Answer key · Unit 28

Machines at work.

Each lesson’s practice questions, their correct answers and the mistakes they spot. Answers read off the Scene use the same setting the learner sees.

Misconception catalogue

Answer key 6

Lesson 1 of 6 · About 40 minutes

Engines

A petrol engine is a heat engine: four strokes every two turns, an efficiency set by the compression ratio, heat that must leave in the exhaust, and power that grows with speed.

Open the lesson

The 10 steps and their answers

Step 1 of 10
Open this step
Four strokes
Practice · Choose
Question
In which of the four strokes does the gas push the piston and give the crankshaft its work?

Correct answer
- Correct The power stroke, just after the spark, as the hot gas expands
Setup, choices and feedback
The Scene at this step
Engine speed
2000 rpm
Throttle (intake pressure)
60 kPa
Compression ratio
10 : 1
Labels
offdefault
Draw the P–V loop
Off
Choices
Shown in a random order.
- Correct The power stroke, just after the spark, as the hot gas expands
- Wrong The compression stroke, as the piston rises
  Feedback
  On compression it is the other way round: the crankshaft does work on the gas, squeezing it. The engine gets that work back, with a profit, on the power stroke, when the gas is far hotter and pushes far harder.
- Wrong The intake stroke, as the mixture comes in
  Feedback
  On intake the piston moves down and draws mixture in through a part-open throttle, against a pressure below the air’s. That costs the engine a little work rather than giving it any.
- Wrong The exhaust stroke, as the burnt gas leaves
  Feedback
  On exhaust the piston rises and pushes the gas out. That costs work too: with the intake it makes the small pumping loop.
Feedback for a right answer
Only on the power stroke does the gas push the piston. On the other three strokes the crankshaft, helped by the flywheel and the other cylinders, moves the piston.

Hint
Watch the piston after the flash: what pushes what?

Worked solution
1. On compression the piston rises and squeezes the mixture: the crankshaft does work on the gas.
2. At the top the plug sparks, and the pressure jumps to tens of atmospheres.
3. On the power stroke the expanding gas pushes the piston down hard: the gas does work on the crankshaft.
4. On intake and exhaust the piston only draws gas in and pushes it out, which costs a little work.
The work on the power stroke is larger than the work put into compression because the gas is far hotter, so its pressure is far higher at the same volume.
Step 2 of 10
Open this step
Power strokes a second
Practice · Calculate
Question
The Engine speed is fixed at 3000 rpm, three thousand turns of the crankshaft a minute. How many power strokes does this cylinder make each second?

Correct answer
25 Hz ± 0.3 Hz
Setup, choices and feedback
The Scene at this step
Engine speed
3000 rpmfixed
Throttle (intake pressure)
60 kPa
Compression ratio
10 : 1
Labels
offdefault
Draw the P–V loop
Off
Accepted within
± 0.3 Hz
Computed from the Scene at this step’s setting
Power strokes each second in this cylinder: one every two turns/nodes/engine/outputs/powerStrokes
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 50 Hz ± 0.3 Hz
  Misconception: Thinks each cylinder of a four-stroke engine fires once every turn of the crankshaft.
  Feedback from the catalogue
  In a four-stroke engine each cylinder goes through four strokes: intake, compression, power and exhaust. Each stroke takes half a turn of the crankshaft, so a whole cycle takes two turns and has one spark. At $n$ ⁠ revolutions a minute each cylinder makes $\dfrac{n}{120}$ ⁠ power strokes a second, not $\dfrac{n}{60}$ ⁠.
Feedback for a right answer
3000 ÷ 60 = 50 turns a second, and one power stroke every two turns: 25 a second. In a four-cylinder engine that is 100 sparks a second.

Hint
How many turns a second? And how many turns does one four-stroke cycle take?

Worked solution
1. Turns a second: $\dfrac{3000}{60} = 50$ ⁠.
2. Each stroke takes half a turn, so the four strokes take two turns.
3. Power strokes a second: $\dfrac{50}{2} = 25$ ⁠.
If you got 50, you counted a spark every turn, as in a two-stroke engine. In a four-stroke engine half of the turns go on exhaust and intake.
Step 3 of 10
Open this step
Compression ratio
Practice · Predict, then look
Question
The ideal efficiency of the Otto cycle, $\eta = 1 - r^{1-\gamma}$ ⁠ with $\gamma = 1.4$ ⁠, is about 0.60 at a compression ratio of 10. What will it be at a Compression ratio of 14? Give a decimal.

Correct answer
0.652 ± 0.01
A number prediction counts as right within ± 0.01
Setup, choices and feedback
The Scene at this step
Engine speed
2000 rpm
Throttle (intake pressure)
60 kPa
Compression ratio
10 : 1fixed
Labels
offdefault
Draw the P–V loop
On
Once the prediction is locked in, the Scene changes
Compression ratio
14 : 1fixed
Draw the P–V loop
On
The Scene then shows: 0.652 · Ideal (air-standard Otto) efficiency, 1 − r^(1−γ): the share of the heat that becomes work/nodes/engine/outputs/efficiency
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$1 - 14^{-0.4} \approx 0.65$ ⁠. Squeezing harder raises the efficiency, but slowly: from 10 to 14 it gains only five percentage points. Squeezing harder still heats the mixture until it lights by itself before the spark, so petrol engines stop at about 10 to 13.

Worked solution
1. $r = 14$ ⁠, $\gamma - 1 = 0.4$ ⁠.
2. $14^{-0.4} = \dfrac{1}{14^{0.4}} \approx \dfrac{1}{2.87} \approx 0.348$ ⁠.
3. $\eta = 1 - 0.348 = 0.652$ ⁠.
Even the ideal engine, with no friction and no heat lost to the walls, throws about a third of the heat away in the exhaust. A real engine’s efficiency is nearer 0.3.
Step 4 of 10
Open this step
Where the heat goes
Practice · Calculate
Question
In one cycle, with the Throttle (intake pressure) at 60 kPa, the burning adds 653 J of heat to the gas, and the ideal efficiency is 0.602. How much heat leaves with the exhaust each cycle?

Correct answer
259.9 J ± 3 J
Setup, choices and feedback
The Scene at this step
Engine speed
2000 rpm
Throttle (intake pressure)
60 kPafixed
Compression ratio
10 : 1fixed
Labels
offdefault
Draw the P–V loop
Off
Accepted within
± 3 J
Computed from the Scene at this step’s setting
Heat thrown away with the exhaust in one cycle: heat in − the power loop’s work/nodes/engine/outputs/heatOut
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0 J ± 1 J
  Misconception: Thinks a good enough engine could turn all of its heat into work.
  Feedback from the catalogue
  A heat engine takes heat from a hot gas, turns part of it into work, and must throw the rest away to colder surroundings, in the exhaust and the cooling. Even an ideal engine with no friction at all does not turn everything into work. The ideal Otto cycle’s efficiency is $1 - r^{1-\gamma}$ ⁠, and a real engine’s is lower still.
Feedback for a right answer
About 260 J: around 40% of the heat leaves in the hot exhaust, even in an ideal engine.

Hint
Conservation of energy: the heat in is the work plus the heat out.

Worked solution
1. The power loop’s work: $W = \eta Q_{\text{in}} = 0.602 \times 653 \approx 393$ ⁠ J.
2. Conservation of energy: $Q_{\text{in}} = W + Q_{\text{out}}$ ⁠.
3. $Q_{\text{out}} = 653 - 393 = 260$ ⁠ J.
If you wrote 0, you assumed a good enough engine turns all its heat into work. A heat engine must give heat out to colder surroundings to get back to the start of its cycle.
Step 5 of 10
Open this step
Power
Practice · Predict, then look
Question
At an Engine speed of 3000 rpm with the throttle wide open the cylinder gives about 16 kW. The work in each cycle does not depend on the speed. Predict the power at 6000 rpm, in kilowatts.

Correct answer
32.61 kW ± 5%
A number prediction counts as right within ± 5%
Setup, choices and feedback
The Scene at this step
Engine speed
3000 rpmfixed
Throttle (intake pressure)
100 kPafixed
Compression ratio
10 : 1
Labels
offdefault
Draw the P–V loop
On
Once the prediction is locked in, the Scene changes
Engine speed
6000 rpmfixed
The Scene then shows: 32.61 kW · This cylinder’s power: net work × power strokes a second/nodes/engine/outputs/power
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
About 33 kW: twice as many power strokes a second, each with the same work, give twice the power. That is why an engine is strongest at high revs, until the valves can no longer fill the cylinder in time.

Worked solution
1. Power is the work in each cycle times the cycles a second: $P = W \cdot \dfrac{n}{120}$ ⁠.
2. At 3000 rpm: 25 cycles a second, and the work is about 652 J, so about 16.3 kW.
3. At 6000 rpm: 50 cycles a second with the same work, $652 \times 50 \approx 32\,600$ ⁠ W.
4. About 33 kW, twice as much.
Step 6 of 10
Open this step
The cylinder
Practice · Calculate
Question
The cylinder is 86 mm across inside, and the piston travels 86 mm each stroke. What volume does the piston sweep each stroke, in cubic centimetres?

Correct answer
499.6 cm³ ± 2%
Setup, choices and feedback
The Scene at this step
Engine speed
2000 rpm
Throttle (intake pressure)
60 kPa
Compression ratio
10 : 1
Labels
offdefault
Draw the P–V loop
Off
Accepted within
± 2%
Computed from the Scene at this step’s setting
Volume the piston sweeps in one stroke/nodes/engine/outputs/displacement
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 1998 cm³ ± 2%
  Misconception: Puts the rotor's diameter into πr², as if it were the radius.
  Feedback from the catalogue
  The area of a circle is $\pi r^2$ ⁠, and $r$ ⁠ is the radius, half the diameter. Put the diameter in place of the radius and the area comes out four times too large, and the power with it.
Feedback for a right answer
About 500 cm³, half a litre: four of these make a two-litre engine.

Hint
The volume is the area of the piston’s circle times the distance it travels. The radius is half the diameter.

Worked solution
1. The radius: 4.3 cm. The piston’s area: $\pi \times 4.3^2 \approx 58.1$ ⁠ square centimetres.
2. The volume: $58.1 \times 8.6 \approx 500$ ⁠ cubic centimetres, half a litre.
3. In an engine with four such cylinders: 2000 cm³, a two-litre engine.
If you got about 2000 cm³ for one cylinder, you put the diameter in place of the radius.
Step 7 of 10
Open this step
Opening the throttle
Practice · Calculate
Question
With the Throttle (intake pressure) at 60 kPa, the burning adds 653 J of heat to the gas each cycle. The throttle is opened wide, to 100 kPa. How much heat does the burning add each cycle now?

Correct answer
1088 J ± 2%
Setup, choices and feedback
The Scene at this step
Engine speed
3000 rpm
Throttle (intake pressure)
100 kPafixed
Compression ratio
10 : 1fixed
Labels
offdefault
Draw the P–V loop
On
Accepted within
± 2%
Computed from the Scene at this step’s setting
Heat the burning charge adds in one cycle/nodes/engine/outputs/heatIn
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 653 J ± 2%
  Misconception: Thinks opening the throttle only makes the engine turn faster, without changing what each cycle burns.
  Feedback from the catalogue
  The throttle sets how much air gets into the cylinder each cycle, and the computer injects fuel in a fixed ratio to the air. So with the throttle open each cycle burns more fuel and gives more work, even at the same engine speed.
Feedback for a right answer
About 1088 J: $\dfrac{100}{60}$ ⁠ times as much, because $\dfrac{100}{60}$ ⁠ times as much air and fuel get in.

Hint
The throttle sets how much air, and with it how much fuel, gets into the cylinder each cycle. At a fixed volume, the amount of gas goes with the pressure.

Worked solution
1. The cylinder fills the same volume each cycle, so the amount of air goes with the intake pressure.
2. Fuel is injected in a fixed ratio to the air, so the heat goes with the pressure too.
3. $653 \times \dfrac{100}{60} \approx 1088$ ⁠ joules.
The throttle does not feed in “more power” directly: it lets the engine breathe more air.
Step 8 of 10
Open this step
The work it gives
Practice · Calculate
Question
The throttle is wide open: the burning adds 1088 J of heat each cycle, and the Otto cycle’s efficiency at a Compression ratio of $10$ ⁠ is $0.60$ ⁠. How much work does the cylinder give each cycle?

Correct answer
652.3 J ± 2%
Setup, choices and feedback
The Scene at this step
Engine speed
3000 rpm
Throttle (intake pressure)
100 kPafixed
Compression ratio
10 : 1fixed
Labels
offdefault
Draw the P–V loop
On
Accepted within
± 2%
Computed from the Scene at this step’s setting
Net work in one cycle: the power loop less the pumping loop/nodes/engine/outputs/work
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 1088 J ± 2%
  Misconception: Thinks a good enough engine could turn all of its heat into work.
  Feedback from the catalogue
  A heat engine takes heat from a hot gas, turns part of it into work, and must throw the rest away to colder surroundings, in the exhaust and the cooling. Even an ideal engine with no friction at all does not turn everything into work. The ideal Otto cycle’s efficiency is $1 - r^{1-\gamma}$ ⁠, and a real engine’s is lower still.
Feedback for a right answer
About 652 J: less than two thirds of the heat. The rest leaves hot down the exhaust.

Hint
The efficiency is the share of the heat that becomes work.

Worked solution
1. $W = \eta \, Q_{\text{in}} = 0.60 \times 1088 \approx 653$ ⁠ joules.
2. The rest, about 435 J, leaves with the burnt gases.
3. At 3000 rpm there are $25$ ⁠ cycles a second: $653 \times 25 \approx 16$ ⁠ kilowatts.
With the throttle partly closed, the engine loses a little more work sucking air in past it. Wide open, that loss almost vanishes.
Step 9 of 10
Open this step
After the spark
Practice · Predict, then look
Question
Throttle (intake pressure) at 60 kPa. Just after the spark the gas reaches about 3300 K. Predict: what will that temperature be when the throttle is opened wide, to 100 kPa?

Correct answer
- Correct Almost the same temperature
Setup, choices and feedback
The Scene at this step
Engine speed
3000 rpm
Throttle (intake pressure)
60 kPa
Compression ratio
10 : 1fixed
Labels
offdefault
Draw the P–V loop
On
Choices
- Wrong Higher, about 5500 K, because more fuel burns
  Misconception: Thinks burning more fuel in a cylinder always makes the gas hotter, even when the air grows in step.
  Feedback from the catalogue
  The temperature depends on the heat each gram of gas gets, not the total. When the throttle opens, more fuel and more air get in in the same ratio, so each gram gets the same heat and reaches the same temperature. What grows is the pressure.
- Correct Almost the same temperature
- Wrong Lower, because more air cools the gas
  No feedback of its own; the learner sees the standard message
  Not yet. Look again and try once more.
Once the prediction is locked in, the Scene changes
Throttle (intake pressure)
100 kPa
The Scene then shows: 3313 K · Gas temperature just after the spark/nodes/engine/outputs/peakTemperature
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The same temperature, about 3300 K: the heat and the gas both grew in the same ratio. The pressure is what grew.

Hint
When more fuel gets in, more air gets in to be heated too. How much heat does each gram of gas get?

Worked solution
1. With the throttle open $\dfrac{100}{60}$ ⁠ times as much air gets in, and with it $\dfrac{100}{60}$ ⁠ times as much fuel.
2. The heat grows $\dfrac{100}{60}$ ⁠ times, but so does the amount of gas it heats.
3. Each gram of gas gets the same heat, so it reaches the same temperature, about 3300 K.
What does grow is the pressure: more gas in the same volume at the same temperature presses harder, so the piston is pushed harder.
Step 10 of 10
Open this step
The highest pressure
Practice · Calculate
Question
With the Throttle (intake pressure) at 60 kPa, the pressure in the cylinder just after the spark is about 6.2 MPa, more than sixty times the air’s. The throttle is opened wide, to 100 kPa, and the temperature after the spark does not change. What is the pressure now?

Correct answer
10.35 MPa ± 2%
Setup, choices and feedback
The Scene at this step
Engine speed
3000 rpm
Throttle (intake pressure)
100 kPafixed
Compression ratio
10 : 1fixed
Labels
offdefault
Draw the P–V loop
On
Accepted within
± 2%
Computed from the Scene at this step’s setting
Pressure just after the spark, at top dead centre/nodes/engine/outputs/peakPressure
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
About 10.4 MPa, more than a hundred times the air pressure outside.

Hint
At the same volume and temperature, the pressure goes with the amount of gas.

Worked solution
1. With the throttle open $\dfrac{100}{60}$ ⁠ times as much gas gets in.
2. At the same volume and temperature, $p = \dfrac{nRT}{V}$ ⁠ grows in the same ratio.
3. $6.2 \times \dfrac{100}{60} \approx 10.4$ ⁠ megapascals.
A higher pressure pushes the piston harder all along the power stroke, so the work each cycle grows.

Lesson 2 of 6 · About 35 minutes

Refrigerators and heat pumps

A fridge does not make cold, it moves heat from the box to the room. How much heat it moves for each joule of electricity, why a freezer costs more, and how the same loop heats a house.

Open the lesson

The 9 steps and their answers

Step 1 of 9
Open this step
A heat pump
Practice · Choose
Question
Why is the black grille on the back of the fridge warm while it runs?

Correct answer
- Correct It gives the room the heat taken out of the food, plus the compressor’s energy
Setup, choices and feedback
The Scene at this step
Inside temperature
277.1 K
Room temperature
295.1 K
Show the heat moved
0.4 (Fridge → Heat moved)
Choices
Shown in a random order.
- Correct It gives the room the heat taken out of the food, plus the compressor’s energy
- Wrong The fridge makes cold and puts it in the box, and the motor that makes it warms up
  Misconception: Thinks a fridge makes cold and puts it in the box, rather than pumping heat out of it.
  Feedback from the catalogue
  There is no such stuff as “cold”. A fridge takes heat out of the box and dumps it into the room: the refrigerant boils inside at a low temperature, soaking up heat, and the compressor squeezes it so that it condenses in the grille on the back and gives the heat out. That is why the back is warm: it gives out the heat from the food plus the compressor’s electrical energy.
- Wrong Heat from the room collects there because the insulation stops it
  Feedback
  The insulation slows heat coming in, but it does not warm the back. The grille is warmer than the room: it has to be, to give heat out to it.
Feedback for a right answer
The fridge is a heat pump: the refrigerant boils in the box, soaking up heat from the food, and the compressor squeezes it so that it condenses in the grille and gives the heat to the room. The compressor’s work comes out of the grille too.

Hint
Drag Show the heat moved and watch the refrigerant: where does it boil and where does it condense?

Worked solution
1. In the box the refrigerant boils at low pressure, at a temperature below the food’s, so heat flows from the food into it.
2. The compressor squeezes the vapour to high pressure, where it condenses at a temperature above the room’s.
3. In the grille the vapour condenses, giving the room the heat it soaked up plus the energy the compressor put in.
4. So the back is warm, and warmer the more heat the fridge moves.
“Cold” is not something you can put in. A fridge only takes heat out.
Step 2 of 9
Open this step
Coefficient of performance
Practice · Calculate
Question
The refrigerant boils in the box at 267.15 K and condenses in the grille at 307.15 K. The ideal (Carnot) cooling coefficient of performance is $\dfrac{T_c}{T_h - T_c}$ ⁠, and a household fridge reaches about 40% of it. What is this fridge’s COP, the joules of heat it takes from the box for each joule of electricity?

Correct answer
2.671 ± 0.05
Setup, choices and feedback
The Scene at this step
Inside temperature
277.1 Kfixed
Room temperature
295.1 Kfixed
Show the heat moved
0.4 (Fridge → Heat moved)
Accepted within
± 0.05
Computed from the Scene at this step’s setting
Cooling coefficient of performance: heat drawn out ÷ electricity in/nodes/fridge/outputs/cop
Feedback for a right answer
About 2.7: for each joule of electricity the fridge takes nearly three joules of heat out of the box. That does not break conservation of energy, because the electricity is not turned into cold: it moves heat.

Hint
$T_h - T_c = 40$ ⁠ kelvin.

Worked solution
1. $T_c = 267.15$ ⁠ K, $T_h = 307.15$ ⁠ K, $T_h - T_c = 40$ ⁠ K.
2. Carnot COP: $\dfrac{267.15}{40} \approx 6.68$ ⁠.
3. The fridge reaches 40%: $0.4 \times 6.68 \approx 2.67$ ⁠.
Work in kelvin. In Celsius $T_c$ ⁠ would come out negative, a sign that the formula needs absolute temperature.
Step 3 of 9
Open this step
A colder box
Practice · Predict, then look
Question
Turn the Inside temperature down from 277.15 K (fridge) to 255.15 K (freezer), with the room unchanged. What will happen to the coefficient of performance?

Correct answer
- Correct It will fall: each joule of electricity moves less heat
Setup, choices and feedback
The Scene at this step
Inside temperature
277.1 Kfixed
Room temperature
295.1 K
Show the heat moved
0.4 (Fridge → Heat moved)
Choices
- Correct It will fall: each joule of electricity moves less heat
- Wrong It will rise: something already cold is easier to cool
  Feedback
  The reverse: the colder the box, the further the pump must lift the heat, from a lower temperature to the same room, and that costs more work for each joule.
- Wrong It will not change: it is the same compressor
  Feedback
  The compressor is the same, but the COP depends on the temperature gap it has to bridge, not only on the machine.
Once the prediction is locked in, the Scene changes
Inside temperature
255.2 Kfixed
The Scene then shows: 1.582 · Cooling coefficient of performance: heat drawn out ÷ electricity in/nodes/fridge/outputs/cop
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The COP falls from about 2.7 to about 1.6. The refrigerant has to boil at 245 K instead of 267 K, and the gap to the grille grows from 40 to 62 kelvin. That is why a freezer costs more to run.

Worked solution
1. The refrigerant boils about 10 kelvin below the inside: now $T_c = 245.15$ ⁠ K.
2. The grille has not changed: $T_h = 307.15$ ⁠ K, and the gap is 62 K.
3. Carnot: $\dfrac{245.15}{62} \approx 3.95$ ⁠, and 40% of that is about 1.58.
4. The COP has fallen, and more heat leaks in through the walls as well, so the freezer runs for longer.
Step 4 of 9
Open this step
Heating a house
Practice · Calculate
Question
Turn the same loop into a heat pump that warms a house: the cold box outside, the hot grille indoors. The cooling COP is 2.67. How many joules of heat come out indoors for each joule of electricity?

Correct answer
3.671 ± 0.05
Setup, choices and feedback
The Scene at this step
Inside temperature
277.1 Kfixed
Room temperature
295.1 Kfixed
Show the heat moved
0.4 (Fridge → Heat moved)
Accepted within
± 0.05
Computed from the Scene at this step’s setting
The same loop used as a heat pump: heat given out ÷ electricity in, one more than the cooling COP/nodes/fridge/outputs/heatingCop
Wrong answers it recognises
- 1 ± 0.05
  Misconception: Thinks no heater can give out more heat than the electricity it uses, so a heat pump's COP is at most 1.
  Feedback from the catalogue
  An electric heater turns each joule of electricity into one joule of heat, and no more. A heat pump does not make its heat, it moves it: it takes heat from the cold air outside and adds the compressor’s work. So it gives the house more heat than the electricity it uses, three or four times as much, and energy is still conserved: $Q_{\text{hot}} = Q_{\text{cold}} + W$ ⁠.
Feedback for a right answer
3.67: the house gets the heat taken from outside plus the electricity. An electric heater gives one joule for each joule, so the heat pump saves about two thirds of the electricity.

Hint
For each joule of electricity: how much heat is taken from outside, and how much comes out indoors?

Worked solution
1. For each joule of electricity, $W = 1$ ⁠, the cold side gives up $Q_{\text{cold}} = 2.67$ ⁠ J.
2. Conservation of energy: $Q_{\text{hot}} = Q_{\text{cold}} + W = 2.67 + 1 = 3.67$ ⁠ J.
3. The heating COP: $\dfrac{Q_{\text{hot}}}{W} = 3.67$ ⁠.
A COP above 1 breaks no law: the heat pump does not make heat, it moves it from outside in.
Step 5 of 9
Open this step
A heatwave
Practice · Predict, then look
Question
Raise the Room temperature from 295.15 K (kitchen) to 311.15 K (heatwave). What will happen to the share of the time the compressor runs?

Correct answer
- Correct It will grow a lot, more than double
Setup, choices and feedback
The Scene at this step
Inside temperature
277.1 K
Room temperature
295.1 Kfixed
Show the heat moved
0.4 (Fridge → Heat moved)
Choices
- Correct It will grow a lot, more than double
- Wrong It will grow a little, in step with the temperature difference
  Feedback
  More heat leaks in, and each joule of electricity also moves less heat, because the grille has to run hotter. The two effects add up.
- Wrong It will not change: the inside temperature has not changed
  Feedback
  The inside temperature is the same, but to hold it the fridge must take out all the heat that leaks through the walls, and a hot room lets in more.
Once the prediction is locked in, the Scene changes
Room temperature
311.1 Kfixed
Show the heat moved
Heat moved
The Scene then shows: 0.3818 · Share of the time it runs to hold the inside temperature, at most 1/nodes/fridge/outputs/duty
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
From about 14% of the time to about 38%, more than two and a half times. The leak grows from 27 W to 51 W, and the COP falls from 2.7 to 1.9.

Worked solution
1. 1.5 W leaks through the walls for each kelvin of difference: $1.5 \times 18 = 27$ ⁠ W in the kitchen, $1.5 \times 34 = 51$ ⁠ W in the heatwave.
2. The grille has to condense at 323 K instead of 307 K, so the COP falls to 1.9.
3. While it runs the fridge takes out $1.9 \times 70 \approx 134$ ⁠ W instead of 187 W.
4. The share of the time: $\dfrac{51}{134} \approx 0.38$ ⁠ instead of $\dfrac{27}{187} \approx 0.14$ ⁠.
Step 6 of 9
Open this step
A freezer
Practice · Calculate
Question
Inside temperature is locked at Freezer, 255.15 K, and the room is the Kitchen. The refrigerant now boils at 245.15 K and condenses at 307.15 K. What is the ideal, Carnot, coefficient of performance, $\dfrac{T_c}{T_h - T_c}$ ⁠?

Correct answer
3.954 ± 2%
Setup, choices and feedback
The Scene at this step
Inside temperature
255.2 Kfixed
Room temperature
295.1 Kfixed
Show the heat moved
0.4 (Fridge → Heat moved)
Accepted within
± 2%
Computed from the Scene at this step’s setting
The ideal (Carnot) cooling COP between the two refrigerant temperatures, T_cold ÷ (T_hot − T_cold)/nodes/fridge/outputs/carnotCop
Wrong answers it recognises
- 6.679 ± 2%
  Misconception: Thinks a fridge’s coefficient of performance is fixed by the machine, whatever the temperatures it works between.
  Feedback from the catalogue
  The coefficient of performance depends on the temperatures the loop works between: $\dfrac{T_c}{T_h - T_c}$ ⁠. The colder the box or the hotter the room, the wider the gap, and the less heat each joule of electricity moves. The same machine is less efficient as a freezer.
Feedback for a right answer
About $3.95$ ⁠, against $6.68$ ⁠ for the fridge: a wider gap, less heat for each joule.

Hint
Replace $T_c$ ⁠ with the new boiling temperature. $T_h$ ⁠ has not changed.

Worked solution
1. $T_h - T_c = 307.15 - 245.15 = 62$ ⁠ kelvin.
2. $\dfrac{245.15}{62} \approx 3.95$ ⁠.
In an ordinary fridge, at 267.15 K, the ratio was $6.68$ ⁠. The wider the gap, the less heat each joule of electricity moves, which is why a freezer uses more electricity than a fridge.
Step 7 of 9
Open this step
The electricity
Practice · Calculate
Question
In the kitchen, while the compressor runs, the loop draws 187 J of heat out of the fridge box each second, and the coefficient of performance is $2.67$ ⁠. How much electricity, in watts, does the compressor draw while it runs?

Correct answer
70 W ± 2%
Setup, choices and feedback
The Scene at this step
Inside temperature
277.1 Kfixed
Room temperature
295.1 Kfixed
Show the heat moved
Heat moved
Accepted within
± 2%
Computed from the Scene at this step’s setting
Electricity the compressor draws while it runs/nodes/fridge/outputs/electric
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 499 W ± 2%
  Misconception: Multiplies the heat moved by the COP to get the electricity, turning the ratio upside down.
  Feedback from the catalogue
  The coefficient of performance is the heat taken out divided by the electricity put in. So the electricity is the heat divided by the COP. A COP of $2.67$ ⁠ means each joule of electricity takes out $2.67$ ⁠ joules of heat: the electricity is less than the heat.
Feedback for a right answer
About 70 W, like an old light bulb: and it moves 2.67 times as much heat.

Hint
The coefficient of performance is the heat taken out over the electricity put in.

Worked solution
1. $\text{COP} = \dfrac{Q_c}{W}$ ⁠, so $W = \dfrac{Q_c}{\text{COP}}$ ⁠.
2. $W = \dfrac{187}{2.67} \approx 70$ ⁠ watts.
3. Into the room go $187 + 70 = 257$ ⁠ watts: the heat from the box plus the electricity.
If you multiplied $187 \times 2.67$ ⁠, you turned the ratio upside down. A good fridge moves more heat than the electricity it uses.
Step 8 of 9
Open this step
Through the walls
Practice · Calculate
Question
In the kitchen, at 295.15 K, 27 W of heat get into the box, at 277.15 K, through the walls and door. The Room temperature rises to a Heatwave, 311.15 K. How much heat will get in each second now?

Correct answer
51 W ± 3%
Setup, choices and feedback
The Scene at this step
Inside temperature
277.1 Kfixed
Room temperature
311.1 Kfixed
Show the heat moved
Heat moved
Accepted within
± 3%
Computed from the Scene at this step’s setting
Heat leaking in through the walls and door/nodes/fridge/outputs/leak
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 28.5 W ± 3%
  Misconception: Scales heat flow through a wall by the absolute temperature, not the temperature difference across it.
  Feedback from the catalogue
  Heat flows from hot to cold at a rate that goes with the difference in temperature across the wall. With no difference there is no flow, even if both sides are very hot. A difference of 34 kelvin instead of 18 almost doubles the flow.
Feedback for a right answer
51 W: almost twice as much, because the difference almost doubled. That is why the compressor works harder in summer.

Hint
Heat flows through the insulation at a rate that goes with the difference in temperature between the room and the box.

Worked solution
1. In the kitchen the difference is $295.15 - 277.15 = 18$ ⁠ kelvin, and $\dfrac{27}{18} = 1.5$ ⁠ watts per kelvin.
2. In the heatwave the difference is $311.15 - 277.15 = 34$ ⁠ kelvin.
3. $1.5 \times 34 = 51$ ⁠ watts.
If you answered about 28.5 W, you scaled by the ratio of the absolute temperatures. The flow depends only on the difference.
Step 9 of 9
Open this step
A day of cooling
Practice · Calculate
Question
In the kitchen the compressor runs only 14.4% of the time, and while it runs it draws 70 W. How much electrical energy, in kilojoules, does the fridge use in a day?

Correct answer
873.2 kJ ± 3%
Setup, choices and feedback
The Scene at this step
Inside temperature
277.1 Kfixed
Room temperature
295.1 Kfixed
Show the heat moved
Heat moved
Accepted within
± 3%
Computed from the Scene at this step’s setting
Electricity used in a day: the share of time it runs × its power × 24 h/nodes/fridge/outputs/energyPerDay
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 6048 kJ ± 3%
  Misconception: Thinks a fridge’s compressor runs all the time.
  Feedback from the catalogue
  The compressor switches on only when the box has warmed a little above its set temperature, and off once it has cooled again. All it has to take out is the heat leaking in, so in the kitchen it runs only about seven minutes in fifty.
Feedback for a right answer
About 873 kJ, a quarter of a kilowatt-hour a day.

Hint
A day has 86,400 seconds. For how many of them does the compressor run?

Worked solution
1. Running time in a day: $0.144 \times 86\,400 \approx 12\,500$ ⁠ seconds, about three and a half hours.
2. The energy: $70 \times 12\,500 \approx 873\,000$ ⁠ joules, about 873 kJ.
3. In kilowatt-hours, $1\ \mathrm{kWh} = 3600$ ⁠ kilojoules: about $0.24$ ⁠ a day.
If you answered 6048 kJ, you assumed the compressor runs all day. It rests most of the time, switching itself on only when the box warms up.

Lesson 3 of 6 · About 20 minutes

Hydraulics

A hydraulic lift raises a car with a hand's push: the pressure is the same throughout the oil, the force grows with the area, and the distance shrinks by the same ratio, so the work is conserved.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
The same pressure
Practice · Choose
Question
The handle pushes the small piston, and the oil lifts the car on the big piston. Where is the pressure in the oil greater?

Correct answer
- Correct It is the same throughout the oil
Setup, choices and feedback
The Scene at this step
Piston area ratio
40 to 1
Push on the handle
350 N
Show the pressure inside
Off
Choices
Shown in a random order.
- Correct It is the same throughout the oil
- Wrong Under the big piston, where the force is larger
  Misconception: Thinks the pressure is larger under the big piston, because the force there is larger.
  Feedback from the catalogue
  In a closed liquid that is not flowing the pressure is the same everywhere at the same height: Pascal’s law. The force on the big piston is larger only because its area is larger, $F = pA$ ⁠, not because the pressure there is higher.
- Wrong Under the small piston, where the push is
  Feedback
  The push makes the pressure, but the oil passes it on without losing any: as long as it is not flowing, the pressure is the same everywhere.
Feedback for a right answer
Pascal’s law: a pressure applied to a closed liquid passes undiminished to every part of it. The big piston gets a larger force only because of its larger area.

Hint
Choose On under Show the pressure inside, and compare the arrows on both sides.

Worked solution
1. A liquid barely compresses, so a push at one end reaches all the oil at once.
2. While the oil is not flowing, its pressure is the same at every point at the same height.
3. The force on each piston is the pressure times its area, $F = pA$ ⁠.
4. So the pressure is the same, and the force on the big piston is larger by the ratio of the areas.
Step 2 of 6
Open this step
The pressure in the oil
Practice · Calculate
Question
The Push on the handle is 350 N, and all of it reaches the small piston. The big piston’s area is 0.1 m², and the Piston area ratio is 40 to 1. What pressure does the push make in the oil?

Correct answer
140 kPa ± 1%
Setup, choices and feedback
The Scene at this step
Piston area ratio
40 to 1fixed
Push on the handle
350 Nfixed
Show the pressure inside
Off
Accepted within
± 1%
Computed from the Scene at this step’s setting
The pressure the push makes in the oil: push ÷ the small piston's area/nodes/lift/outputs/oilPressure
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
140 kPa, a little more than the pressure of the atmosphere. That pressure reaches all of the oil, the big piston too.

Hint
The small piston’s area is $\dfrac{0.1}{40}$ ⁠ square metres, and pressure is $p = \dfrac{F}{A}$ ⁠.

Worked solution
1. The small piston’s area: $A = \dfrac{0.1}{40} = 0.0025$ ⁠ m².
2. $p = \dfrac{F}{A} = \dfrac{350}{0.0025} = 140\,000$ ⁠ Pa.
3. That is 140 kPa.
If you got 3.5 kPa, you divided by the big area. The hand’s push acts on the small piston.
Step 3 of 6
Open this step
The force on the ram
Practice · Calculate
Question
The same push, 350 N, and the same Piston area ratio, 40 to 1. With what force does the oil push the big piston up?

Correct answer
14000 N ± 1%
Setup, choices and feedback
The Scene at this step
Piston area ratio
40 to 1fixed
Push on the handle
350 Nfixed
Show the pressure inside
Off
Accepted within
± 1%
Computed from the Scene at this step’s setting
The oil's push on the big piston while the hand pushes: push × ratio/nodes/lift/outputs/liftForce
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
14,000 N, more than the car’s weight of 11,800 N, so each push lifts it a little.

Hint
$F = pA$ ⁠, and the area is 40 times larger.

Worked solution
1. The pressure in the oil: 140,000 Pa.
2. The force on the big piston: $F = pA = 140\,000 \times 0.1 = 14\,000$ ⁠ N.
3. Or in short: $350 \times 40 = 14\,000$ ⁠ N; the force ratio equals the area ratio.
Step 4 of 6
Open this step
One hand
Practice · Reach the target
Question
The Push on the handle is fixed at 150 N, one hand. Change the Piston area ratio until that push lifts the car, then press Check.

Goal
- 2 ± 0.1
  At the start of the step: 1
Setup, choices and feedback
The Scene at this step
Piston area ratio
40 to 1
Push on the handle
150 Nfixed
Show the pressure inside
Off
Read from the Scene at
0 no push, 1 too small to lift the car, 2 lifts it/nodes/lift/outputs/regime
One setting that reaches it
Piston area ratio
79 to 1
Feedback for a right answer
From a ratio of 79 up, one hand’s push lifts the car. But each push lifts it only about half a centimetre.

Feedback for a wrong answer
Not yet: the force on the big piston is still less than the car’s weight. Raise the ratio.

Hint
The car weighs 11,772 N. How many times must 150 N be multiplied?

Worked solution
1. The weight of the car and platform: $1200 \times 9.81 = 11\,772$ ⁠ N.
2. You need $150 \times r \ge 11\,772$ ⁠.
3. $r \ge \dfrac{11\,772}{150} \approx 78.5$ ⁠, so a ratio of at least 79.
At a ratio of 79 each 40 cm push lifts the car only $\dfrac{40}{79} \approx 0.5$ ⁠ cm.
Step 5 of 6
Open this step
No free work
Practice · Calculate
Question
At a Piston area ratio of 40 to 1 each push moves the small piston 40 cm. How far does the car rise in one push?

Correct answer
1 cm ± 0.05 cm
Setup, choices and feedback
The Scene at this step
Piston area ratio
40 to 1fixed
Push on the handle
350 N
Show the pressure inside
Off
Accepted within
± 0.05 cm
Computed from the Scene at this step’s setting
How far one stroke lifts the car/nodes/lift/outputs/rise
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 40 cm ± 0.5 cm
  Misconception: Thinks a hydraulic lift multiplies the work done as well as the force.
  Feedback from the catalogue
  The lift multiplies the force, not the energy. The oil the small piston pushes fills only a thin layer under the big piston, so when the force grows 40 times the distance shrinks 40 times. The work, force times distance, is the same on both sides, exactly as with a lever.
Feedback for a right answer
One centimetre. The force grows 40 times and the distance shrinks 40 times: the hand’s work, $294 \times 0.4 \approx 118$ ⁠ J, equals the work on the car, $11\,772 \times 0.01 \approx 118$ ⁠ J.

Hint
The volume of oil that leaves the small cylinder enters the big one.

Worked solution
1. The volume of oil pushed: $A_{\text{small}} \times 0.4$ ⁠.
2. The same volume fills a layer under the big piston: $A_{\text{big}} \times h$ ⁠.
3. $h = 0.4 \times \dfrac{A_{\text{small}}}{A_{\text{big}}} = \dfrac{0.4}{40} = 0.01$ ⁠ m, one centimetre.
4. Check with work: the push that balances the car is $\dfrac{11\,772}{40} \approx 294$ ⁠ N, and $294 \times 0.4 \approx 118$ ⁠ J, exactly $11\,772 \times 0.01$ ⁠.
If you wrote 40 cm, you assumed the lift multiplies the force and keeps the distance. That would make energy out of nothing.
Step 6 of 6
Open this step
Holding the car up
Practice · Calculate
Question
The car has a mass of 1200 kg, and the big piston’s area is 0.1 m². What pressure in the oil holds the car up? Take g = 9.81 m/s².

Correct answer
117.7 kPa ± 2%
Setup, choices and feedback
The Scene at this step
Piston area ratio
40 to 1fixed
Push on the handle
350 Nfixed
Show the pressure inside
On
Accepted within
± 2%
Computed from the Scene at this step’s setting
The pressure that holds the car up: its weight ÷ the big piston's area/nodes/lift/outputs/holdPressure
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 12 kPa ± 2%
  Misconception: Uses a mass in kilograms as if it were a weight in newtons, leaving out g.
  Feedback from the catalogue
  Mass and weight are different things. Mass is measured in kilograms, and weight is a force, in newtons: $W = mg$ ⁠, with $g \approx 9.8$ ⁠ newtons per kilogram. A 40 t lorry, 40,000 kg, weighs about 392,000 N.
Feedback for a right answer
About 118 kPa: less than the pressure in the same car’s tyres.

Hint
Pressure is force over area. The force is the car’s weight, not its mass.

Worked solution
1. The car’s weight: $1200 \times 9.81 = 11\,772$ ⁠ newtons.
2. $p = \dfrac{F}{A} = \dfrac{11\,772}{0.1} = 117\,720$ ⁠ pascals.
3. About 118 kPa, a little more than the air’s pressure.
Any pressure in the oil above this lifts the car. So the push on the handle, times the area ratio, has to be more than the car’s weight.

Lesson 4 of 6 · About 25 minutes

Flight

A wing lifts a plane because it turns air downward. How much lift each square metre gives, why it goes with the square of the speed, and what happens at too steep an angle.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
Turning the air
Practice · Choose
Question
What makes the wing lift?

Correct answer
- Correct The wing turns the airstream downward, and the air pushes it upward
Setup, choices and feedback
The Scene at this step
Angle of attack
5 °
Airspeed
50 m/s
Show the pressure
Off
Choices
Shown in a random order.
- Correct The wing turns the airstream downward, and the air pushes it upward
- Wrong Air over the wing must meet the air from underneath at the back, so it speeds up
  Misconception: Thinks air over the top of a wing must meet the air from underneath at the back edge at the same moment, and that is why the wing lifts.
  Feedback from the catalogue
  No rule makes the air meet again at the back. In experiments, and in the smoke in the Scene, the air over the wing reaches the back edge well before the air underneath. The wing lifts because it turns the airstream downward: by Newton’s third law the air pushes the wing up. The same force shows in the pressure: low above the wing, high below it.
- Wrong Air hits the underside and bounces off it, like a ball off a wall
  Feedback
  The underside does turn some air, but most of the lift comes from the top: air follows the curved upper surface and leaves heading down, and the pressure there is low. A wing that worked only from below would give less than half the lift.
Feedback for a right answer
The wing turns air downward, and by Newton’s third law the air pushes it upward. Show the pressure shows the same force: low pressure above the wing, high below it.

Hint
Watch the smoke lines behind the wing: where do they go?

Worked solution
1. In the smoke the air leaving the wing moves slightly downward.
2. To change the air’s motion, the wing pushes down on it.
3. By the third law, the air pushes on the wing with an equal and opposite force: lift, upward.
4. In pressure it looks like this: above the wing the pressure is below the air’s, and underneath it is above.
The “equal transit time” explanation is common in books, but wrong: in the smoke the upper air reaches the back edge before the lower air.
Step 2 of 5
Open this step
Lift on a square metre
Practice · Calculate
Question
An Airspeed of 50 m/s and an Angle of attack of 5°, where the wing’s lift coefficient is $C_L \approx 0.705$ ⁠. The air’s density is 1.225 kg/m³. What is the lift on each square metre of wing? (Newtons per square metre are pascals.)

Correct answer
1080 Pa ± 2%
Setup, choices and feedback
The Scene at this step
Angle of attack
5 °fixed
Airspeed
50 m/sfixed
Show the pressure
Off
Accepted within
± 2%
Computed from the Scene at this step’s setting
Wing: lift on each square metre of wing, ½ρv²·C_L (N/m², the same unit as Pa)/nodes/wing/outputs/lift
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 21.6 Pa ± 3%
  Misconception: Thinks lift grows in proportion to airspeed, so half the speed gives half the lift.
  Feedback from the catalogue
  Lift goes with the square of the speed: $L = \tfrac12 \rho v^2 C_L A$ ⁠. At twice the speed each kilogram of air meets the wing twice as fast, and twice as many kilograms pass each second, so the lift grows four times. At half the speed it falls to a quarter, which is why a plane lands at a high angle of attack with its flaps out.
Feedback for a right answer
About 1080 N on each square metre: enough to hold up about 110 kg. A light aircraft’s wing, about 16 m², carries over a tonne and a half at this speed.

Hint
$L = \tfrac12 \rho v^2 C_L$ ⁠ for each square metre.

Worked solution
1. The dynamic pressure: $\tfrac12 \rho v^2 = \tfrac12 \times 1.225 \times 50^2 \approx 1531$ ⁠ Pa.
2. The lift on each square metre: $1531 \times 0.705 \approx 1080$ ⁠ N.
If you got about 2160, you left out the half. If you got something small like 22, you used $v$ ⁠ rather than $v^2$ ⁠.
Step 3 of 5
Open this step
Half the speed
Practice · Predict, then look
Question
Bring the Airspeed down from 50 m/s to 25 m/s, with the Angle of attack still 5°. What will happen to the lift?

Correct answer
- Correct It will fall to a quarter
Setup, choices and feedback
The Scene at this step
Angle of attack
5 °fixed
Airspeed
50 m/sfixed
Show the pressure
On
Choices
- Correct It will fall to a quarter
- Wrong It will fall to half
  Misconception: Thinks lift grows in proportion to airspeed, so half the speed gives half the lift.
  Feedback from the catalogue
  Lift goes with the square of the speed: $L = \tfrac12 \rho v^2 C_L A$ ⁠. At twice the speed each kilogram of air meets the wing twice as fast, and twice as many kilograms pass each second, so the lift grows four times. At half the speed it falls to a quarter, which is why a plane lands at a high angle of attack with its flaps out.
- Wrong It will not change, because the angle has not
  Feedback
  The angle sets the lift coefficient, but the lift itself is $\tfrac12 \rho v^2 C_L$ ⁠: it depends on the speed too.
Once the prediction is locked in, the Scene changes
Airspeed
25 m/sfixed
The Scene then shows: 269.9 Pa · Wing: lift on each square metre of wing, ½ρv²·C_L (N/m², the same unit as Pa)/nodes/wing/outputs/lift
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
From 1080 to about 270 newtons per square metre: a quarter. To land slowly, a pilot raises the angle of attack and lowers the flaps, which both raise $C_L$ ⁠.

Worked solution
1. The lift on each square metre: $L = \tfrac12 \rho v^2 C_L$ ⁠, and $C_L$ ⁠ has not changed.
2. The speed halved, so $v^2$ ⁠ fell 4 times.
3. $\dfrac{1080}{4} = 270$ ⁠ newtons per square metre.
Step 4 of 5
Open this step
The stall
Practice · Predict, then look
Question
Raise the Angle of attack from 14° to 18° at the same speed. What will happen to the lift?

Correct answer
- Correct It will fall, and the drag will soar
Setup, choices and feedback
The Scene at this step
Angle of attack
14 °fixed
Airspeed
50 m/s
Show the pressure
Off
Choices
- Correct It will fall, and the drag will soar
- Wrong It will keep growing: a steeper angle turns more air down
  Feedback
  Up to a point that is true. Beyond it the air can no longer follow the curved top, breaks away from it and swirls: a stall.
- Wrong It will stay about the same
  Feedback
  In a stall the lift does not hold: the flow tears away from the top, where most of the lift is made.
Once the prediction is locked in, the Scene changes
Angle of attack
18 °fixed
The Scene then shows: 1.154 · Wing: lift coefficient C_L at this angle of attack/nodes/wing/outputs/liftCoefficient
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The lift coefficient falls from about 1.6 to about 1.15, and the drag grows more than twelve times. That is a stall: the flow breaks away from the top of the wing.

Worked solution
1. Up to about 15° a larger angle turns more air down, and the lift grows.
2. Beyond that the air cannot follow the curved top: it breaks away and swirls.
3. The lift falls, and the drag, the force that slows the wing, grows hugely.
So a pilot cannot keep raising the angle to fly slower: below a certain speed the wing stalls.
Step 5 of 5
Open this step
Coming in to land
Practice · Calculate
Question
On the approach to land: an Airspeed of 30 m/s and an Angle of attack of 10°, where $C_L \approx 1.21$ ⁠. How many kilograms does each square metre of wing hold up?

Correct answer
67.81 ± 2%
Setup, choices and feedback
The Scene at this step
Angle of attack
10 °fixed
Airspeed
30 m/sfixed
Show the pressure
Off
Accepted within
± 2%
Computed from the Scene at this step’s setting
Wing: kilograms each square metre of wing holds up, lift ÷ g/nodes/wing/outputs/holds
Wrong answers it recognises
- 665 ± 2%
  Misconception: Uses a mass in kilograms as if it were a weight in newtons, leaving out g.
  Feedback from the catalogue
  Mass and weight are different things. Mass is measured in kilograms, and weight is a force, in newtons: $W = mg$ ⁠, with $g \approx 9.8$ ⁠ newtons per kilogram. A 40 t lorry, 40,000 kg, weighs about 392,000 N.
Feedback for a right answer
About 68 kg for each square metre. So a 1000 kg plane needs a wing of about 15 m², roughly the size of a light aircraft’s wing.

Hint
Work out the lift in newtons first, then divide by $g$ ⁠.

Worked solution
1. The dynamic pressure: $\tfrac12 \times 1.225 \times 30^2 \approx 551$ ⁠ Pa.
2. The lift on each square metre: $551 \times 1.21 \approx 665$ ⁠ N.
3. The mass that lift holds up: $\dfrac{665}{9.81} \approx 68$ ⁠ kg.
If you wrote 665, you stopped at newtons. A weight of 665 N belongs to a mass of 68 kg.

Lesson 5 of 6 · About 25 minutes

Wind turbines

A wind turbine takes part of the kinetic energy of the air passing through it. The power grows with the cube of the wind speed and the square of the diameter, and it has a limit.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
The power in the wind
Practice · Calculate
Question
A Wind speed of 10 m/s and a Rotor diameter of 120 m. The air’s density is 1.225 kg/m³, and the turbine takes $C_p = 0.44$ ⁠ of the power in the wind. What is its electrical power?

Correct answer
3.048 MW ± 2%
Setup, choices and feedback
The Scene at this step
Wind speed
10 m/sfixed
Rotor diameter
120 mfixed
Show the power
Off
Accepted within
± 2%
Computed from the Scene at this step’s setting
Turbine: electrical power out/nodes/turbine/outputs/power
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 6.93 MW ± 2%
  Misconception: Thinks a perfect wind turbine could take all the power in the wind.
  Feedback from the catalogue
  To take all of the wind’s energy the turbine would have to stop the air completely, but still air behind the rotor would block the new air arriving. The air has to keep flowing. Betz’s calculation shows that even a perfect turbine takes at most $\tfrac{16}{27}$ ⁠, about 59%, and large turbines reach about 45%.
- 12.19 MW ± 2%
  Misconception: Puts the rotor's diameter into πr², as if it were the radius.
  Feedback from the catalogue
  The area of a circle is $\pi r^2$ ⁠, and $r$ ⁠ is the radius, half the diameter. Put the diameter in place of the radius and the area comes out four times too large, and the power with it.
Feedback for a right answer
About 3 MW, enough for some two thousand homes on average. In this wind about 6.9 MW passes through the rotor, and the turbine takes less than half.

Hint
$P = C_p \cdot \tfrac12 \rho A v^3$ ⁠, with $A = \pi r^2$ ⁠.

Worked solution
1. The area the blades sweep: $A = \pi \times 60^2 \approx 11\,310$ ⁠ m².
2. The power in the wind: $\tfrac12 \times 1.225 \times 11\,310 \times 10^3 \approx 6.93 \times 10^6$ ⁠ W.
3. The electrical power: $0.44 \times 6.93 \approx 3.05$ ⁠ MW.
If you got 6.9 MW, you took all of the wind’s power. A turbine cannot stop all the air.
Step 2 of 5
Open this step
Twice the wind
Practice · Predict, then look
Question
The Wind speed rises from 5 m/s to 10 m/s. How many times will the turbine’s power grow?

Correct answer
- Correct 8 times
Setup, choices and feedback
The Scene at this step
Wind speed
5 m/sfixed
Rotor diameter
120 mfixed
Show the power
Off
Choices
- Correct 8 times
- Wrong 4 times
  Feedback
  Each kilogram of air does carry 4 times the kinetic energy, but twice as many kilograms also pass the rotor each second.
- Wrong 2 times
  Misconception: Thinks a wind turbine's power grows in proportion to the wind speed.
  Feedback from the catalogue
  The wind’s power grows with the cube of its speed: $P = \tfrac12 \rho A v^3$ ⁠. Wind twice as fast carries twice as much air through the rotor each second, and each kilogram carries four times the kinetic energy, $\tfrac12 v^2$ ⁠. Together that is eight times.
Once the prediction is locked in, the Scene changes
Wind speed
10 m/sfixed
Show the power
On
The Scene then shows: 3.048 MW · Turbine: electrical power out/nodes/turbine/outputs/power
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
From about 0.38 MW to about 3.05 MW: 8 times, $2^3$ ⁠. That is why turbines go on ridges and out at sea, where the wind is only a little stronger.

Worked solution
1. $P = C_p \cdot \tfrac12 \rho A v^3$ ⁠, and $C_p$ ⁠, $\rho$ ⁠ and $A$ ⁠ have not changed.
2. The speed doubled, so $v^3$ ⁠ grew $2^3 = 8$ ⁠ times.
3. Twice the mass a second, each kilogram with four times the kinetic energy: together 8 times.
Step 3 of 5
Open this step
The Betz limit
Practice · Choose
Question
Why can even a perfect turbine not take all the power in the wind?

Correct answer
- Correct To take it all it would have to stop the air, and still air would block the air coming after it
Setup, choices and feedback
The Scene at this step
Wind speed
9 m/s
Rotor diameter
120 m
Show the power
On
Choices
Shown in a random order.
- Correct To take it all it would have to stop the air, and still air would block the air coming after it
- Wrong Because of friction in the generator and gearbox
  Feedback
  Friction takes a little more, but the limit holds even with no friction at all. It comes from the air having to keep flowing.
- Wrong It can: a perfect turbine takes all the power
  Misconception: Thinks a perfect wind turbine could take all the power in the wind.
  Feedback from the catalogue
  To take all of the wind’s energy the turbine would have to stop the air completely, but still air behind the rotor would block the new air arriving. The air has to keep flowing. Betz’s calculation shows that even a perfect turbine takes at most $\tfrac{16}{27}$ ⁠, about 59%, and large turbines reach about 45%.
Feedback for a right answer
The air slows behind the rotor and spreads out, but keeps flowing. The kinetic energy left in it is the price. Betz’s limit is $\tfrac{16}{27} \approx 0.59$ ⁠, reached when the rotor slows the wind to a third of its speed behind it.

Hint
Choose On under Show the power and watch the stream tube behind the rotor: does the air stop there?

Worked solution
1. The turbine takes energy by slowing the air.
2. If it stopped the air completely, the still air would block the flow, and no new air would pass through the rotor.
3. If it did not slow it at all, it would take nothing.
4. Between the two lies a best: taking at most $\tfrac{16}{27}$ ⁠ of the power, about 59%. Large turbines reach about 45%.
Step 4 of 5
Open this step
A smaller rotor
Practice · Predict, then look
Question
At a Wind speed of 8 m/s, shrink the Rotor diameter from 120 m to 60 m. What will happen to the power?

Correct answer
- Correct It will fall to a quarter
Setup, choices and feedback
The Scene at this step
Wind speed
8 m/sfixed
Rotor diameter
120 mfixed
Show the power
Off
Choices
- Correct It will fall to a quarter
- Wrong It will fall to half
  Feedback
  The power depends on the area the blades sweep, $\pi r^2$ ⁠, not on the diameter itself. Half the diameter is a quarter of the area.
- Wrong It will fall to an eighth
  Feedback
  The cube belongs to the wind speed. The power depends on the diameter through the area, as its square.
Once the prediction is locked in, the Scene changes
Rotor diameter
60 mfixed
The Scene then shows: 0.3901 MW · Turbine: electrical power out/nodes/turbine/outputs/power
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
From about 1.56 MW to about 0.39 MW: a quarter. That is why new turbines’ blades are so long, over 100 m each at sea.

Worked solution
1. $P \propto A = \pi \left(\dfrac{D}{2}\right)^2$ ⁠, so $P \propto D^2$ ⁠.
2. The diameter halved, so the area and the power fell 4 times.
3. $\dfrac{1.56}{4} \approx 0.39$ ⁠ MW.
Step 5 of 5
Open this step
A gale
Practice · Predict, then look
Question
The Wind speed rises from 13 m/s to 20 m/s. What will happen to the electrical power?

Correct answer
- Correct It will stay as it is
Setup, choices and feedback
The Scene at this step
Wind speed
13 m/sfixed
Rotor diameter
120 m
Show the power
On
Choices
- Correct It will stay as it is
- Wrong It will grow nearly 4 times, as $v^3$ ⁠ does
  Feedback
  The power in the wind does grow like that, but the generator can give only up to its rated power. Above that the blades turn on their axes and spill the extra.
- Wrong The turbine will stop so as not to break
  Feedback
  It stops only in a storm, above about 25 m/s. At 20 m/s it still runs, at full power.
Once the prediction is locked in, the Scene changes
Wind speed
20 m/sfixed
Show the power
On
The Scene then shows: 4.524 MW · Turbine: electrical power out/nodes/turbine/outputs/power
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The power stays at the rated power, about 4.5 MW. The blades are pitched so that the turbine takes only about 8% of the wind’s power, so as not to overload the generator.

Worked solution
1. Above about 11.4 m/s the turbine already gives its rated power, 400 W for each square metre of rotor: about 4.5 MW.
2. The generator is not built for more, so the blades turn on their axes and take less of the wind.
3. At 13 m/s and at 20 m/s the power is the same. Only above 25 m/s does the turbine stop.

Lesson 6 of 6 · About 30 minutes

Structures and bridges

A bridge hands the lorry's weight on to the banks. Where it is squeezed and where stretched, how hard the supports push, and why an arch pushes the banks out while cables pull them in.

Open the lesson

The 9 steps and their answers

Step 1 of 9
Open this step
The truss
Practice · Choose
Question
In the Scene’s truss the road hangs from the bottom. Which part is squeezed, pushed along its length?

Correct answer
- Correct The top chord
Setup, choices and feedback
The Scene at this step
Bridge
Truss
Lorry
40 t
Show the forces
On
Choices
Shown in a random order.
- Correct The top chord
- Wrong The bottom chord, which carries the road
  Feedback
  The bottom chord is stretched: the truss bends a little like a beam, and the underside of a bent beam stretches.
- Wrong The diagonals
  Feedback
  In this kind of truss the diagonals slope down towards the middle, so the load pulls on them: they are in tension.
Feedback for a right answer
The top chord is squeezed and the bottom one stretched, just like the top and bottom of a bent beam. The truss holds them far apart with triangles, which makes it stiff and light.

Hint
Choose On under Show the forces: orange is compression and blue is tension.

Worked solution
1. The whole truss behaves like a long beam on two bearings, and the load bends it slightly downward.
2. In a bent beam the top shortens and the bottom lengthens.
3. So the top chord is in compression, and the bottom chord in tension.
4. The triangles between the chords carry the load and hold the shape, because a triangle cannot change shape without changing the length of its sides.
Step 2 of 9
Open this step
The supports
Practice · Calculate
Question
A 40 t Lorry drives onto the bridge. By how much does the total upward force of the supports on the bridge grow?

Correct answer
392400 N ± 2%
Setup, choices and feedback
The Scene at this step
Bridge
Truss
Lorry
40 tfixed
Show the forces
On
Accepted within
± 2%
Computed from the Scene at this step’s setting
Weight of the lorry and its cargo/nodes/bridge/outputs/lorryWeight
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 40000 N ± 2%
  Misconception: Uses a mass in kilograms as if it were a weight in newtons, leaving out g.
  Feedback from the catalogue
  Mass and weight are different things. Mass is measured in kilograms, and weight is a force, in newtons: $W = mg$ ⁠, with $g \approx 9.8$ ⁠ newtons per kilogram. A 40 t lorry, 40,000 kg, weighs about 392,000 N.
Feedback for a right answer
By exactly the lorry’s weight, about 392 kN. The bridge stays at rest, so every newton the lorry adds, the supports give back.

Hint
The bridge does not move: the forces on it add up to zero.

Worked solution
1. The bridge is at rest, so the vertical forces on it add up to zero: the supports push up with exactly all the weight on it.
2. The lorry adds its weight: $W = mg = 40\,000 \times 9.81 \approx 392\,000$ ⁠ N.
3. So the supports’ force grows by 392 kN, whatever the bridge.
If you wrote 40,000 N, you used the mass as if it were a weight.
Step 3 of 9
Open this step
The beam
Practice · Choose
Question
In the beam bridge, where are the heavy girders stretched?

Correct answer
- Correct Along the bottom edge
Setup, choices and feedback
The Scene at this step
Bridge
Beam
Lorry
40 t
Show the forces
On
Choices
Shown in a random order.
- Correct Along the bottom edge
- Wrong Along the top edge, which the lorry presses on
  Feedback
  The top edge is squeezed: as the girder bends down, its top shortens.
- Wrong Nowhere: a beam is only squeezed
  Feedback
  A bent beam is both squeezed and stretched, each side its own way. That is why concrete, strong when squeezed and weak when stretched, gets steel bars along its bottom.
Feedback for a right answer
As the girder bends down, its bottom edge lengthens and its top shortens. Halfway up there is almost no stress, which is why steel girders are I-shaped: plenty of steel at the edges and little in the middle.

Hint
Choose On under Show the forces and compare the colour of the top and bottom edges.

Worked solution
1. The lorry bends the girder downward between its two bearings.
2. In the bent curve the outer side, the bottom, is longer than the inner side, the top.
3. What lengthens is stretched, and what shortens is squeezed.
4. So the bottom edge is in tension, and the top edge in compression.
Step 4 of 9
Open this step
The arch
Practice · Predict, then look
Question
Choose the Arch bridge. Its rib is only squeezed. Which way does the arch push on the banks?

Correct answer
- Correct Outward and down, into the banks
Setup, choices and feedback
The Scene at this step
Bridge
Trussfixed
Lorry
40 tfixed
Show the forces
Off
Choices
- Correct Outward and down, into the banks
- Wrong Only down, like a beam on bearings
  Feedback
  An arch is squeezed along its length, and near its feet it slopes. So the force it hands the bank slopes too: it has a sideways part, outward.
- Wrong Inward: it pulls the banks together
  Feedback
  Pulling belongs to cables, which are stretched. The arch rib is squeezed, so it pushes.
Once the prediction is locked in, the Scene changes
Bridge
Archfixed
Show the forces
On
The Scene then shows: 753000 N · Sideways force of the structure on each bank: positive pushes the bank outwards (an arch), negative pulls it inwards (cables)/nodes/bridge/outputs/bankPush
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
With a 40 t lorry the arch pushes each bank outward with about 750 kN, nearly twice the lorry’s weight. An arch needs rocky banks that will not move, or it flattens and falls.

Worked solution
1. The arch rib is squeezed along its length, and at each point the force in it runs along the rib.
2. At the feet the rib slopes, so the force it puts on the bank points down and outward.
3. The bank pushes back, up and inward: that is how it holds the arch.
4. The flatter the arch, the larger the sideways push.
Step 5 of 9
Open this step
The suspension bridge
Practice · Predict, then look
Question
Now choose the Suspension bridge. The cables pass over the towers and are anchored in the banks. Which way does the bridge act on the banks?

Correct answer
- Correct The cables pull the banks inward, towards each other
Setup, choices and feedback
The Scene at this step
Bridge
Archfixed
Lorry
40 tfixed
Show the forces
On
Choices
- Correct The cables pull the banks inward, towards each other
- Wrong The bridge pushes the banks outward, like an arch
  Feedback
  A cable cannot push: it can only be stretched and pull. A suspension bridge is an upside-down arch, and its forces are reversed too.
- Wrong No sideways force: the cables only hold the road up
  Feedback
  A cable holding a weight in its middle must be pulled sideways at its ends too, or it would drop straight down. The anchors hold that sideways pull.
Once the prediction is locked in, the Scene changes
Bridge
Suspensionfixed
Show the forces
On
The Scene then shows: -592700 N · Sideways force of the structure on each bank: positive pushes the bank outwards (an arch), negative pulls it inwards (cables)/nodes/bridge/outputs/bankPush
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The cables pull each bank inward with about 590 kN (the reading comes out negative: a pull, not a push). That is why a suspension bridge has heavy anchors at its ends, sunk deep in the ground.

Worked solution
1. The cable is stretched along its length, and at its ends it is pulled towards the middle and up.
2. The cables pass over the towers, which are pushed down, and run on to the anchors.
3. The anchors are pulled towards the river, inward, and the ground holds them.
4. It is the reverse of the arch: a squeezed arch pushes the banks out, and a stretched cable pulls them in.
Step 6 of 9
Open this step
The least steel
Practice · Reach the target
Question
All four bridges have the same span and the same road. Choose the Bridge with the smallest weight of its own, and press Check.

Goal
- 711100 ± 2000
  At the start of the step: 924100
Setup, choices and feedback
The Scene at this step
Bridge
Beam
Lorry
40 tfixed
Show the forces
On
Read from the Scene at
Weight of the bridge itself/nodes/bridge/outputs/bridgeWeight
One setting that reaches it
Bridge
Truss
Feedback for a right answer
Truss: about 711 kN, less than any other. The triangles work without bending.

Hint
Which shape carries the load only through bars stretched or squeezed along their length, with no part bending?

Worked solution
1. In a beam bridge the top of the girder is squeezed and the bottom stretched, but the steel in the middle hardly works. So the girders are heavy: about 924 kN.
2. In a truss every bar is only stretched or only squeezed, and the triangles together stand in for a solid girder: about 711 kN.
3. The arch and the suspension bridge also need towers, anchors or a thick rib: about 900 to 1000 kN.
Over a short span a truss is the bridge that uses least steel. Over spans of hundreds of metres only a suspension bridge can carry itself.
Step 7 of 9
Open this step
All of the weight
Practice · Calculate
Question
The beam bridge weighs 924,000 N. A 40 t Lorry stands in the middle. With what total force do the supports push the bridge up? Take g = 9.81 m/s².

Correct answer
1317000 N ± 2%
Setup, choices and feedback
The Scene at this step
Bridge
Beamfixed
Lorry
40 tfixed
Show the forces
On
Accepted within
± 2%
Computed from the Scene at this step’s setting
Total upward push of all the supports, with the lorry at midspan/nodes/bridge/outputs/supportForce
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 392400 N ± 2%
  Misconception: Counts only the traffic on a bridge and forgets the weight of the bridge itself.
  Feedback from the catalogue
  The supports carry everything on the bridge, the bridge itself included. In steel and concrete bridges the bridge’s own weight is usually more than the traffic on it, which is why engineers first design a bridge that can hold itself up.
Feedback for a right answer
About 1316 kN: more than two thirds of it is the bridge’s own weight.

Hint
The bridge is at rest: every downward force, the bridge’s own weight included, is balanced by the supports’ push.

Worked solution
1. The lorry’s weight: $40\,000 \times 9.81 \approx 392$ ⁠ kilonewtons.
2. The supports carry the bridge too: $924 + 392 = 1316$ ⁠ kilonewtons.
If you answered 392 kN, you forgot the bridge’s own weight. In big bridges it is most of the load.
Step 8 of 9
Open this step
The sag
Practice · Predict, then look
Question
A 40 t Lorry in the middle of the beam bridge lowers the middle of the road by 6.4 cm. Predict: what will happen to the sag when you choose the Suspension bridge?

Correct answer
- Correct It will sag more
Setup, choices and feedback
The Scene at this step
Bridge
Beam
Lorry
40 tfixed
Show the forces
On
Choices
- Wrong It will sag less, because the cables are very strong
  Misconception: Thinks a structure that is strong, hard to break, must also bend little under load.
  Feedback from the catalogue
  Strength is how much load a part can carry before it breaks. Stiffness is how much it changes shape under the load. A steel cable is very strong in tension, but a hanging cable changes its curve easily, so a suspension bridge sags more than a girder.
- Wrong About the same, because the lorry is the same
  No feedback of its own; the learner sees the standard message
  Not yet. Look again and try once more.
- Correct It will sag more
Once the prediction is locked in, the Scene changes
Bridge
Suspension
The Scene then shows: 20.44 cm · How far the middle of the road sinks with the lorry at midspan (real, not exaggerated)/nodes/bridge/outputs/sag
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
About 20 cm, three times the girder’s: strength in tension is not stiffness.

Hint
A cable does not resist bending. When the lorry presses on it, it changes shape.

Worked solution
1. A girder resists bending through its whole depth, so it sags little.
2. A hanging cable is very strong in tension, but it is not stiff: a weight in the middle changes its curve, and the road hung from it goes down with it.
3. So the suspension bridge sags about 20 cm, roughly three times as much as the girder.
That is why long suspension bridges get a stiff truss under the road, which spreads the load over many cables.
Step 9 of 9
Open this step
Top and bottom
Practice · Predict, then look
Question
Bridge: Beam, with Show the forces on. At the bottom of the girders the steel is stretched at a stress of up to 84 MPa. Predict: how hard is it squeezed at the top?

Correct answer
- Correct About the same, about 84 MPa
Setup, choices and feedback
The Scene at this step
Bridge
Beamfixed
Lorry
40 tfixed
Show the forces
Off
Choices
- Wrong Much less, because the weight hangs mostly at the bottom
  No feedback of its own; the learner sees the standard message
  Not yet. Look again and try once more.
- Correct About the same, about 84 MPa
- Wrong Much more, because the lorry presses from above
  No feedback of its own; the learner sees the standard message
  Not yet. Look again and try once more.
Once the prediction is locked in, the Scene changes
Show the forces
On
The Scene then shows: 84.39 MPa · The largest squeezing stress in any member/nodes/bridge/outputs/maxCompression
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The same, about 84 MPa: squeezed at the top, stretched at the bottom, and almost nothing in the middle.

Hint
A girder that bends gets shorter on one side and longer on the other. Where in the middle does it not change at all?

Worked solution
1. As the girder bends, the top gets shorter and the bottom longer. Through the middle runs a line whose length does not change.
2. The girder is symmetric about that line, so the squeeze at the top equals the stretch at the bottom.
3. The largest squeezing stress: about 84 MPa.
That is why steel girders have an I-shaped section: most of the steel at the top and bottom, where it works hardest, and little in the middle.

Across the unit · 20 misconceptions

Misconception catalogue

Each common mistake this unit recognises, in a few words. Open one for what the learner reads when an answer matches it, and where it is asked.

Index

Engines

Four strokes

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

Power strokes a second

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Compression ratio

Question

Correct answer

The Scene at this step

Once the prediction is locked in, the Scene changes

Worked solution

Where the heat goes

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Power

Question

Correct answer

The Scene at this step

Once the prediction is locked in, the Scene changes

Worked solution

The cylinder

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Opening the throttle

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

The work it gives

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

After the spark

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Hint

Worked solution

The highest pressure

Question

Correct answer

The Scene at this step

Hint

Worked solution

Refrigerators and heat pumps

A heat pump

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

Coefficient of performance

Question

Correct answer