Answer key · Unit 6

Heat and temperature.

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 3

Lesson 1 of 3 · About 25 minutes

Heat and temperature

What temperature measures and what heat is: particles that jiggle, kelvin and degrees, conduction along a bar of copper and of wood, why metal feels cold, specific heat, convection and radiation.

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The 7 steps and their answers

Step 1 of 7
Open this step
Particles that jiggle
Practice · Predict, then look
Question
The Heater is off, at 293 K, and the whole bar is at room temperature. Now the burner is turned up to 600 K. What will the particles next to the flame do?

Correct answer
- Correct Vibrate harder about their places
Setup, choices and feedback
The Scene at this step
Heater
293 K
Bar made of
Copper
Show the heat map
0.6 (Bar → Heat map)
Choices
- Correct Vibrate harder about their places
- Wrong Flow along the bar to the cold end, carrying the heat with them
  Feedback
  The particles of a solid are held in place. What travels along the bar is the vibration itself, from each particle to its neighbour, not the particles.
- Wrong Grow bigger
  Feedback
  The particles themselves don’t change size. They only move harder, and so take up a little more room: that is why a bar gets longer when it is heated.
- Wrong They were still until now, and will start to move
  Feedback
  The particles vibrate at every temperature above absolute zero, and room temperature is almost 300 K above it. The flame only makes the vibration stronger.
Once the prediction is locked in, the Scene changes
Heater
600 K
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Next to the flame the particles rattle hardest. Even at room temperature they were never still, only shaking less. Temperature measures how hard that motion is.

Worked solution
1. The flame hands energy to the particles at the end of the bar.
2. In a solid each particle is held in its place, so that energy becomes a harder vibration about the same spot.
3. A higher temperature means a harder vibration.
If you thought the particles were still before: they vibrate at every temperature above absolute zero, in the room too.
Step 2 of 7
Open this step
Kelvin and degrees
Practice · Calculate
Question
The Heater is set to 600 K. What is its temperature in degrees Celsius?

Correct answer
326.9 °C ± 1 °C
Setup, choices and feedback
The Scene at this step
Heater
600 Kfixed
Bar made of
Copper
Show the heat map
0.6 (Bar → Heat map)
Accepted within
± 1 °C
Computed from the Scene at this step’s setting
Heater temperature/nodes/bar/props/heaterTemperature
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 873 °C ± 1.5 °C
  Misconception: Adds 273 when converting kelvin to degrees Celsius, or subtracts it the other way.
  Feedback from the catalogue
  The kelvin scale starts at absolute zero, −273.15 °C, so the kelvin number is always 273.15 more than the degrees: $T_{\mathrm{K}} = T_{^\circ\mathrm{C}} + 273.15$ ⁠. From kelvin to Celsius you subtract. A quick check: water freezes at 0 °C, which is 273 K.
Feedback for a right answer
600 − 273 ≈ 327 °C: hot enough to melt lead.

Hint
$T_{\mathrm{K}} = T_{^\circ\mathrm{C}} + 273.15$ ⁠.

Worked solution
1. Given: T = 600 K.
2. From kelvin to Celsius, subtract 273.15: $600 - 273.15 = 326.85$ ⁠.
3. The heater is at about 327 °C.
If you got 873, you added 273 instead of subtracting it. The kelvin number is always the larger of the two.
Step 3 of 7
Open this step
Conduction
Practice · Predict, then look
Question
The Copper bar is hot, and the heat map shows colour reaching the far end. Now Bar made of is changed to Wood. What will the heat map show?

Correct answer
- Correct The heat stays close to the flame, and the far end stays cool
Setup, choices and feedback
The Scene at this step
Heater
600 K
Bar made of
Copper
Show the heat map
Heat map
Choices
- Correct The heat stays close to the flame, and the far end stays cool
- Wrong The same picture: heat spreads along any bar at the same pace
  Feedback
  Materials conduct heat at very different rates: copper conducts thousands of times better than wood.
- Wrong The far end warms faster, because wood burns
  Feedback
  Wood does char next to the flame, but it is a poor conductor, so the heat hardly moves along it.
Once the prediction is locked in, the Scene changes
Bar made of
Wood
The Scene then shows: 0.15 · Thermal conductivity of the material, W/(m·K)/nodes/bar/outputs/conductivity
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Wood has no free electrons, and the vibrations pass slowly from neighbour to neighbour. Its thermal conductivity is about 0.15 W/(m·K), against 401 for copper. That is why a wooden handle stays cool on a hot pan.

Worked solution
1. Conduction is the vibration passing from each particle to its neighbour through a solid.
2. In a metal free electrons carry energy too, so it conducts fast.
3. Wood has no free electrons and is full of tiny pockets of air, so it conducts badly.
4. The heat stays near the flame, and the far end hardly warms.
Step 4 of 7
Open this step
Why metal feels cold
Practice · Choose
Question
A copper bar and a wooden bar lie all night in a room at 20 °C. In the morning you hold both, and the copper feels colder. Why?

Correct answer
- Correct Both are at 20 °C, but the copper draws heat out of your hand faster
Setup, choices and feedback
The Scene at this step
Heater
600 K
Bar made of
Copper
Show the heat map
Heat map
Choices
Shown in a random order.
- Correct Both are at 20 °C, but the copper draws heat out of your hand faster
- Wrong Metal is always colder than wood in the same room
  Misconception: Thinks metal left in a room is colder than wood beside it, because it feels colder.
  Feedback from the catalogue
  After an hour in the room the metal and the wood are at exactly the same temperature, the room’s. Metal feels cold because it conducts well: it draws heat out of your fingers quickly, and your hand feels that flow of heat, not the metal’s temperature. Wood conducts badly, so the skin touching it stays almost as warm as before.
- Wrong Cold flows from the copper into your hand
  Feedback
  There is no such thing as “cold” that flows. Only heat flows, always from hot to cold: from your hand into the copper.
- Wrong The copper stores more cold during the night
  Misconception: Thinks heat and temperature are the same thing, or that a hot thing contains heat.
  Feedback from the catalogue
  Temperature measures how hard a thing’s particles are moving; heat is energy passing from a hotter thing to a colder one because of the temperature difference. A cup of tea doesn’t hold “heat”: it has a temperature, and heat leaves it into the air until the two match. A bucket of lukewarm water can hold more energy than a red-hot spark, though its temperature is far lower.
Feedback for a right answer
Your hand doesn’t measure temperature but the rate heat flows out of the skin. Copper conducts heat thousands of times better than wood, so it cools your fingers fast.

Hint
After a night in the room, what temperature is everything in it at?

Worked solution
1. After a night in the room everything in it has reached the room’s temperature, 20 °C.
2. Skin is warmer, about 33 °C, so heat flows from it into both bars.
3. Copper conducts well and carries the heat away from where you touch it fast. Wood conducts badly, and the skin on it stays almost warm.
4. What you feel is the flow of heat, so the copper feels cold.
Step 5 of 7
Open this step
Specific heat
Practice · Calculate
Question
How much energy does it take to warm one kilogram of Copper from room temperature, 293 K, to the Heater’s 600 K? The specific heat of copper is 385 J/(kg·K).

Correct answer
118.2 kJ ± 2%
Setup, choices and feedback
The Scene at this step
Heater
600 Kfixed
Bar made of
Copperfixed
Show the heat map
Heat map
Accepted within
± 2%
Computed from the Scene at this step’s setting
Energy to warm 1 kg of the material from the room to the heater, c · ΔT/nodes/bar/outputs/heatToWarm
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 231 kJ ± 2%
  Misconception: Uses the final temperature in Q = mcΔT instead of the change in temperature.
  Feedback from the catalogue
  The formula $Q = mc\Delta T$ ⁠ has the change in temperature, $\Delta T = T_{\text{end}} - T_{\text{start}}$ ⁠, not the final temperature. The heat it takes to warm something depends on how far up the scale it goes, not on where it stops. A change of 1 kelvin is also a change of 1 degree, so $\Delta T$ ⁠ is the same in both scales.
- 112.8 kJ ± 1%
  Misconception: Uses the final temperature in Q = mcΔT instead of the change in temperature.
  Feedback from the catalogue
  The formula $Q = mc\Delta T$ ⁠ has the change in temperature, $\Delta T = T_{\text{end}} - T_{\text{start}}$ ⁠, not the final temperature. The heat it takes to warm something depends on how far up the scale it goes, not on where it stops. A change of 1 kelvin is also a change of 1 degree, so $\Delta T$ ⁠ is the same in both scales.
Feedback for a right answer
$Q = 1 \times 385 \times 307 \approx 118\,000$ ⁠ joules, about 118 kJ.

Hint
$Q = mc\Delta T$ ⁠, and $\Delta T$ ⁠ is the change in temperature.

Worked solution
1. Given: m = 1 kg, c = 385 J/(kg·K), from 293 K to 600 K.
2. The change in temperature: $\Delta T = 600 - 293 = 307$ ⁠ kelvin.
3. $Q = mc\Delta T = 1 \times 385 \times 307 = 118\,195$ ⁠ joules.
4. About 118 kJ.
If you got 231 kJ, you multiplied by 600, the final temperature, instead of the change.
Step 6 of 7
Open this step
Convection
Practice · Choose
Question
A heater stands by one wall of a room, and soon the whole room is warm, even the far corner. How does the heat mainly get there?

Correct answer
- Correct Air warmed by the heater expands, rises and spreads round the room, and cooler air sinks to take its place
Setup, choices and feedback
The Scene at this step
Heater
600 K
Bar made of
Copper
Show the heat map
Heat map
Choices
Shown in a random order.
- Correct Air warmed by the heater expands, rises and spreads round the room, and cooler air sinks to take its place
- Wrong The air conducts the heat from molecule to molecule, like the bar
  Feedback
  Air is a very poor conductor, worse even than wood. By conduction alone the heat would creep across the room very slowly.
- Wrong Heat itself goes up, because heat always rises
  Misconception: Thinks heat and temperature are the same thing, or that a hot thing contains heat.
  Feedback from the catalogue
  Temperature measures how hard a thing’s particles are moving; heat is energy passing from a hotter thing to a colder one because of the temperature difference. A cup of tea doesn’t hold “heat”: it has a temperature, and heat leaves it into the air until the two match. A bucket of lukewarm water can hold more energy than a red-hot spark, though its temperature is far lower.
Feedback for a right answer
This is convection: warm air is less dense, so it rises, carrying its heat with it. In liquids and gases it is the fast way heat moves.

Worked solution
1. Air by the heater warms, expands and becomes less dense.
2. Less dense air rises, like a cork in water, and cool, denser air sinks to take its place.
3. A circulating current forms and carries the heat round the room: convection.
Heat doesn’t “rise” by itself. What rises is warm air, and the heat with it.
Step 7 of 7
Open this step
Radiation
Practice · Calculate
Question
Every warm body radiates, and by the Stefan–Boltzmann law the radiation grows as $T^4$ ⁠, in kelvin. The Heater is set to 600 K and the room is at 293 K. How many times more does the hot end of the bar radiate than at room temperature?

Correct answer
17.58 ± 3%
Setup, choices and feedback
The Scene at this step
Heater
600 Kfixed
Bar made of
Steel
Show the heat map
Bar
Accepted within
± 3%
Computed from the Scene at this step’s setting
How many times more the hot end radiates than at room temperature, (T_heater / T_room)⁴/nodes/bar/outputs/radiationRatio
Wrong answers it recognises
- 2.05 ± 0.05
  Misconception: Assumes anything that grows with temperature grows in step with it: twice the kelvin, twice the swing or twice the glow.
  Feedback from the catalogue
  Not everything that grows with temperature grows in step with it. A particle’s average energy grows in proportion to $T$ ⁠ in kelvin, but the size of its swing grows only as $\sqrt{T}$ ⁠, and the radiation a body gives off grows as $T^4$ ⁠: twice the absolute temperature is 1.4 times the swing and 16 times the radiation.
Feedback for a right answer
$\left(\dfrac{600}{293}\right)^4 \approx 2.05^4 \approx 17.6$ ⁠: twice the temperature, 17.6 times the radiation.

Hint
Divide the two temperatures in kelvin, and raise the result to the power 4.

Worked solution
1. The ratio of the temperatures: $\dfrac{600}{293} \approx 2.048$ ⁠.
2. Radiation grows as $T^4$ ⁠, so $2.048^4 \approx 17.6$ ⁠.
3. The hot end radiates about 17.6 times as much.
If you answered 2.05, you assumed radiation grows in step with temperature. With the fourth power, doubling the temperature multiplies the radiation by 16.

Lesson 2 of 3 · About 25 minutes

Entropy and heat engines

A Stirling engine between a flame and cooling water: why it needs two sides at different temperatures, the books of heat in, work and heat out, entropy and the second law, and the Carnot limit no engine can pass.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
Two sides
Practice · Predict, then look
Question
The Hot end temperature is turned down to 313 K, and Warm water is chosen for the Cooling water temperature, also 313 K. What will the engine do?

Correct answer
- Correct Stop: with no temperature difference there is no work
Setup, choices and feedback
The Scene at this step
Hot end temperature
573.1 K
Cooling water temperature
293.1 K
Show the Carnot limit
Engine
Choices
- Correct Stop: with no temperature difference there is no work
- Wrong Keep turning slowly, because it still has heat
  Misconception: Thinks heat and temperature are the same thing, or that a hot thing contains heat.
  Feedback from the catalogue
  Temperature measures how hard a thing’s particles are moving; heat is energy passing from a hotter thing to a colder one because of the temperature difference. A cup of tea doesn’t hold “heat”: it has a temperature, and heat leaves it into the air until the two match. A bucket of lukewarm water can hold more energy than a red-hot spark, though its temperature is far lower.
- Wrong Turn as before: it has a flame
  Feedback
  The flame only keeps the hot end at its temperature. If the cold side is just as warm, the heat has nowhere to flow.
Once the prediction is locked in, the Scene changes
Hot end temperature
313.1 K
Cooling water temperature
313.1 K
The Scene then shows: 0 · Turns per second as drawn, in proportion to the temperature difference/nodes/engine/outputs/turnRate
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
With both sides at one temperature, the gas pushes just as hard on the hot strokes as on the cold ones, and no work is left over. A heat engine runs on a temperature difference, not on heat.

Worked solution
1. The engine gets work because the gas pushes hard while most of it is on the hot side, and weakly while most of it is on the cold side.
2. With both sides at 313 K, the pressure is the same on both strokes.
3. The work on one stroke cancels the work on the other, and the wheel stops.
Step 2 of 6
Open this step
The Carnot limit
Practice · Calculate
Question
The hot end is at 300 °C and the cooling water at 20 °C. What is the Carnot limit: the largest share of the heat that can become work? Write it as a decimal, such as 0.25.

Correct answer
0.4885 ± 0.01
Setup, choices and feedback
The Scene at this step
Hot end temperature
573.1 Kfixed
Cooling water temperature
293.1 Kfixed
Show the Carnot limit
Enginefixed
Accepted within
± 0.01
Computed from the Scene at this step’s setting
The Carnot limit, 1 − T_cold / T_hot: the largest share of the heat that can become work/nodes/engine/outputs/efficiency
Wrong answers it recognises
- 0.933 ± 0.01
  Misconception: Puts temperatures in degrees Celsius into Q/T or the Carnot limit instead of kelvin.
  Feedback from the catalogue
  In $\dfrac{Q}{T}$ ⁠ and in the Carnot limit the temperatures must be in kelvin, because the formulas divide by them: zero on the Celsius scale is only where water freezes, and dividing by it means nothing. Add 273.15 to every temperature in degrees before you substitute.
- 0.512 ± 0.008
  Misconception: Thinks a well-enough built engine, with no friction, could turn all its heat into work.
  Feedback
  You worked out $\dfrac{T_{\text{cold}}}{T_{\text{hot}}}$ ⁠, the share that must go to the cold side. The limit on work is what is left: 1 minus that.
Feedback for a right answer
$1 - \dfrac{293}{573} \approx 0.49$ ⁠: almost half the heat can become work, and more than half must go into the water.

Hint
$1 - \dfrac{T_{\text{cold}}}{T_{\text{hot}}}$ ⁠, with both temperatures in kelvin.

Worked solution
1. Into kelvin: $T_{\text{hot}} = 300 + 273 = 573$ ⁠, $T_{\text{cold}} = 20 + 273 = 293$ ⁠.
2. $\dfrac{T_{\text{cold}}}{T_{\text{hot}}} = \dfrac{293}{573} \approx 0.51$ ⁠.
3. The Carnot limit: $1 - 0.51 = 0.49$ ⁠.
If you got 0.93, you put in degrees Celsius: $1 - \dfrac{20}{300}$ ⁠. The formula divides by the temperatures, so they must be in kelvin.
Step 3 of 6
Open this step
The books of heat and work
Practice · Predict, then look
Question
Hot water is chosen for the Hot end temperature, 363 K, and the cooling water stays Tap water, 293 K. Predict: at most how many joules of work will the engine give for every 100 J of heat?

Correct answer
19.28 J ± 2 J
A number prediction counts as right within ± 2 J
Setup, choices and feedback
The Scene at this step
Hot end temperature
573.1 K
Cooling water temperature
293.1 Kfixed
Show the Carnot limit
Enginefixed
Once the prediction is locked in, the Scene changes
Hot end temperature
363.1 K
Show the Carnot limit
0.3 (Engine → Carnot limit)fixed
The Scene then shows: 19.28 J · The most work for every 100 J of heat taken in/nodes/engine/outputs/work
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$100 \times \left(1 - \dfrac{293}{363}\right) \approx 19$ ⁠ joules. The work arrow has shrunk, and the arrow into the water has grown to 81 J.

Worked solution
1. $T_{\text{hot}} = 363$ ⁠ kelvin, $T_{\text{cold}} = 293$ ⁠ kelvin.
2. The Carnot limit: $1 - \dfrac{293}{363} \approx 0.19$ ⁠.
3. Of 100 J of heat, at most 19 J becomes work, and 81 J goes into the water.
Step 4 of 6
Open this step
Entropy
Practice · Calculate
Question
100 J of heat come in from the hot end, at 300 °C, and bring entropy $\dfrac{Q}{T}$ ⁠ with them. The heat that goes into the water, at 20 °C, must carry all of it away. At least how much heat must go into the water?

Correct answer
51.15 J ± 1 J
Setup, choices and feedback
The Scene at this step
Hot end temperature
573.1 Kfixed
Cooling water temperature
293.1 Kfixed
Show the Carnot limit
Enginefixed
Accepted within
± 1 J
Computed from the Scene at this step’s setting
The least heat that must go to the cold side for every 100 J taken in/nodes/engine/outputs/heatOut
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 6.7 J ± 0.5 J
  Misconception: Puts temperatures in degrees Celsius into Q/T or the Carnot limit instead of kelvin.
  Feedback from the catalogue
  In $\dfrac{Q}{T}$ ⁠ and in the Carnot limit the temperatures must be in kelvin, because the formulas divide by them: zero on the Celsius scale is only where water freezes, and dividing by it means nothing. Add 273.15 to every temperature in degrees before you substitute.
- 0 J ± 0.5 J
  Misconception: Thinks a well-enough built engine, with no friction, could turn all its heat into work.
  Feedback from the catalogue
  Even a perfect engine, with no friction and no leaks, must give some of its heat to the cold side. The heat taken in brings entropy, $\dfrac{Q}{T}$ ⁠, and work carries none away. The only way out for it is heat given to the cold side, so the efficiency can never pass the Carnot limit, $1 - \dfrac{T_{\text{cold}}}{T_{\text{hot}}}$ ⁠. That is a law of nature, not an engineering problem.
Feedback for a right answer
$\dfrac{100}{573} \approx 0.175$ ⁠ joules per kelvin come in, and $0.175 \times 293 \approx 51$ ⁠ joules must go into the water. Only 49 J are left for work: exactly the Carnot limit.

Hint
The entropy coming in: $\dfrac{100}{T_{\text{hot}}}$ ⁠. The heat going out carries $\dfrac{Q_{\text{out}}}{T_{\text{cold}}}$ ⁠, and that must be at least as much.

Worked solution
1. Into kelvin: $T_{\text{hot}} = 573$ ⁠, $T_{\text{cold}} = 293$ ⁠.
2. The entropy coming in: $\Delta S = \dfrac{100}{573} \approx 0.175$ ⁠ joules per kelvin.
3. The heat going out must carry at least that away: $\dfrac{Q_{\text{out}}}{293} \ge 0.175$ ⁠.
4. $Q_{\text{out}} \ge 0.175 \times 293 \approx 51$ ⁠ joules.
If you got about 7 J, you put in degrees Celsius: $100 \times \dfrac{20}{300}$ ⁠. If you answered 0, that is the perfect engine the second law forbids.
Step 5 of 6
Open this step
Seventy per cent
Practice · Reach the target
Question
Change the two temperatures until the Carnot limit is 70%.

Goal
- 0.7 ± 0.01
  At the start of the step: 0.4885
Setup, choices and feedback
The Scene at this step
Hot end temperature
573.1 K
Cooling water temperature
293.1 K
Show the Carnot limit
0.7 (Engine → Carnot limit)
Read from the Scene at
The Carnot limit, 1 − T_cold / T_hot: the largest share of the heat that can become work/nodes/engine/outputs/efficiency
One setting that reaches it
Hot end temperature
881.1 K
Cooling water temperature
273.1 K
Feedback for a right answer
For 70% to become work, the cold side must be 0.3 times as hot as the hot side, in kelvin: tap water at 293 K against an end at about 977 K, or ice water against about 910 K.

Hint
$1 - \dfrac{T_{\text{cold}}}{T_{\text{hot}}} = 0.7$ ⁠ means $T_{\text{hot}} = \dfrac{T_{\text{cold}}}{0.3}$ ⁠.

Worked solution
1. $1 - \dfrac{T_{\text{cold}}}{T_{\text{hot}}} = 0.7$ ⁠, so $\dfrac{T_{\text{cold}}}{T_{\text{hot}}} = 0.3$ ⁠.
2. With Tap water, 293 K: $T_{\text{hot}} = \dfrac{293}{0.3} \approx 977$ ⁠ kelvin.
3. With Ice water, 273 K: $T_{\text{hot}} \approx 910$ ⁠ kelvin.
4. Any pair whose ratio is 0.3 will do.
Step 6 of 6
Open this step
Real engines
Practice · Choose
Question
A power station heats steam to 600 °C and condenses it at 30 °C. Its Carnot limit is 65%, and the best stations reach about 45%. Which is true?

Correct answer
- Correct No engineering will take it above 65%; friction and leaks explain only the gap down to 45%
Setup, choices and feedback
The Scene at this step
Hot end temperature
1073 K
Cooling water temperature
273.1 K
Show the Carnot limit
0.7 (Engine → Carnot limit)
Choices
Shown in a random order.
- Correct No engineering will take it above 65%; friction and leaks explain only the gap down to 45%
- Wrong Without friction it could reach 100%
  Misconception: Thinks a well-enough built engine, with no friction, could turn all its heat into work.
  Feedback from the catalogue
  Even a perfect engine, with no friction and no leaks, must give some of its heat to the cold side. The heat taken in brings entropy, $\dfrac{Q}{T}$ ⁠, and work carries none away. The only way out for it is heat given to the cold side, so the efficiency can never pass the Carnot limit, $1 - \dfrac{T_{\text{cold}}}{T_{\text{hot}}}$ ⁠. That is a law of nature, not an engineering problem.
- Wrong A better fuel would raise the limit without changing the temperatures
  Feedback
  The Carnot limit depends only on the two temperatures. A better fuel helps only if it allows hotter steam.
Feedback for a right answer
The limit is set by the second law of thermodynamics and the two temperatures alone. To raise it you need hotter steam or colder water.

Worked solution
1. Into kelvin: 873 K and 303 K.
2. The Carnot limit: $1 - \dfrac{303}{873} \approx 0.65$ ⁠.
3. Even an ideal engine cannot pass it, because the heat out must carry away the entropy the heat in brought.
4. Friction, leaks and the temperature differences heat needs in order to flow bring a real station down to about 45%.

Lesson 3 of 3 · About 20 minutes

Brownian motion and random walks

A speck that never rests in still water: what pushes it, why a big speck wanders less, why in a random walk getting twice as far takes four times as long, and Einstein's formula that proved molecules exist.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
A speck that never rests
Practice · Choose
Question
The speck in the water never stops for a moment, even when the water is perfectly still. What moves it?

Correct answer
- Correct Water molecules hit it from every side, and the hits never quite balance
Setup, choices and feedback
The Scene at this step
Water temperature
293.1 K
Speck size
1 µm
Zoom to the molecules
One speck
Choices
Shown in a random order.
- Correct Water molecules hit it from every side, and the hits never quite balance
- Wrong It is alive and swimming
  Misconception: Thinks the speck in Brownian motion moves because it is alive, or because currents in the water carry it.
  Feedback from the catalogue
  Brown himself first thought the specks were alive, then saw the same motion in soot and powdered glass. Currents don’t explain it either: a current would sweep all the specks the same way together, and here each one wanders its own way. What pushes the speck is the water’s molecules, striking it from every side, and the hits never quite balance.
- Wrong Tiny currents in the water carry it
  Misconception: Thinks the speck in Brownian motion moves because it is alive, or because currents in the water carry it.
  Feedback from the catalogue
  Brown himself first thought the specks were alive, then saw the same motion in soot and powdered glass. Currents don’t explain it either: a current would sweep all the specks the same way together, and here each one wanders its own way. What pushes the speck is the water’s molecules, striking it from every side, and the hits never quite balance.
- Wrong The microscope’s light heats it and pushes it
  Feedback
  The motion goes on in very dim light too, at the same pace. The light only lets you see it.
Feedback for a right answer
This is Brownian motion. At any moment a few more molecules happen to strike one side than the other, and the speck is nudged that way. A moment later the imbalance points somewhere else.

Worked solution
1. Water is made of molecules thousands of times smaller than the speck, always moving.
2. A vast number of them hit the speck every second, from every direction.
3. The hits almost balance, but not quite: the chance difference pushes the speck, and its direction changes every moment.
A current would sweep all the specks along together; here each wanders its own way.
Step 2 of 5
Open this step
The speck's size
Practice · Predict, then look
Question
Speck size is raised from 1 µm to 5 µm. Predict: what will happen to its wandering?

Correct answer
- Correct It will wander less
Setup, choices and feedback
The Scene at this step
Water temperature
293.1 Kfixed
Speck size
1 µm
Zoom to the molecules
One speck
Choices
- Correct It will wander less
- Wrong It will wander more, because more molecules hit it
  Misconception: Thinks a bigger speck wanders more, because more molecules hit it.
  Feedback from the catalogue
  A big speck is indeed hit by more molecules, but that is exactly why it moves less: the more hits at once, the better they cancel, and the water also drags on it more. By Einstein, $D = \dfrac{kT}{6\pi\eta r}$ ⁠: the radius is underneath, so a speck five times as wide spreads five times as slowly.
- Wrong The same: it is the same water
  Feedback
  The water is the same, but the big speck is hit by more molecules at once, and they cancel better.
Once the prediction is locked in, the Scene changes
Speck size
5 µm
The Scene then shows: 1.852 µm · Typical distance wandered in 10 s, √(4D·10 s)/nodes/grain/outputs/spreadIn10s
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
A speck five times as wide spreads five times as slowly: $D = \dfrac{kT}{6\pi\eta r}$ ⁠, with the radius underneath. In ten seconds it gets only about 1.9 µm, instead of about 4.1 µm.

Worked solution
1. A big speck is hit by more molecules at once, so the random pushes cancel better.
2. The water also drags harder on a big speck.
3. By Einstein $D \propto \dfrac{1}{r}$ ⁠: five times the radius, a fifth of $D$ ⁠.
4. The distance grows as $\sqrt{D}$ ⁠, so the speck gets $\sqrt{5} \approx 2.2$ ⁠ times less far.
Step 3 of 5
Open this step
Four times as long
Practice · Calculate
Question
In water at room temperature, a 1 µm speck gets on average about 4.1 µm from its start in ten seconds. How far will it get, on average, after 40 s?

Correct answer
8.282 µm ± 4%
Setup, choices and feedback
The Scene at this step
Water temperature
293.1 Kfixed
Speck size
1 µmfixed
Zoom to the molecules
Many specks
Accepted within
± 4%
Computed from the Scene at this step’s setting
Typical distance wandered in 10 s, √(4D·10 s)/nodes/grain/outputs/spreadIn10s
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 16.56 µm ± 4%
  Misconception: Thinks a random walk gets away in proportion to time, as a straight walk does.
  Feedback from the catalogue
  In a random walk many steps undo each other. After $N$ ⁠ steps the typical distance is only $\sqrt{N}$ ⁠ steps, so it grows as the square root of the time: $\langle r^2 \rangle = 4Dt$ ⁠. To get twice as far takes four times as long, and twice the time takes you only 1.4 times as far.
Feedback for a right answer
Four times the time is only $\sqrt{4} = 2$ ⁠ times the distance: about 8.3 µm.

Hint
$\langle r^2 \rangle = 4Dt$ ⁠: the distance grows as $\sqrt{t}$ ⁠.

Worked solution
1. In a random walk $\langle r^2 \rangle = 4Dt$ ⁠, so the typical distance grows as $\sqrt{t}$ ⁠.
2. $\dfrac{40}{10} = 4$ ⁠ times as long.
3. $\sqrt{4} = 2$ ⁠, so the distance doubles: $2 \times 4.1 \approx 8.3$ ⁠ micrometres.
If you answered about 16.6 µm, you assumed the speck gets away in proportion to time, as in a straight walk.
Step 4 of 5
Open this step
Einstein's formula
Practice · Calculate
Question
The water is boiling, 373 K, and its viscosity is 0.28 × 10⁻³ Pa·s. The speck is 1 µm across, and Boltzmann’s constant is k = 1.38 × 10⁻²³ J/K. Work out $D$ ⁠, and then: how far will the speck get on average in ten seconds?

Correct answer
8.853 µm ± 4%
Setup, choices and feedback
The Scene at this step
Water temperature
373.1 Kfixed
Speck size
1 µmfixed
Zoom to the molecules
One speck
Accepted within
± 4%
Computed from the Scene at this step’s setting
Typical distance wandered in 10 s, √(4D·10 s)/nodes/grain/outputs/spreadIn10s
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 6.26 µm ± 4%
  Misconception: Puts the diameter where a formula needs the radius.
  Feedback from the catalogue
  In the Stokes–Einstein formula $r$ ⁠ is the radius, half the diameter. A speck 1 µm across has a radius of 0.5 µm, which is $0.5 \times 10^{-6}$ ⁠ metres. Put the diameter in and $D$ ⁠ comes out half as big.
Feedback for a right answer
$D \approx 1.95 \times 10^{-12}$ ⁠ square metres per second, and in ten seconds $\sqrt{4D \times 10} \approx 8.8$ ⁠ micrometres: more than twice as far as at room temperature.

Hint
$D = \dfrac{kT}{6\pi\eta r}$ ⁠, and $r$ ⁠ is the radius. Then $r_{10} = \sqrt{4D \times 10}$ ⁠.

Worked solution
1. The radius: $r = 0.5 \times 10^{-6}$ ⁠ metres.
2. The top: $kT = 1.38 \times 10^{-23} \times 373 \approx 5.15 \times 10^{-21}$ ⁠ joules.
3. The bottom: $6\pi\eta r = 6\pi \times 0.28 \times 10^{-3} \times 0.5 \times 10^{-6} \approx 2.64 \times 10^{-9}$ ⁠.
4. $D \approx \dfrac{5.15 \times 10^{-21}}{2.64 \times 10^{-9}} \approx 1.95 \times 10^{-12}$ ⁠ square metres per second.
5. $r_{10} = \sqrt{4 \times 1.95 \times 10^{-12} \times 10} \approx 8.8 \times 10^{-6}$ ⁠ metres, about 8.8 µm.
If you got about 6.2 µm, you put in the diameter instead of the radius.
Step 5 of 5
Open this step
Faster
Practice · Reach the target
Question
Change the Water temperature and the Speck size until the speck gets on average 10 µm in ten seconds.

Goal
- 10 µm ± 0.4 µm
  At the start of the step: 4.141 µm
Setup, choices and feedback
The Scene at this step
Water temperature
293.1 K
Speck size
1 µm
Zoom to the molecules
One speck
Read from the Scene at
Typical distance wandered in 10 s, √(4D·10 s)/nodes/grain/outputs/spreadIn10s
One setting that reaches it
Water temperature
338.1 K
Speck size
0.5 µm
Feedback for a right answer
To get 10 µm in ten seconds takes $D = \dfrac{(10^{-5})^2}{40} = 2.5 \times 10^{-12}$ ⁠ square metres per second, six times the room-temperature value: water near boiling and a speck under a micrometre.

Hint
You need a big $D$ ⁠: hot water, which is also runnier, and a small speck.

Worked solution
1. The distance in ten seconds is $\sqrt{4D \times 10}$ ⁠. For it to be $10^{-5}$ ⁠ metres, $D = \dfrac{10^{-10}}{40} = 2.5 \times 10^{-12}$ ⁠.
2. At room temperature a 1 µm speck gives only $4.3 \times 10^{-13}$ ⁠, so you need about six times the $D$ ⁠.
3. Heating the water helps twice: $T$ ⁠ on top grows, and the viscosity underneath falls up to fourfold.
4. Boiling water, 373 K, and a speck of about 0.8 µm reach the goal.

Across the unit · 11 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

Heat and temperature

Particles that jiggle

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Kelvin and degrees

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Conduction

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Why metal feels cold

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

Specific heat

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Convection

Question

Correct answer

The Scene at this step

Choices

Worked solution

Radiation

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Entropy and heat engines

Two sides

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

The Carnot limit

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

The books of heat and work

Question

Correct answer

The Scene at this step

Once the prediction is locked in, the Scene changes

Worked solution

Entropy

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Seventy per cent

Question

Goal