,\text{MJ}$, time $t = 1\,\text{h}$ and efficiency $\eta = 20\,\%$
Find
the average electrical power of the panel, that is, the useful power $P_\text{a}$ in watts
Principle
The electrical energy is the useful energy of the panel, and the efficiency tells what share of the energy that hits the panel becomes electricity. Power is energy divided by time, so the useful power is found from the useful energy and the time.
Formula
$E_\text{a} = \eta \cdot E_\text{o}$ and $P_\text{a} = \dfrac{E_\text{a}}{t}$, so
Conversions
$20\,\% = 20 \cdot 0.01 = 0.20$.
Substitution
Answer
$250\,\text{W}$
Or $2.5 \cdot 10^{2}\,\text{W}$, two significant figures, as in the energy $4.5\,\text{MJ}$ and the efficiency $20\,\%$.
Reasonableness
The average power hitting the panel is $4.5 \cdot 10^{6}\,\text{J} : 3600\,\text{s} = 1250\,\text{W}$, and one fifth of it is $250\,\text{W}$. The power is below the power hitting the panel, as it must be.
Check your understanding
Exit ticket
Answer on your own first and only then open the answer.
- A device takes energy of $500\,\text{J}$ and gives useful energy of $350\,\text{J}$ from it. What is its efficiency, and where does the rest of the energy go?
Answer
$\eta$$= \dfrac{350\,\text{J}}{500\,\text{J}}$$= 0.70$$= 0.70 \cdot 100\,\%$$= 70\,\%$The remaining $500\,\text{J} - 350\,\text{J} = 150\,\text{J}$ is losses, usually heat that spreads into the surroundings. No energy has disappeared, because $350\,\text{J} + 150\,\text{J} = 500\,\text{J}$.
- The efficiency of a device is $25\,\%$ and it takes a power of $200\,\text{W}$. How large are the useful power and the loss power?
Answer
The useful power is
$P_\text{a}$$= 0.25 \cdot 200\,\text{W}$$= 50\,\text{W}$The loss power is
$P_\text{h}$$= 200\,\text{W} - 50\,\text{W}$$= 150\,\text{W}$ - It is claimed: “A high-power device always has a better efficiency than a low-power device.” Is the claim true? Justify your answer.
Answer
It is not. Efficiency is the share of the useful power in the power taken, not the power itself. For example, $100\,\text{W}$ and $60\,\text{W}$ incandescent lamps have the same efficiency of about $5\,\%$, even though the high-power lamp takes more power.
1. Concept
2. When heat is the benefit
3. Calculation
An electric motor takes electrical energy of $2.5\,\text{kJ}$, and its losses are $0.50\,\text{kJ}$. What is the efficiency of the motor? Give the answer as a percentage.
Hint 1: a guiding question
What you know: the energy taken and the losses. You need the useful energy for the efficiency. Where can you get it?
Hint 2: what is the connection
Energy is conserved, so $E_\text{o} = E_\text{a} + E_\text{h}$. Solve $E_\text{a}$ from it and then calculate $\eta = E_\text{a} : E_\text{o}$.
Hint 3: an intermediate step
so
Show the solution
Given
energy taken $E_\text{o} = 2.5\,\text{kJ}$ and losses $E_\text{h} = 0.50\,\text{kJ}$
Find
the efficiency $\eta$ of the motor as a percentage
Principle
Efficiency is the share of the useful energy in the energy taken. The useful energy was not given, but energy is conserved, so it is found as the difference between the energy taken and the losses.
Formula
$E_\text{a} = E_\text{o} - E_\text{h}$ and $\eta = \dfrac{E_\text{a}}{E_\text{o}}$, so
Conversions
Not needed: all energies are in kilojoules, so the units cancel. The ratio is converted to a percentage by multiplying it by $100\,\%$.
Substitution
so
Answer
$80\,\%$
Two significant figures, as in the given energies.
Reasonableness
The losses are $0.50 : 2.5 = 20\,\%$ of the energy taken, and $80\,\% + 20\,\% = 100\,\%$. The result is below $100\,\%$ and close to the typical $90\,\%$ of an electric motor.
Suggestion: one answer does not show everything you can do, but review the theory summary and do the core exercises.
Optional 4. Explain
The efficiency of an incandescent lamp stays clearly below $100\,\%$ even though energy does not vanish anywhere. Explain why.
Model answer
Energy is conserved, so the energy taken by an incandescent lamp divides into useful energy, that is, light, and losses. In the lamp an electric current heats the filament very hot, and only a small part of the energy leaves as light. For example, of a $60\,\text{W}$ lamp about $3\,\text{W}$ is light and the remaining $57\,\text{W}$ is heat that spreads into the surroundings. The heat is a loss here, because the purpose of the lamp is to give light. The efficiency $3\,\text{W} : 60\,\text{W} = 5\,\%$ is below $100\,\%$, because the useful energy is only a part of the energy taken. No energy is lost: $3\,\text{W} + 57\,\text{W} = 60\,\text{W}$.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Review the chapter's concepts
Cards you mark Review later collect in your review deck.
- Efficiency
- The fraction of the energy taken in that is put to use.$$\eta = \dfrac{E_{\text{a}}}{E_{\text{o}}}$$
- Efficiency from power
- The ratio of output power to input power.$$\eta = \dfrac{P_{\text{a}}}{P_{\text{o}}}$$
- Input energy
- The total energy a device takes in to operate.
- Output energy
- The energy a device transfers or transforms for its intended purpose.
- Transmission loss
- In transmission part of the energy is transformed into an unused form, usually heat.
- Energy of combustion
- The energy released by a fuel can be calculated from its mass and heating value.$$E = Hm$$
- Energy efficiency
- The desired benefit is achieved with as little energy and as few losses as possible.
Exercises
Level 1 · basic practice 8 exercises
One formula or one step of reasoning. Core exercises first, the rest is extra practice.
- Exercise 4-1Core
An electric motor takes electrical energy of $400\,\text{J}$ and gives $340\,\text{J}$ of kinetic energy. What is the efficiency of the motor? Give the answer as a percentage.
Answer
$85\,\%$
Show the solution
Given
energy taken by the device $E_\text{o} = 400\,\text{J}$ and useful energy (kinetic energy) $E_\text{a} = 340\,\text{J}$
Find
the efficiency $\eta$ of the motor as a percentage
Principle
Efficiency tells what share of the energy taken by the device becomes useful. Both energies are known, so the efficiency is their ratio.
Formula
$\eta = \dfrac{E_\text{a}}{E_\text{o}}$
Conversions
Not needed: both energies are in joules, so the units cancel. The result is converted to a percentage by substituting $1 = 100\,\%$.
Substitution
$\eta$$= \dfrac{340\,\text{J}}{400\,\text{J}}$$= 0.85$$= 0.85 \cdot 100\,\%$$= 85\,\%$Answer
$85\,\%$
Two significant figures, because the energies were given to two figures.
Reasonableness
The result is below $100\,\%$, as it should be for a motor, because part of the energy becomes heat in friction and in the windings. The losses are $400\,\text{J} - 340\,\text{J} = 60\,\text{J}$, that is, $60 : 400 = 15\,\%$, and $85\,\% + 15\,\% = 100\,\%$. The typical efficiency of an electric motor is about $90\,\%$, so the result fits.
- Exercise 4-2Core
An LED lamp takes a power of $8.0\,\text{W}$ from the electricity grid and gives a light power of $2.4\,\text{W}$.
What is the efficiency of the lamp? Give the answer as a percentage.
Answer
$30\,\%$
Show the solution
Given
power taken by the device $P_\text{o} = 8.0\,\text{W}$ and useful power (light) $P_\text{a} = 2.4\,\text{W}$
Find
the efficiency $\eta$ of the LED lamp as a percentage
Principle
Efficiency can be calculated as a ratio of powers, because the device runs at constant power: in a time $t$ the energies are $P_\text{a} \cdot t$ and $P_\text{o} \cdot t$, and the time cancels.
Formula
$\eta = \dfrac{P_\text{a}}{P_\text{o}}$
Conversions
Not needed: both powers are in watts. The result is converted to a percentage by substituting $1 = 100\,\%$.
Substitution
$\eta$$= \dfrac{2.4\,\text{W}}{8.0\,\text{W}}$$= 0.30$$= 0.30 \cdot 100\,\%$$= 30\,\%$Answer
$30\,\%$
Two significant figures, as in the powers $2.4\,\text{W}$ and $8.0\,\text{W}$.
Reasonableness
About a third of the electrical power becomes light, and the rest, $8.0\,\text{W} - 2.4\,\text{W} = 5.6\,\text{W}$, becomes heat. The typical efficiency of an LED lamp is about $30\,\%$, so the result is reasonable.
- Exercise 4-3
The efficiency of an incandescent lamp is $5.0\,\%$, and the lamp takes a power of $60\,\text{W}$ from the electricity grid.
How large is the light power given by the lamp? Give the answer in watts.
Answer
$3.0\,\text{W}$
Show the solution
Given
efficiency $\eta = 5.0\,\%$ and power taken $P_\text{o} = 60\,\text{W}$
Find
the light power, that is, the useful power $P_\text{a}$ in watts
Principle
Efficiency is the share of the useful power in the power taken. When the efficiency and the power taken are known, the useful power is found by multiplying.
Formula
$\eta = \dfrac{P_\text{a}}{P_\text{o}}$, so $P_\text{a} = \eta \cdot P_\text{o}$
Conversions
A percentage is converted to a number by substituting $1\,\% = 0.01$: $5.0\,\% = 5.0 \cdot 0.01 = 0.050$.
Substitution
$P_\text{a}$$= 0.050 \cdot 60\,\text{W}$$= 3.0\,\text{W}$Answer
$3.0\,\text{W}$
Two significant figures, as in the efficiency $5.0\,\%$.
Reasonableness
Only about one twentieth becomes light, so the result is much smaller than $60\,\text{W}$. The loss power is $60\,\text{W} - 3.0\,\text{W} = 57\,\text{W}$, which explains why an incandescent lamp gets hot.
- Exercise 4-4
The efficiency of an electric motor is $90\,\%$, and it gives useful energy of $1800\,\text{J}$.
How much electrical energy did the motor take? Give the answer in joules.
Answer
$2000\,\text{J}$
Show the solution
Given
efficiency $\eta = 90\,\%$ and useful energy $E_\text{a} = 1800\,\text{J}$
Find
the electrical energy $E_\text{o}$ taken by the motor in joules
Principle
The efficiency tells what share of the energy taken becomes useful energy. So the energy taken is larger than the useful energy, and it is found by dividing the useful energy by the efficiency.
Formula
$\eta = \dfrac{E_\text{a}}{E_\text{o}}$, so $E_\text{o} = \dfrac{E_\text{a}}{\eta}$
Conversions
A percentage is converted to a number by substituting $1\,\% = 0.01$: $90\,\% = 90 \cdot 0.01 = 0.90$.
Substitution
$E_\text{o}$$= \dfrac{1800\,\text{J}}{0.90}$$= 2000\,\text{J}$Answer
$2000\,\text{J}$
Or $2.0 \cdot 10^{3}\,\text{J}$, two significant figures, as in the efficiency $90\,\%$.
Reasonableness
The energy taken is larger than the useful energy, as it must be. Check: $0.90 \cdot 2000\,\text{J} = 1800\,\text{J}$, and the losses are $200\,\text{J}$, that is, $10\,\%$ of the energy taken.
- Exercise 4-5
Radiation energy of $5.0\,\text{kJ}$ hits a solar panel, and the efficiency of the panel is $20\,\%$. How much energy becomes losses? Give the answer in kilojoules.
Answer
$4.0\,\text{kJ}$
Show the solution
Given
radiation energy hitting the panel $E_\text{o} = 5.0\,\text{kJ}$ and efficiency $\eta = 20\,\%$
Find
the energy that becomes losses, $E_\text{h}$, in kilojoules
Principle
Energy is conserved: the energy taken divides into useful energy and losses, $E_\text{o} = E_\text{a} + E_\text{h}$. The useful energy is found from the efficiency, and the rest is losses.
Formula
$E_\text{a} = \eta \cdot E_\text{o}$ and $E_\text{h} = E_\text{o} - E_\text{a}$
Conversions
A percentage is converted to a number by substituting $1\,\% = 0.01$: $20\,\% = 20 \cdot 0.01 = 0.20$. The energies stay in kilojoules.
Substitution
$E_\text{a}$$= 0.20 \cdot 5.0\,\text{kJ}$$= 1.0\,\text{kJ}$so
$E_\text{h}$$= 5.0\,\text{kJ} - 1.0\,\text{kJ}$$= 4.0\,\text{kJ}$Answer
$4.0\,\text{kJ}$
Two significant figures, as in the given values.
Reasonableness
The efficiency is $20\,\%$, so the share of the losses is $80\,\%$ and $0.80 \cdot 5.0\,\text{kJ} = 4.0\,\text{kJ}$. Most of the energy of the panel becomes heat, as everyday experience also tells: a solar panel warms up in the sun.
- Exercise 4-6
An inventor claims to have built a device with an efficiency of $120\,\%$. What can be said about the claim?
Answer
The claim is impossible, because a device cannot give more useful energy than is put into it.
Show the solution
Find
whether the efficiency of a device can be $120\,\%$.
Principle
Efficiency is the share of useful energy in the energy taken by the device: $\eta = \dfrac{E_\text{a}}{E_\text{o}}$. Energy is conserved, and the energy taken is divided into useful energy and losses, $E_\text{o} = E_\text{a} + E_\text{h}$. Losses are not negative, so the useful energy is at most as large as the energy taken.
Model answer
The claim is impossible. An efficiency of $120\,\%$ would mean that $E_\text{a} = 1.2 \cdot E_\text{o}$, that is, the device would give more useful energy than is put into it, and the extra energy would appear out of nothing. Efficiency is at most $100\,\%$. In real devices it is usually below $100\,\%$, because part of the energy taken does not end up as useful energy. Careful building or a long running time does not change this.
- Exercise 4-7
A cordless drill takes energy of $36\,\text{kJ}$ from its battery and gives $27\,\text{kJ}$ of kinetic energy. How much energy becomes losses? Give the answer in kilojoules.
Answer
$9.0\,\text{kJ}$
Show the solution
Given
energy taken $E_\text{o} = 36\,\text{kJ}$ and useful energy (kinetic energy) $E_\text{a} = 27\,\text{kJ}$
Find
the energy that becomes losses, $E_\text{h}$, in kilojoules
Principle
Energy is conserved, so the energy taken by the device is divided into useful energy and losses. The losses are found by subtracting the useful energy from the energy taken.
Formula
$E_\text{o} = E_\text{a} + E_\text{h}$, so $E_\text{h} = E_\text{o} - E_\text{a}$
Conversions
Not needed: both energies are in kilojoules.
Substitution
$E_\text{h}$$= 36\,\text{kJ} - 27\,\text{kJ}$$= 9.0\,\text{kJ}$Answer
$9.0\,\text{kJ}$
Two significant figures, as in the given energies.
Reasonableness
The losses are $9.0 : 36 = 0.25 = 25\,\%$ of the energy taken, so the efficiency is $27 : 36 = 75\,\%$ and $75\,\% + 25\,\% = 100\,\%$. The losses are smaller than the energy taken, as they should be, and they are mostly heat.
- Exercise 4-8
The efficiency of a generator is $90\,\%$, and it gives an electrical power of $180\,\text{kW}$.
How large a power does the generator take from the turbine? Give the answer in kilowatts.
Answer
$200\,\text{kW}$
Show the solution
Given
efficiency $\eta = 90\,\%$ and useful power (electrical power) $P_\text{a} = 180\,\text{kW}$
Find
the power $P_\text{o}$ that the generator takes from the turbine in kilowatts
Principle
The useful power is the share of the power taken given by the efficiency. So the power taken is larger than the useful power, and it is found by dividing the useful power by the efficiency.
Formula
$\eta = \dfrac{P_\text{a}}{P_\text{o}}$, so $P_\text{o} = \dfrac{P_\text{a}}{\eta}$
Conversions
A percentage is converted to a number by substituting $1\,\% = 0.01$: $90\,\% = 90 \cdot 0.01 = 0.90$. The power does not need converting, because the answer is requested in kilowatts.
Substitution
$P_\text{o}$$= \dfrac{180\,\text{kW}}{0.90}$$= 200\,\text{kW}$Answer
$200\,\text{kW}$
Or $2.0 \cdot 10^{2}\,\text{kW}$, two significant figures, as in the efficiency $90\,\%$.
Reasonableness
The power taken is larger than the useful power, as it should be. Check: $0.90 \cdot 200\,\text{kW} = 180\,\text{kW}$, and the loss power is $200\,\text{kW} - 180\,\text{kW} = 20\,\text{kW}$, that is, $10\,\%$ of the power taken.
Level 2 · applied practice 7 exercises
Two steps, working with measured data, a unit conversion or rearranging a formula. Core exercises first, the rest is extra practice.
- Exercise 4-9Core
An electric motor takes a power of $1.0\,\text{kW}$ from the electricity grid, and its efficiency is $80\,\%$. The motor runs for $2.0\,\text{min}$.
How much kinetic energy does the motor give? Give the answer in kilojoules.
Answer
$96\,\text{kJ}$
Show the solution
Given
power taken $P_\text{o} = 1.0\,\text{kW}$, efficiency $\eta = 80\,\%$ and running time $t = 2.0\,\text{min}$
Find
the kinetic energy given by the motor, that is, the useful energy $E_\text{a}$ in kilojoules
Principle
The motor runs at constant power, so the energy taken is power multiplied by time. The efficiency tells what share of the energy taken is useful energy.
Formula
$E_\text{o} = P_\text{o} \cdot t$ and $E_\text{a} = \eta \cdot E_\text{o}$, so $E_\text{a} = \eta \cdot P_\text{o} \cdot t$
Conversions
$1.0\,\text{kW}$$= 1.0 \cdot 1000\,\text{W}$$= 1000\,\text{W}$$2.0\,\text{min}$$= 2.0 \cdot 60\,\text{s}$$= 120\,\text{s}$$80\,\% = 80 \cdot 0.01 = 0.80$.
Substitution
$E_\text{o}$$= 1000\,\text{W} \cdot 120\,\text{s}$$= 120\,000\,\text{J}$so
$E_\text{a}$$= 0.80 \cdot 120\,000\,\text{J}$$= 96\,000\,\text{J}$$= 96\,\text{kJ}$Answer
$96\,\text{kJ}$
Two significant figures, as in the given values.
Reasonableness
The energy taken is $120\,\text{kJ}$ and $80\,\%$ of it is useful, so the result is smaller than $120\,\text{kJ}$. In the time $120\,\text{s}$ the useful power $0.80 \cdot 1000\,\text{W} = 800\,\text{W}$ gives $800\,\text{W} \cdot 120\,\text{s} = 96\,\text{kJ}$ in the same way.
- Exercise 4-10Core
An LED lamp with an efficiency of $30\,\%$ takes a power of $9.0\,\text{W}$. An incandescent lamp with an efficiency of $5.0\,\%$ gives the same light power.
How large a power does the incandescent lamp take from the electricity grid? Give the answer in watts.
Answer
$54\,\text{W}$
Show the solution
Given
LED lamp: $\eta_\text{LED} = 30\,\%$ and $P_\text{o,LED} = 9.0\,\text{W}$; incandescent lamp: $\eta_\text{inc} = 5.0\,\%$; the light power of both is the same
Find
the power taken by the incandescent lamp, $P_\text{o,inc}$, in watts
Principle
When the light power is the same, the useful powers of the lamps are equal. First calculate the useful power of the LED lamp from its efficiency, and then the power that the incandescent lamp needs to give the same useful power.
Formula
$P_\text{a}$$= \eta_\text{LED} \cdot P_\text{o,LED}$$P_\text{o,inc}$$= \dfrac{P_\text{a}}{\eta_\text{inc}}$Conversions
Percentages are converted to numbers by substituting $1\,\% = 0.01$: $30\,\% = 0.30$ and $5.0\,\% = 0.050$.
Substitution
$P_\text{a}$$= 0.30 \cdot 9.0\,\text{W}$$= 2.7\,\text{W}$so
$P_\text{o,inc}$$= \dfrac{2.7\,\text{W}}{0.050}$$= 54\,\text{W}$Answer
$54\,\text{W}$
Two significant figures, as in the efficiencies.
Reasonableness
The efficiency of the incandescent lamp is one sixth of the efficiency of the LED lamp, so its power is six times as large: $6 \cdot 9.0\,\text{W} = 54\,\text{W}$. The result matches everyday experience: an LED lamp of about $10\,\text{W}$ shines like a $60\,\text{W}$ incandescent lamp.
- Exercise 4-11
An incandescent lamp with an electrical power of $60\,\text{W}$ and an efficiency of $5.0\,\%$ burns for $4.0\,\text{h}$. How much electrical energy becomes heat? Give the answer in kilowatt-hours.
Answer
$0.23\,\text{kWh}$
Show the solution
Given
power taken $P_\text{o} = 60\,\text{W}$, efficiency $\eta = 5.0\,\%$ and burning time $t = 4.0\,\text{h}$
Find
the electrical energy that becomes heat, that is, the loss energy $E_\text{h}$ in kilowatt-hours
Principle
The lamp burns at constant power, so the energy taken is power multiplied by time. The efficiency tells what share becomes light, and the rest is losses, that is, heat, because energy is conserved.
Formula
$E_\text{o}$$= P_\text{o} \cdot t$$E_\text{h}$$= E_\text{o} - \eta \cdot E_\text{o}$$= (1 - \eta) \cdot E_\text{o}$Conversions
Power to kilowatts:
$60\,\text{W}$$= 60 \cdot 0.001\,\text{kW}$$= 0.060\,\text{kW}$When the power is in kilowatts and the time in hours, the energy is obtained directly in kilowatt-hours. In addition, $5.0\,\% = 5.0 \cdot 0.01 = 0.050$.
Substitution
$E_\text{o}$$= 0.060\,\text{kW} \cdot 4.0\,\text{h}$$= 0.24\,\text{kWh}$so
$E_\text{h}$$= (1 - 0.050) \cdot 0.24\,\text{kWh}$$= 0.228\,\text{kWh}$Answer
$0.23\,\text{kWh}$
Two significant figures, as in the given values.
Reasonableness
The lamp turns about $95\,\%$ of the energy taken into heat, so the loss energy is almost as large as the energy taken, $0.24\,\text{kWh}$, but slightly smaller. The useful energy is $0.012\,\text{kWh}$, and $0.228\,\text{kWh} + 0.012\,\text{kWh} = 0.24\,\text{kWh}$.
- Exercise 4-12
The efficiency of the petrol engine of a car is $25\,\%$, and the engine gives a power of $15\,\text{kW}$ to move the car.
How large is the loss power, that is, the power that becomes heat? Give the answer in kilowatts.
Answer
$45\,\text{kW}$
Show the solution
Given
efficiency $\eta = 25\,\%$ and useful power $P_\text{a} = 15\,\text{kW}$
Find
the loss power $P_\text{h}$ in kilowatts
Principle
The power taken by the petrol engine divides into useful power and loss power, because energy is conserved: $P_\text{o} = P_\text{a} + P_\text{h}$. The power taken is found from the useful power and the efficiency, and the loss power is the difference.
Formula
$P_\text{o} = \dfrac{P_\text{a}}{\eta}$ and $P_\text{h} = P_\text{o} - P_\text{a}$
Conversions
A percentage is converted to a number by substituting $1\,\% = 0.01$: $25\,\% = 0.25$. The powers stay in kilowatts.
Substitution
$P_\text{o}$$= \dfrac{15\,\text{kW}}{0.25}$$= 60\,\text{kW}$so
$P_\text{h}$$= 60\,\text{kW} - 15\,\text{kW}$$= 45\,\text{kW}$Answer
$45\,\text{kW}$
Two significant figures, as in the given values.
Reasonableness
An efficiency of $25\,\%$ means that the losses are $75\,\%$ of the power taken, and $0.75 \cdot 60\,\text{kW} = 45\,\text{kW}$. The loss power is three times the useful power, which explains why the engine and the exhaust gases get very hot.
- Exercise 4-13
In an example measurement, the electrical power $P_\text{o}$ of different lamps and the light power $P_\text{a}$ they give were measured. Calculate the efficiency of each lamp.
How many times as large is the efficiency of the LED lamp compared with that of the halogen lamp?
Example measurement: electrical power and light power of lamps Lamp $P_\text{o}\,(\text{W})$ $P_\text{a}\,(\text{W})$ Incandescent lamp 60 3.0 Halogen lamp 40 2.8 Fluorescent lamp 18 3.6 LED lamp 9.0 2.7 Answer
$4.3$
Show the solution
Given
from the table, the power taken $P_\text{o}$ and the light power $P_\text{a}$ of the lamps: incandescent lamp $60\,\text{W}$ and $3.0\,\text{W}$, halogen lamp $40\,\text{W}$ and $2.8\,\text{W}$, fluorescent lamp $18\,\text{W}$ and $3.6\,\text{W}$, LED lamp $9.0\,\text{W}$ and $2.7\,\text{W}$
Find
the ratio of the efficiency of the LED lamp to the efficiency of the halogen lamp, that is, how many times as large the efficiency of the LED is
Principle
The lamps run at constant power, so the efficiency is the ratio of the useful power to the power taken. It is calculated separately for each lamp, and the comparison is made by dividing the efficiency of the LED by the efficiency of the halogen lamp.
Formula
$\eta = \dfrac{P_\text{a}}{P_\text{o}}$
Conversions
Not needed: all powers are in watts, so the units cancel. The efficiencies are calculated as ratios, and they are converted to percentages by substituting $1 = 100\,\%$.
Substitution
Incandescent lamp
$\eta$$= \dfrac{3.0\,\text{W}}{60\,\text{W}}$$= 0.050$$= 5.0\,\%$halogen lamp
$\eta$$= \dfrac{2.8\,\text{W}}{40\,\text{W}}$$= 0.070$$= 7.0\,\%$fluorescent lamp
$\eta$$= \dfrac{3.6\,\text{W}}{18\,\text{W}}$$= 0.20$$= 20\,\%$and LED lamp
$\eta$$= \dfrac{2.7\,\text{W}}{9.0\,\text{W}}$$= 0.30$$= 30\,\%$The ratio of the LED to the halogen lamp is $\dfrac{0.30}{0.070} = 4.3$.
Answer
The efficiency of the LED lamp is about $4.3$ times the efficiency of the halogen lamp. The result is given to two significant figures, as the values in the table.
Reasonableness
The efficiencies grow in the order incandescent lamp, halogen, fluorescent lamp, LED, and all are below $100\,\%$. For the light power $2.7\,\text{W}$ the LED needs only $9.0\,\text{W}$, while a halogen lamp would need $2.7\,\text{W} : 0.070 = 39\,\text{W}$ to give the same light power, and $39 : 9.0 = 4.3$. Note that the fluorescent lamp gives the most light ($3.6\,\text{W}$), but it does not have the best efficiency: a large useful power does not mean a high efficiency.
- Exercise 4-14
In an example measurement, the running of an electric motor was followed for $40\,\text{s}$. An electricity meter showed the electrical energy $E_\text{o}$ taken by the motor and a sensor the kinetic energy $E_\text{a}$ given by the motor, while the motor runs steadily. Calculate the efficiency from each row that is not zero and, based on them, determine the efficiency of the motor as a percentage.
Example measurement: energy taken and given by the motor $t\,(\text{s})$ $E_\text{o}\,(\text{kJ})$ $E_\text{a}\,(\text{kJ})$ 0 0 0 10 30 24 20 61 48 30 89 71 40 120 95 Answer
$79\,\%$
Show the solution
Given
from the table, the electrical energy $E_\text{o}$ taken by the motor and the kinetic energy $E_\text{a}$ given by the motor as a function of time $t$; on the last row $t = 40\,\text{s}$, $E_\text{o} = 120\,\text{kJ}$ and $E_\text{a} = 95\,\text{kJ}$
Find
the efficiency $\eta$ of the motor as a percentage
Principle
Efficiency is the share of the useful energy in the energy taken. The energies have been measured from the start (at $t = 0$ both are zero), so the efficiency is obtained directly from each row. The efficiency of a steadily running motor does not depend on how long a period is examined, so the values of the rows are the same within the accuracy of the measurement, and the row of the whole period gives the result.
Formula
$\eta = \dfrac{E_\text{a}}{E_\text{o}}$
Conversions
Not needed: both energies are in kilojoules, so the units cancel. The result is converted to a percentage by substituting $1 = 100\,\%$.
Substitution
The rows:
$\dfrac{24\,\text{kJ}}{30\,\text{kJ}}$$= 0.80$$\dfrac{48\,\text{kJ}}{61\,\text{kJ}}$$= 0.79$$\dfrac{71\,\text{kJ}}{89\,\text{kJ}}$$= 0.80$$\dfrac{95\,\text{kJ}}{120\,\text{kJ}}$$= 0.792$All are about $0.79$–$0.80$, and from the row of the whole period $\eta = 0.792 \cdot 100\,\% = 79\,\%$.
Answer
The efficiency of the motor is about $79\,\%$, two significant figures. There is a little scatter in the measurement points, so a more exact number would be misleading.
Reasonableness
All rows give the same efficiency. The motor takes a power of $120\,\text{kJ} : 40\,\text{s} = 3.0\,\text{kW}$ and gives a useful power of $95\,\text{kJ} : 40\,\text{s} = 2.4\,\text{kW}$; the losses are $120\,\text{kJ} - 95\,\text{kJ} = 25\,\text{kJ}$, that is, about $21\,\%$. For an electric motor the result is reasonable.
- Exercise 4-15
Motor A takes a power of $2.0\,\text{kW}$ and gives a useful power of $1.5\,\text{kW}$. Motor B takes a power of $500\,\text{W}$ and gives a useful power of $400\,\text{W}$.
Which statement about the efficiencies of the motors is correct?
Answer
The efficiency of motor B is greater: it turns $80\,\%$ of the power taken into useful power, motor A only $75\,\%$.
Show the solution
Find
which motor has the greater efficiency.
Principle
Efficiency is the share of the useful power in the power taken, $\eta = \dfrac{P_\text{a}}{P_\text{o}}$. It is a ratio, so it can be used to compare devices of different sizes. The size of the useful power or of the loss power alone does not decide the comparison.
Model answer
The efficiency of motor A is $\eta_\text{A} = \dfrac{1.5\,\text{kW}}{2.0\,\text{kW}} = 0.75 = 75\,\%$ and that of motor B is $\eta_\text{B} = \dfrac{400\,\text{W}}{500\,\text{W}} = 0.80 = 80\,\%$. So the efficiency of motor B is greater, even though A gives the greater useful power. The loss power of A is $2.0\,\text{kW} - 1.5\,\text{kW} = 0.5\,\text{kW}$ and that of B is $500\,\text{W} - 400\,\text{W} = 100\,\text{W}$, but what matters is the share of the loss in the power taken: $25\,\%$ for A and $20\,\%$ for B.
Level 3 · challenging 5 exercises
Optional: application, two ideas together or a new situation where you reason for yourself.
- Exercise 4-16
Device A takes energy of $200\,\text{J}$ and gives useful energy of $150\,\text{J}$. Device B takes energy of $4.0\,\text{kJ}$, and its losses are $2.0\,\text{kJ}$.
How many times as large is the efficiency of device A compared with the efficiency of device B?
Answer
$1.5$
Show the solution
Given
device A: $E_\text{o,A} = 200\,\text{J}$ and $E_\text{a,A} = 150\,\text{J}$; device B: $E_\text{o,B} = 4.0\,\text{kJ}$ and $E_\text{h,B} = 2.0\,\text{kJ}$
Find
the ratio of the efficiencies $\eta_\text{A} : \eta_\text{B}$, that is, how many times as large $\eta_\text{A}$ is compared with $\eta_\text{B}$
Principle
The efficiency of device A is found directly from the useful energy. For device B the losses were given, so the useful energy is found from conservation of energy: $E_\text{a} = E_\text{o} - E_\text{h}$. The comparison is made by dividing the efficiencies by each other.
Formula
$\eta_\text{A}$$= \dfrac{E_\text{a,A}}{E_\text{o,A}}$$\eta_\text{B}$$= \dfrac{E_\text{o,B} - E_\text{h,B}}{E_\text{o,B}}$and the comparison $\eta_\text{A} : \eta_\text{B}$
Conversions
Efficiencies are ratios, so the unit does not affect them as long as the energies of each device are in the same unit. So the kilojoules of device B do not need to be converted to joules.
Substitution
$\eta_\text{A}$$= \dfrac{150\,\text{J}}{200\,\text{J}}$$= 0.75$$\eta_\text{B}$$= \dfrac{4.0\,\text{kJ} - 2.0\,\text{kJ}}{4.0\,\text{kJ}}$$= 0.50$so
$\eta_\text{A} : \eta_\text{B}$$= 0.75 : 0.50$$= 1.5$Answer
$1.5$
Times, two significant figures, as in the given values.
Reasonableness
Device A turns $75\,\%$ and device B $50\,\%$ of the energy taken into useful energy, so A is better and the ratio is greater than one. Check: $1.5 \cdot 50\,\% = 75\,\%$.
- Exercise 4-17
A radiation power of $1.2\,\text{kW}$ hits a solar panel, and the efficiency of the panel is $20\,\%$.
How long does the panel need to produce $0.50\,\text{MJ}$ of electrical energy? Give the answer in seconds.
Answer
$2083\,\text{s}$
Show the solution
Given
power taken $P_\text{o} = 1.2\,\text{kW}$, efficiency $\eta = 20\,\%$ and electrical energy to be produced $E_\text{a} = 0.50\,\text{MJ}$
Find
the time $t$ in seconds
Principle
The efficiency tells what share of the power taken becomes the useful power of the panel. At constant power the useful energy is useful power multiplied by time, so the time is found by dividing the useful energy by the useful power.
Formula
$P_\text{a} = \eta \cdot P_\text{o}$ and $E_\text{a} = P_\text{a} \cdot t$, so
$t$$= \dfrac{E_\text{a}}{\eta \cdot P_\text{o}}$Conversions
$1.2\,\text{kW}$$= 1.2 \cdot 1000\,\text{W}$$= 1200\,\text{W}$$0.50\,\text{MJ}$$= 0.50 \cdot 10^{6}\,\text{J}$$= 500\,000\,\text{J}$$20\,\% = 20 \cdot 0.01 = 0.20$.
Substitution
$t$$= \dfrac{500\,000\,\text{J}}{0.20 \cdot 1200\,\text{W}}$$= \dfrac{500\,000\,\text{J}}{240\,\text{W}}$$= 2083\,\text{s}$Answer
$2083\,\text{s}$
Or about $2.1 \cdot 10^{3}\,\text{s}$, two significant figures, as in the given values.
Reasonableness
The useful power is only $240\,\text{W}$, so $0.50\,\text{MJ}$ takes more than half an hour: $2083\,\text{s} : 60 = 35\,\text{min}$. If the efficiency were $100\,\%$, it would take $500\,000\,\text{J} : 1200\,\text{W} = 417\,\text{s}$, and a time five times as long corresponds to an efficiency of $20\,\%$.
- Exercise 4-18
The figure shows the energy flow diagram of an electric motor on squared paper. The thicknesses of the arrows are drawn to scale, and the energy entering the motor is marked in the figure.
How much energy becomes losses in the motor? Give the answer in joules.
Answer
$180\,\text{J}$
Show the solution
Given
from the figure, the electrical energy entering the motor $E_\text{o} = 240\,\text{J}$, and the thicknesses of the arrows in squares: incoming arrow $d_\text{o} = 8$ squares, useful energy arrow $d_\text{a} = 2$ squares and loss arrow $d_\text{h} = 6$ squares
Find
the energy that becomes losses, $E_\text{h}$, in joules
Principle
In an energy flow diagram drawn to scale, the thickness of an arrow is proportional to energy. So one square always corresponds to the same energy, and the thickness and the energy of the incoming arrow tell how much. Energy is conserved, so the thicknesses of the outgoing arrows add up to the thickness of the incoming arrow.
Reading the graph
The incoming arrow spans eight squares from top to bottom. The green arrow of useful energy on the right edge is two squares thick and the orange loss arrow is six squares thick; the thickness is read from the straight part of the arrow, before the head. Check: $2 + 6 = 8$ squares.
Formula
$\dfrac{E_\text{h}}{E_\text{o}}$$= \dfrac{d_\text{h}}{d_\text{o}}$so
$E_\text{h}$$= \dfrac{d_\text{h}}{d_\text{o}} \cdot E_\text{o}$Conversions
Not needed: the thicknesses are in the same unit (squares), so they cancel, and the energy is in joules.
Substitution
$E_\text{h}$$= \dfrac{6}{8} \cdot 240\,\text{J}$$= 0.75 \cdot 240\,\text{J}$$= 180\,\text{J}$Answer
$180\,\text{J}$
Or $1.8 \cdot 10^{2}\,\text{J}$, two significant figures, as in the incoming energy $240\,\text{J}$.
Reasonableness
One square corresponds to the energy $240\,\text{J} : 8 = 30\,\text{J}$, so the useful energy is $2 \cdot 30\,\text{J} = 60\,\text{J}$ and the losses $6 \cdot 30\,\text{J} = 180\,\text{J}$; together $240\,\text{J}$, as conservation of energy requires. The efficiency is $60\,\text{J} : 240\,\text{J} = 25\,\%$, which is possible for a small toy motor, for example.
- Exercise 4-19
An electric car takes energy of $17\,\text{kWh}$ from its battery over a distance of one hundred kilometres, and the combined efficiency of its motor and transmission is $90\,\%$. A petrol car uses $6.0\,\textsf{l}$ of petrol per hundred kilometres, and the energy content of petrol is $32\,\text{MJ/l}$. The efficiency of the petrol engine is $25\,\%$.
Convert the energy that the petrol car takes from its tank into kilowatt-hours and calculate how many times as large it is compared with the energy the electric car takes from its battery. Also calculate the useful energy of each car per hundred kilometres and use it to explain why the energies taken by the cars differ so much. The examination covers only the cars' own use of energy: the production of electricity and petrol is not taken into account.
Answer
$3.1$
Show the solution
Given
electric car: energy taken from the battery $E_\text{e} = 17\,\text{kWh}$ per hundred kilometres and efficiency $\eta_\text{e} = 90\,\%$; petrol car: petrol $V = 6.0\,\textsf{l}$ per hundred kilometres, energy content $h = 32\,\text{MJ/l}$ and efficiency $\eta_\text{p} = 25\,\%$
Find
how many times as large the energy $E_\text{p}$ that the petrol car takes from its tank is compared with the energy $E_\text{e}$ that the electric car takes from its battery; also the useful energy of each car and an explanation for the difference
Principle
Cars can be compared only when the energies are in the same unit. The energy of petrol is calculated from the volume and the energy content, and it is converted to kilowatt-hours by substituting $1\,\text{kWh} = 1000\,\text{W} \cdot 3600\,\text{s} = 3.6\,\text{MJ}$. The useful energy is the efficiency multiplied by the energy taken. Energy is conserved: the energy taken is useful energy and losses (heat).
Formula
$E_\text{p} = V \cdot h$, $\eta = \dfrac{E_\text{a}}{E_\text{o}}$, so $E_\text{a} = \eta \cdot E_\text{o}$, and the comparison $\dfrac{E_\text{p}}{E_\text{e}}$
Conversions
$1\,\text{kWh}$$= 1000\,\text{W} \cdot 3600\,\text{s}$$= 3.6\,\text{MJ}$so
$1\,\text{MJ}$$= \dfrac{1}{3.6}\,\text{kWh}$Efficiencies: $90\,\% = 0.90$ and $25\,\% = 0.25$.
Substitution
The energy of the petrol is
$E_\text{p}$$= 6.0 \text{ litres} \cdot 32\,\text{MJ/l}$$= 192\,\text{MJ}$In kilowatt-hours
$E_\text{p}$$= \dfrac{192\,\text{MJ}}{3.6\,\text{MJ/kWh}}$$= 53.3\,\text{kWh}$The ratio is
$\dfrac{E_\text{p}}{E_\text{e}}$$= \dfrac{53.3\,\text{kWh}}{17\,\text{kWh}}$$= 3.1$The useful energies are
$E_\text{a,e}$$= 0.90 \cdot 17\,\text{kWh}$$= 15.3\,\text{kWh}$for the electric car and
$E_\text{a,p}$$= 0.25 \cdot 53.3\,\text{kWh}$$= 13.3\,\text{kWh}$for the petrol car.
Answer
The petrol car takes about $3.1$ times as much energy from its tank as the electric car takes from its battery, two significant figures. The useful energies of cars of similar size are almost equal, $15\,\text{kWh}$ and $13\,\text{kWh}$, because both have to drive the same distance. The difference in the energy taken is due to the efficiency: the electric car uses $90\,\%$ of the energy taken, the petrol car only $25\,\%$, and the remaining $75\,\%$, about $40\,\text{kWh}$, becomes heat.
Reasonableness
The ratio of the efficiencies $0.90 : 0.25 = 3.6$ and the ratio of the useful energies $15.3 : 13.3 = 1.15$ give $3.6 : 1.15 = 3.1$, as they should. The examination covers only the cars' own use of energy, and the losses in the production of electricity and petrol are not included, so the result alone does not decide which is better for the environment.
- Exercise 4-20
In an example measurement, a student followed the changes of energy in an LED lamp. An electricity meter showed the electrical energy $E_\text{o}$ taken by the lamp, and a heat measurement gave the thermal energy $E_\text{h}$ transferred from the lamp to its surroundings. The lamp runs at constant power. The student reasoned: “On the last row $\eta = \dfrac{E_\text{h}}{E_\text{o}} = \dfrac{1510\,\text{J}}{2160\,\text{J}} = 0.70 = 70\,\%$, so the efficiency of the lamp is $70\,\%$.”
Find the student's error and calculate the correct efficiency of the lamp as a percentage. The lamp is on at home for $3.0\,\text{h}$ a day all year round ($365$ days). Using the table, determine the power taken by the lamp and calculate how large an amount of electrical energy becomes heat in a year. Give the answer in kilowatt-hours. Also calculate the cost of the electricity that becomes heat when the price of electricity is $0.20\,\text{€/kWh}$. Explain why the student's $70\,\%$ is still a useful number for this calculation.
Example measurement: electrical energy taken by an LED lamp and energy that became heat $t\,(\text{s})$ $E_\text{o}\,(\text{J})$ $E_\text{h}\,(\text{J})$ 0 0 0 60 540 380 120 1080 750 180 1620 1130 240 2160 1510 Answer
$6.9\,\text{kWh}$
Show the solution
Given
from the table, the electrical energy $E_\text{o}$ taken by the LED lamp and the energy $E_\text{h}$ that became heat; on the last row $t = 240\,\text{s}$, $E_\text{o} = 2160\,\text{J}$ and $E_\text{h} = 1510\,\text{J}$; the student's result $\eta = 70\,\%$; operating time $3.0\,\text{h}$ a day, $365$ days; price $0.20\,\text{€/kWh}$
Find
the student's error, the correct efficiency $\eta$ of the lamp, the electrical energy $E_\text{h,y}$ that becomes heat in a year in kilowatt-hours, and its cost
Principle
Efficiency is the share of the useful energy in the energy taken. The measured heat is loss energy, not useful energy: the useful output of an LED lamp is light. Energy is conserved, so $E_\text{o} = E_\text{a} + E_\text{h}$. Power is $P = E : t$, the energy in a year is the power multiplied by the operating time, and the share of heat in it is the share of the losses in the energy taken.
Reading the graph
The ratio $E_\text{o} : t$ is the same on every row, $540\,\text{J} : 60\,\text{s} = 9.0\,\text{W}$, so the power taken is $9.0\,\text{W}$. The share of the loss is about $0.70$ on every row.
Formula
$E_\text{a} = E_\text{o} - E_\text{h}$, $\eta = \dfrac{E_\text{a}}{E_\text{o}}$, $P = \dfrac{E}{t}$,
$E_\text{o,y}$$= P \cdot t_\text{y}$$E_\text{h,y}$$= \dfrac{E_\text{h}}{E_\text{o}} \cdot E_\text{o,y}$Conversions
The operating time in a year is
$t_\text{y}$$= 3.0\,\text{h} \cdot 365$$= 1095\,\text{h}$The power is $9.0\,\text{W} = 0.0090\,\text{kW}$.
Substitution
The student's error: they divided the loss energy by the energy taken, so $0.70$ is the share of the losses and not the efficiency. The light energy is
$E_\text{a}$$= 2160\,\text{J} - 1510\,\text{J}$$= 650\,\text{J}$and
$\eta$$= \dfrac{650\,\text{J}}{2160\,\text{J}}$$= 0.301$$= 30\,\%$In a year the lamp takes the energy
$E_\text{o,y}$$= 0.0090\,\text{kW} \cdot 1095\,\text{h}$$= 9.855\,\text{kWh}$The share $\dfrac{1510}{2160} = 0.699$ of it becomes heat, that is,
$E_\text{h,y}$$= 0.699 \cdot 9.855\,\text{kWh}$$= 6.89\,\text{kWh}$The cost is $6.89\,\text{kWh} \cdot 0.20\,\text{€/kWh} = 1.38\,\text{€}$.
Answer
The correct efficiency is $30\,\%$. The student's $70\,\%$ is the share of the losses, not the efficiency. Of the electricity taken by the lamp, about $6.9\,\text{kWh}$ becomes heat in a year, and its cost is about $1.4\,\text{€}$. The student's $70\,\%$ is the right number for this calculation, because the question is about the part that becomes heat: $30\,\% + 70\,\% = 100\,\%$.
Reasonableness
Light takes $0.30 \cdot 9.855\,\text{kWh} = 2.96\,\text{kWh}$, and light and heat together are $2.96\,\text{kWh} + 6.89\,\text{kWh} = 9.85\,\text{kWh}$, as they should be. The typical efficiency of an LED lamp is about $30\,\%$. The electricity for the whole lamp costs only $1.97\,\text{€}$ a year, so $1.4\,\text{€}$ is reasonable.