Answer key · Unit 23

Ice and climate.

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

Glaciers and ice sheets

Snow that outlasts the summer piles up into ice, and the ice flows like a very slow river. See how the summer sets an ice sheet's size, where the ice flows fastest, how far it pushes the ground down and how much water it takes from the sea.

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The 6 steps and their answers

Step 1 of 6
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Snow that outlasts the summer
Practice · Predict, then look
Question
The Summer temperature on the coast is at Today, 6 °C, and the Snowfall 0.35 m a year: there is no ice sheet, because every summer melts the winter’s snow. Move the summer temperature to Ice age, −2 °C, without changing the snow. What will happen?

Correct answer
- Correct An ice sheet thousands of kilometres wide will grow
Setup, choices and feedback
The Scene at this step
Summer temperature
279.1 K
Snowfall
0.35 m/yr
Look under the ice
Surface
Choices
- Correct An ice sheet thousands of kilometres wide will grow
- Wrong Nothing, because what builds ice is a cold winter, not a cool summer
  Misconception: Thinks ice sheets grow because winters get colder, so summers do not matter.
  Feedback from the catalogue
  Near the pole winter is always cold enough to snow. What decides whether snow stays is the summer: if it is too weak to melt all the winter’s snow, a little is left each year, and it piles up into ice. So ice sheets grow when northern summers cool, even if the winters are milder.
- Wrong Only a thin layer of snow, because the snowfall has not changed
  Feedback
  Snow piles up year after year. When the summer does not melt it all, what is left builds up over thousands of years into ice kilometres thick.
Once the prediction is locked in, the Scene changes
Summer temperature
271.1 K
The Scene then shows: 2036 km · Width of the ice sheet across the section, margin to coast/nodes/ice/outputs/width
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
A sheet about 2,000 km wide: the cool summer no longer melts the winter’s snow.

Worked solution
1. A glacier starts as snow that does not melt. Where more snow falls in winter than the summer can melt, a little is left each year.
2. With a summer of −2 °C on the coast, the summer on the high ground is too cold to melt all the snow.
3. The snow piles up, is squeezed into firn and then ice, and the sheet grows until it balances: about 2,000 km wide and more than 3 km thick.
The snow did not change, only the summer. That is how ice sheets grow: from weak summers, not harsh winters.
Step 2 of 6
Open this step
A budget of snow
Practice · Reach the target
Question
The Snowfall is locked at 0.35 m a year. Find the Summer temperature at which the ice sheet is about 1,000 km wide, and press Check.

Goal
- 1000 km ± 60 km
  At the start of the step: 1768 km
Setup, choices and feedback
The Scene at this step
Summer temperature
272.1 K
Snowfall
0.35 m/yrfixed
Look under the ice
Surface
Read from the Scene at
Width of the ice sheet across the section, margin to coast/nodes/ice/outputs/width
One setting that reaches it
Summer temperature
275.1 K
Feedback for a right answer
About 2 °C on the coast: three degrees warmer than an ice age, and a sheet half as wide.

Feedback for a wrong answer
Not yet. If the sheet is too wide, warm the summer; if too narrow, cool it.

Hint
A warmer summer raises the equilibrium line and shrinks the sheet.

Worked solution
1. Every ice sheet keeps a budget: in the high accumulation zone snowfall wins, and in the low ablation zone the summer melts more than fell.
2. A warmer summer raises the equilibrium line between them, and the sheet shrinks until it balances again.
3. A width of about 1,000 km comes with a summer of about 2 °C on the coast.
A little more warmth melts a lot: a few degrees more, and the sheet’s surface sinks into warmer air, and it disappears.
Step 3 of 6
Open this step
Solid ice flows
Practice · Choose
Question
Look under the ice is at Flow inside. The arrows give the ice’s speed at five depths in a few places, and the dots are parcels of snow moving with the ice. Where does the ice flow fastest?

Correct answer
- Correct Near the surface, and especially near the edges
Setup, choices and feedback
The Scene at this step
Summer temperature
272.1 K
Snowfall
0.35 m/yr
Look under the ice
Flow inside
Choices
Shown in a random order.
- Correct Near the surface, and especially near the edges
- Wrong Nowhere: ice is a solid, and it stays where the snow fell
  Misconception: Thinks the ice in a glacier or ice sheet stays where it fell, because ice is solid.
  Feedback from the catalogue
  Ice is a solid, but under its own weight it creeps: its crystals slide along layers inside them, like a very stiff paste. A pile of ice more than about fifty metres thick spreads outward, so an ice sheet flows from its high middle towards its edges, faster near its surface and near the edges.
- Wrong At the bottom, under the ice divide, where the ice is thickest
  Feedback
  Near the divide the ice hardly moves, and at the bottom the ground holds it back. Each layer slides over the one beneath, so the top one is fastest.
Feedback for a right answer
At the top and near the edges: the ice flows like a very slow river.

Worked solution
1. Under its own weight ice creeps, and an ice sheet spreads from its high middle, the ice divide, towards its edges.
2. Each layer slides over the one beneath it, and the ground holds back the bottom layer, so the ice is fastest near its surface.
3. Near the edges all the snow from upstream has to pass, so the ice is faster there: hundreds of metres a year, and kilometres in ice streams.
Snow that falls near the divide sinks deep and moves slowly along the bottom for a hundred thousand years. That is why the oldest ice lies deep under the middle, where ice cores are drilled.
Step 4 of 6
Open this step
The ground sinks
Practice · Calculate
Question
Look under the ice is at Under the ice, and the climate is locked. The dashed line is where the ground would be with no ice. The ice over the divide is about 3.1 km thick, and its density about 917 kg/m³ against about 3300 kg/m³ for the mantle. By how many metres has the ground sunk under the divide?

Correct answer
835 m ± 8%
Setup, choices and feedback
The Scene at this step
Summer temperature
272.1 Kfixed
Snowfall
0.35 m/yrfixed
Look under the ice
Under the ice
Accepted within
± 8%
Computed from the Scene at this step’s setting
How far the crust has sunk under the divide/nodes/ice/outputs/sag
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0 m ± 20 m
  Misconception: Thinks the ground under an ice sheet stays put, because rock is solid.
  Feedback from the catalogue
  The Earth’s crust floats on the mantle, which over thousands of years flows like a very thick fluid. Under three kilometres of ice it sinks, like a boat being loaded, until the mantle pushed aside weighs as much as the ice: about a quarter of the ice’s thickness. When the ice melts, the ground rises again, slowly.
Feedback for a right answer
About 840 m: about a quarter of the ice’s thickness.

Hint
The mantle pushed aside weighs as much as the ice: $\rho_{\text{ice}} h_{\text{ice}} = \rho_{\text{mantle}} h_{\text{sag}}$ ⁠.

Worked solution
1. The crust floats on the mantle, and like a loaded boat it sinks until the mantle pushed aside weighs as much as the load.
2. $h_{\text{sag}} = h_{\text{ice}} \cdot \dfrac{\rho_{\text{ice}}}{\rho_{\text{mantle}}} = 3100 \times \dfrac{917}{3300} \approx 860$ ⁠ metres.
3. In the Scene the ground has sunk about 835 m: about a quarter of the ice’s thickness, a little less, because the stiff crust bends over a wide area.
When the ice melts the ground rises, slowly: around Hudson Bay it is still rising by about a centimetre a year.
Step 5 of 6
Open this step
Water missing from the sea
Practice · Calculate
Question
Ice age, and the climate is locked. Every tonne of ice on land is a tonne of water taken from the sea. The chart beside the Scene shows how much water the sheet takes. By how many metres does it lower the sea level across the whole world?

Correct answer
20.78 m ± 5%
Setup, choices and feedback
The Scene at this step
Summer temperature
271.1 Kfixed
Snowfall
0.35 m/yrfixed
Look under the ice
Surface
Accepted within
± 5%
Computed from the Scene at this step’s setting
Change in global sea level from the water held in the ice (negative: lower)/nodes/ice/outputs/seaLevel
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
About 21 m: water moved from the sea to the land.

Hint
Read the fall in sea level on the chart, at the current summer temperature.

Worked solution
1. The ice-age sheet is about 2,000 km wide and about 3.4 km thick in the middle.
2. Its water, spread over all the oceans, an area of about 360 million km², is a layer about 21 m deep.
3. So the sea falls by about 21 m.
At the height of the last ice age, about $20{,}000$ ⁠ years ago, several such sheets together lowered the sea by about 120 m, and you could walk from Siberia to Alaska.
Step 6 of 6
Open this step
Floating ice
Practice · Choose
Question
The floating sea ice of the Arctic, and the floating ice shelves at the edge of Antarctica, are melting. What does that do to the sea level, compared with melting the same amount of ice resting on land?

Correct answer
- Correct Floating ice hardly raises the sea: it already displaces its own weight of water
Setup, choices and feedback
The Scene at this step
Summer temperature
271.1 K
Snowfall
0.35 m/yr
Look under the ice
Surface
Choices
Shown in a random order.
- Correct Floating ice hardly raises the sea: it already displaces its own weight of water
- Wrong The same: ice is ice, and when it melts it adds water
  Misconception: Thinks melting sea ice or floating ice shelves raises the sea level as much as melting ice on land.
  Feedback from the catalogue
  Floating ice already displaces its own weight of water, by Archimedes’ principle. When it melts, the water it makes fills exactly the volume it displaced, so the sea level hardly changes, like ice cubes melting in a glass. Only ice resting on land, such as the ice sheets of Greenland and Antarctica, raises the sea when it melts.
- Wrong It lowers the sea, because water takes up less room than ice
  Feedback
  Water does take up less room than the ice, but exactly the volume of the part below the water, so the sea does not fall.
Feedback for a right answer
Almost nothing: only ice on land raises the sea.

Worked solution
1. By Archimedes’ principle, a floating body displaces its own weight of water.
2. When floating ice melts, the water it makes weighs the same, so it fills exactly the volume it displaced.
3. Ice on land displaced no water. When it melts and runs into the sea, it adds its whole weight to the sea.
So what raises the sea is the ice of Greenland, about 7 m if it all melted, and of Antarctica, about 58 m more.

Lesson 2 of 3 · About 25 minutes

Ice ages and Milankovitch cycles

Three slow changes in the Earth's orbit and axis change the summer sunlight in the far north. Work out the distance to the Sun and the gap in sunlight, see why the summer and not the winter decides the ice, and line summer sunlight up with the ice ages of the last 400,000 years.

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The 7 steps and their answers

Step 1 of 7
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An oval orbit
Practice · Calculate
Question
The Eccentricity is locked at today’s value, $e = 0.0167$ ⁠. The Earth’s mean distance from the Sun, the orbit’s semi-major axis, is about 149.6 million kilometres. How close to the Sun is the Earth at its closest point, perihelion?

Correct answer
147100000 km ± 0.2%
Setup, choices and feedback
The Scene at this step
Eccentricity
0.0167fixed
Axial tilt
23.44 °
Precession: closest point
283 °
Years of ice
The orbit
Accepted within
± 0.2%
Computed from the Scene at this step’s setting
Earth–Sun distance at the closest point, for an orbit 1 AU across its long half-axis/nodes/orbit/outputs/perihelionDistance
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
About 147.1 million kilometres, in early January.

Hint
At the closest point the distance is $a(1-e)$ ⁠.

Worked solution
1. In the Unit on the Earth, the Moon and the Sun we saw that the Earth’s orbit is an ellipse, with the Sun at one focus.
2. At the closest point: $r = a(1 - e) = 149.6 \times (1 - 0.0167) \approx 147.1$ ⁠ million kilometres.
3. At the farthest point: $a(1 + e) \approx 152.1$ ⁠ million.
Today the Earth is closest in early January and farthest in early July: a difference of only about 3%.
Step 2 of 7
Open this step
Closest in winter
Practice · Choose
Question
The Earth is closest to the Sun in early January. Why is January the middle of winter in the northern hemisphere?

Correct answer
- Correct Because the seasons come from the axial tilt: in January the north leans away from the Sun, and the distance changes the light by only about 7%
Setup, choices and feedback
The Scene at this step
Eccentricity
0.0167
Axial tilt
23.44 °
Precession: closest point
283 °
Years of ice
The orbit
Choices
Shown in a random order.
- Correct Because the seasons come from the axial tilt: in January the north leans away from the Sun, and the distance changes the light by only about 7%
- Wrong That is a mistake: the Earth is in fact farthest in January, which is why it is cold
  Misconception: Thinks the seasons come from the Earth being nearer to or farther from the Sun.
  Feedback from the catalogue
  The Earth is closest to the Sun in early January, in the middle of the northern winter. The distance changes by only about 3%, and the sunlight by about 7%. The seasons come from the axial tilt: in summer your hemisphere leans towards the Sun, it stands high in the sky and the day is long. That is why it is winter in the south when it is summer in the north.
- Wrong Because the Sun’s heat takes half a year to arrive
  Feedback
  Sunlight reaches us in about eight minutes. The oceans do delay the seasons, by a few weeks, not half a year.
Feedback for a right answer
The tilt: it sets the seasons, not the distance.

Worked solution
1. The change in distance is small: between 147.1 and 152.1 million kilometres. The light changes as the square of the distance, by about 7%.
2. The axial tilt, 23.44°, matters far more: in summer the Sun is high in the sky, its rays fall almost straight down, and the day is long.
3. In January the northern hemisphere leans away from the Sun, so it is winter, even though the Earth is close.
That is why in the south, which then leans towards the Sun, January is midsummer.
Step 3 of 7
Open this step
The most oval orbit
Practice · Calculate
Question
The Eccentricity is locked at $e = 0.057$ ⁠, the most oval the orbit gets. How many times more sunlight does the Earth receive at its closest point than at its farthest?

Correct answer
1.256 ± 1%
Setup, choices and feedback
The Scene at this step
Eccentricity
0.057fixed
Axial tilt
23.44 °
Precession: closest point
283 °
Years of ice
The orbit
Accepted within
± 1%
Computed from the Scene at this step’s setting
Sunlight at the closest point, times that at the farthest/nodes/orbit/outputs/perihelionBoost
Wrong answers it recognises
- 1.121 ± 0.8%
  Misconception: Takes sunlight as falling in proportion to distance, not to the square of the distance.
  Feedback from the catalogue
  Sunlight spreads out over an ever larger sphere, and a sphere’s area grows with the square of its radius. So the light’s intensity falls as $\dfrac{1}{r^2}$ ⁠: at twice the distance, a quarter of the light. The ratio of the sunlight at the closest point to that at the farthest is $\left(\dfrac{1+e}{1-e}\right)^2$ ⁠, not $\dfrac{1+e}{1-e}$ ⁠.
- 1.057 ± 0.8%
  Misconception: Takes sunlight as falling in proportion to distance, not to the square of the distance.
  Feedback from the catalogue
  Sunlight spreads out over an ever larger sphere, and a sphere’s area grows with the square of its radius. So the light’s intensity falls as $\dfrac{1}{r^2}$ ⁠: at twice the distance, a quarter of the light. The ratio of the sunlight at the closest point to that at the farthest is $\left(\dfrac{1+e}{1-e}\right)^2$ ⁠, not $\dfrac{1+e}{1-e}$ ⁠.
Feedback for a right answer
About $1.26$ ⁠: a quarter more sunlight at the closest pass.

Hint
The light falls as $\dfrac{1}{r^2}$ ⁠, and the distances are $a(1-e)$ ⁠ and $a(1+e)$ ⁠.

Worked solution
1. The distances: $a(1 - 0.057) = 0.943a$ ⁠ and $a(1 + 0.057) = 1.057a$ ⁠.
2. The light’s intensity falls as $\dfrac{1}{r^2}$ ⁠, so the ratio is $\left(\dfrac{1.057}{0.943}\right)^2$ ⁠.
3. $\left(1.121\right)^2 \approx 1.26$ ⁠: about 26% more sunlight.
An answer of $1.12$ ⁠ forgot the square: the light spreads over an area that grows as the square of the distance.
Step 4 of 7
Open this step
Northern summers
Practice · Predict, then look
Question
Everything at today’s values: the Earth is closest in January, and midsummer sunlight at 65° north is about 478 W/m². Move Precession: closest point to June. What will happen to the northern ice cap?

Correct answer
- Correct It will shrink: northern summers will get more sunlight and melt more snow
Setup, choices and feedback
The Scene at this step
Eccentricity
0.0167
Axial tilt
23.44 °
Precession: closest point
283 °
Years of ice
The orbit
Choices
- Correct It will shrink: northern summers will get more sunlight and melt more snow
- Wrong It will grow: the northern winter will be at the far point and grow colder
  Misconception: Thinks ice sheets grow because winters get colder, so summers do not matter.
  Feedback from the catalogue
  Near the pole winter is always cold enough to snow. What decides whether snow stays is the summer: if it is too weak to melt all the winter’s snow, a little is left each year, and it piles up into ice. So ice sheets grow when northern summers cool, even if the winters are milder.
- Wrong Nothing, because the total sunlight in a year does not change
  Feedback
  The total hardly changes, but the share between the seasons does, and ice sheets depend on the strength of the summer.
Once the prediction is locked in, the Scene changes
Precession: closest point
90 °
The Scene then shows: 510.5 W/m² · Daily-mean sunlight at the top of the air at 65° N on the June solstice/nodes/orbit/outputs/summerSunlight
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Shrink: about 511 W/m² in summer, enough to melt the snow.

Worked solution
1. When the closest point falls in June, the northern summer happens while the Earth is near the Sun.
2. Midsummer sunlight at 65° north rises from about 478 to about 511 W/m².
3. A strong summer melts the winter’s snow, and the ice cap retreats north.
The northern winter will be farther from the Sun, but near the pole winter is cold enough to snow anyway. What decides is the summer.
Step 5 of 7
Open this step
A weak summer
Practice · Reach the target
Question
Precession: closest point is locked at 270°, in the northern winter. Change the Eccentricity and the Axial tilt so that midsummer sunlight at 65° north falls to about 430 W/m², a weak summer that starts an ice age, and press Check.

Goal
- 430 W/m² ± 10 W/m²
  At the start of the step: 477.5 W/m²
Setup, choices and feedback
The Scene at this step
Eccentricity
0.0167
Axial tilt
23.44 °
Precession: closest point
270 °fixed
Years of ice
The orbit
Read from the Scene at
Daily-mean sunlight at the top of the air at 65° N on the June solstice/nodes/orbit/outputs/summerSunlight
One setting that reaches it
Eccentricity
0.0336
Axial tilt
22 °
Feedback for a right answer
A weak summer: the ice cap spreads south.

Feedback for a wrong answer
Not yet. To weaken the summer, lower the tilt and raise the eccentricity; to strengthen it, the reverse.

Hint
Less tilt lowers the summer Sun over the north. A more oval orbit takes the summer farther from the Sun, when the closest point is in winter.

Worked solution
1. Smaller Axial tilt: the Sun stands lower in summer over the far north.
2. Large Eccentricity, with the closest point in winter: in summer the Earth is far from the Sun.
3. For example an eccentricity of $0.057$ ⁠ and a tilt of 23° give about 435 W/m², and a tilt of 22.1° about 422 W/m².
The three cycles have to line up: on a nearly round orbit, the closest point hardly matters.
Step 6 of 7
Open this step
The record
Practice · Choose
Question
Choose 400,000 years of ice under Years of ice. The chart shows summer sunlight at 65° north beside the temperature from Antarctic ice cores. What is true of the ice ages?

Correct answer
- Correct Each warm spell begins as northern summer sunlight peaks, and feedbacks, such as melting ice and rising carbon dioxide, amplify the nudge
Setup, choices and feedback
The Scene at this step
Eccentricity
0.0167
Axial tilt
23.44 °
Precession: closest point
283 °
Years of ice
400,000 years of ice
Choices
Shown in a random order.
- Correct Each warm spell begins as northern summer sunlight peaks, and feedbacks, such as melting ice and rising carbon dioxide, amplify the nudge
- Wrong In ice ages the whole Earth is farther from the Sun, and gets far less sunlight in a year
  Misconception: Thinks ice ages happen because the whole Earth moves farther from the Sun for thousands of years.
  Feedback from the catalogue
  The orbit’s average size hardly changes, and Milankovitch’s changes alter the total sunlight the Earth receives in a year by less than a fifth of a per cent. What they change is how the light is shared between the seasons and latitudes, above all the strength of the far-northern summer. Feedbacks, such as ice reflecting sunlight and carbon dioxide falling, amplify that small nudge.
- Wrong Ice ages begin when the northern winters are coldest
  Misconception: Thinks ice sheets grow because winters get colder, so summers do not matter.
  Feedback from the catalogue
  Near the pole winter is always cold enough to snow. What decides whether snow stays is the summer: if it is too weak to melt all the winter’s snow, a little is left each year, and it piles up into ice. So ice sheets grow when northern summers cool, even if the winters are milder.
Feedback for a right answer
The northern summer sets the beat, and the feedbacks the size.

Worked solution
1. The cycles change the year’s total sunlight by less than a fifth of a per cent, but the northern summer sunlight by tens of watts per square metre.
2. The warm spells, the interglacials, begin as northern summer sunlight climbs to a peak: about $11{,}000$ ⁠, $128{,}000$ ⁠, $240{,}000$ ⁠ and $335{,}000$ ⁠ years ago.
3. The nudge is small, and feedbacks do most of the work: melting ice reflects less sunlight, and warmer oceans release carbon dioxide, from about 180 to about 280 ppm.
That is why the orbit was called the “pacemaker of the ice ages”: it sets the beat, and the climate amplifies it.
Step 7 of 7
Open this step
Farthest out
Practice · Calculate
Question
Eccentricity is locked at $e = 0.057$ ⁠, the most oval the orbit gets. The Earth’s mean distance from the Sun, the semi-major axis, is 1 AU. How far is it from the Sun at its farthest point, $a(1 + e)$ ⁠?

Correct answer
1.057 AU ± 0.2%
Setup, choices and feedback
The Scene at this step
Eccentricity
0.057fixed
Axial tilt
23.44 °
Precession: closest point
283 °
Years of ice
The orbit
Accepted within
± 0.2%
Computed from the Scene at this step’s setting
Earth–Sun distance at the farthest point/nodes/orbit/outputs/aphelionDistance
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.943 AU ± 0.2%
  Misconception: Mixes up the nearest and farthest points of an orbit: subtracts ae for the farthest.
  Feedback from the catalogue
  In an elliptical orbit the Sun is not in the middle but at one focus, $ae$ ⁠ from the centre. So the farthest point, aphelion, is $a + ae$ ⁠ from the Sun, and the nearest, perihelion, $a - ae$ ⁠.
Feedback for a right answer
1.057 AU, about 158 million kilometres: 6 million kilometres farther than today’s farthest point.

Hint
At the farthest point you add $ae$ ⁠ to the semi-major axis. At the nearest you take it away.

Worked solution
1. $a = 1$ ⁠ and $e = 0.057$ ⁠.
2. $a(1 + e) = 1.057$ ⁠ astronomical units, about 158 million kilometres.
3. At the nearest point: $a(1 - e) = 0.943$ ⁠.
The ratio of the two distances is $\dfrac{1.057}{0.943} \approx 1.12$ ⁠, and sunlight goes with the square of the distance: at the nearest point 1.26 times as much light arrives. Today the difference is only about 7%.

Lesson 3 of 3 · About 25 minutes

Climate

Sunlight comes in, heat goes out, and a few gases in the air decide how much of it escapes. Work out how cold the Earth would be with no greenhouse gas, how much heat the extra carbon dioxide holds back, why each doubling adds the same, and how much warmer that makes us.

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The 6 steps and their answers

Step 1 of 6
Open this step
The balance
Practice · Calculate
Question
Greenhouse gas is at None, and Sunlight is locked at 100%. The Earth absorbs on average about 238 W/m² of sunlight. With no greenhouse gas, the ground has to radiate all of it straight to space: $\sigma T^4 = 238\ \mathrm{W/m^2}$ ⁠, with $\sigma = 5.67 \times 10^{-8}\ \mathrm{W/(m^2\,K^4)}$ ⁠. What will its temperature be?

Correct answer
-18.57 °C ± 2 °C
Setup, choices and feedback
The Scene at this step
Greenhouse gas
0 ppm
Sunlight
100 %fixed
Show the energy
On
Accepted within
± 2 °C
Computed from the Scene at this step’s setting
The ground temperature that would radiate the absorbed sunlight straight to space, with no greenhouse gas/nodes/sky/outputs/bareTemperature
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 15 °C ± 2 °C
  Misconception: Thinks a gas that makes up a few hundredths of a per cent of the air is too little to matter.
  Feedback from the catalogue
  Nitrogen and oxygen, more than 99% of the air, are transparent to both sunlight and infrared, so they play no part. Carbon dioxide and water vapour do a job nothing else in the air does. Like a thin blanket on a bed with nothing else on it: it is thin, but it is the only thing holding the heat in.
Feedback for a right answer
About −18 °C: a frozen ball. The greenhouse effect adds about 33 °C.

Hint
$T = \left(\dfrac{238}{\sigma}\right)^{1/4}$ ⁠, in kelvin. Then subtract $273$ ⁠.

Worked solution
1. The Sun gives 1361 W/m² to each square metre facing it. The Earth catches it on a disc but spreads it over a sphere four times larger, and about 30% is reflected to space: $\dfrac{1361}{4} \times 0.7 \approx 238$ ⁠.
2. $T^4 = \dfrac{238}{5.67 \times 10^{-8}} \approx 4.2 \times 10^{9}$ ⁠.
3. $T \approx 255\ \mathrm{K}$ ⁠, that is about −18 °C.
The real average temperature is about 15 °C: the difference, about 33 °C, is the greenhouse effect.
Step 2 of 6
Open this step
The trap
Practice · Choose
Question
Show the energy is On, and Greenhouse gas at Pre-industrial. Short yellow waves are sunlight, and long red waves are infrared given off by the warm ground. What do the carbon dioxide molecules do?

Correct answer
- Correct Let sunlight through, and catch some of the infrared and give it off again in every direction, down too
Setup, choices and feedback
The Scene at this step
Greenhouse gas
280 ppm
Sunlight
100 %
Show the energy
On
Choices
Shown in a random order.
- Correct Let sunlight through, and catch some of the infrared and give it off again in every direction, down too
- Wrong Catch sunlight on its way in and warm the air
  Misconception: Thinks greenhouse gases warm the Earth by trapping sunlight on its way in.
  Feedback from the catalogue
  Greenhouse gases are almost transparent to sunlight, so it passes through them and warms the ground. They are not transparent to the infrared the warm ground gives off upwards: they absorb some of it and give it off again in every direction, down too. If they blocked sunlight, they would cool the ground.
- Wrong Almost nothing: there are too few of them, less than a tenth of a per cent of the air
  Misconception: Thinks a gas that makes up a few hundredths of a per cent of the air is too little to matter.
  Feedback from the catalogue
  Nitrogen and oxygen, more than 99% of the air, are transparent to both sunlight and infrared, so they play no part. Carbon dioxide and water vapour do a job nothing else in the air does. Like a thin blanket on a bed with nothing else on it: it is thin, but it is the only thing holding the heat in.
Feedback for a right answer
Transparent to sunlight, not to infrared: that is the greenhouse effect.

Worked solution
1. Sunlight, mostly visible light in short waves, passes through the gas and warms the ground.
2. The warm ground glows in infrared, in waves about twenty times longer. At those waves carbon dioxide is not transparent.
3. A molecule that catches a packet flashes and gives it off in a random direction, so some of the heat goes back to the ground.
Nitrogen and oxygen, more than 99% of the air, are transparent to both kinds of light. That is why a rare gas doing a job nothing else does matters so much.
Step 3 of 6
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Radiative forcing
Practice · Calculate
Question
Greenhouse gas is locked at Today, 420 ppm, against 280 ppm before the industrial revolution. Radiative forcing is how many watts per square metre fewer escape to space because of the extra gas. For carbon dioxide $\Delta F = 5.35 \ln\dfrac{C}{C_0}\ \mathrm{W/m^2}$ ⁠. What is the forcing today?

Correct answer
2.169 W/m² ± 3%
Setup, choices and feedback
The Scene at this step
Greenhouse gas
420 ppmfixed
Sunlight
100 %
Show the energy
On
Accepted within
± 3%
Computed from the Scene at this step’s setting
Radiative forcing of the gas against pre-industrial 280 ppm, 5.35 ln(C/280) (Myhre et al. 1998); NaN with no gas/nodes/sky/outputs/forcing
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 2.675 W/m² ± 2%
  Misconception: Thinks each extra part per million of CO₂ adds the same forcing, so the effect grows in proportion to the amount.
  Feedback from the catalogue
  The centre of carbon dioxide’s absorption band is already saturated: there the air stops almost all the infrared. Extra gas works mainly at the edges of the band, so each doubling of the amount adds about the same, about 3.7 W/m²: $\Delta F = 5.35 \ln\dfrac{C}{C_0}$ ⁠. Going from 560 to 1120 ppm adds as much as going from 280 to 560, even though it is twice as much gas.
Feedback for a right answer
About 2.2 W/m²: one per cent of the energy coming in.

Hint
$\dfrac{C}{C_0} = \dfrac{420}{280} = 1.5$ ⁠.

Worked solution
1. $\dfrac{C}{C_0} = \dfrac{420}{280} = 1.5$ ⁠.
2. $\ln 1.5 \approx 0.405$ ⁠.
3. $\Delta F = 5.35 \times 0.405 \approx 2.17\ \mathrm{W/m^2}$ ⁠.
That is about one per cent of the 238 W/m² the Earth absorbs, and it is enough to warm it by more than a degree. An answer of $2.7$ ⁠ worked out $5.35 \times 0.5$ ⁠, as if the forcing grew in proportion to the amount of gas.
Step 4 of 6
Open this step
Each doubling
Practice · Predict, then look
Question
Greenhouse gas is at 560 ppm, double the amount before the industrial revolution: the forcing is about 3.7 W/m². Raise it to 1120 ppm, another doubling. What will the forcing be?

Correct answer
- Correct About 7.4 W/m²: each doubling adds another 3.7 or so
Setup, choices and feedback
The Scene at this step
Greenhouse gas
560 ppm
Sunlight
100 %
Show the energy
On
Choices
- Correct About 7.4 W/m²: each doubling adds another 3.7 or so
- Wrong About 11 W/m²: we added three times as much gas as in the first doubling
  Misconception: Thinks each extra part per million of CO₂ adds the same forcing, so the effect grows in proportion to the amount.
  Feedback from the catalogue
  The centre of carbon dioxide’s absorption band is already saturated: there the air stops almost all the infrared. Extra gas works mainly at the edges of the band, so each doubling of the amount adds about the same, about 3.7 W/m²: $\Delta F = 5.35 \ln\dfrac{C}{C_0}$ ⁠. Going from 560 to 1120 ppm adds as much as going from 280 to 560, even though it is twice as much gas.
- Wrong About 3.7 W/m²: the gas already absorbs all it can
  Misconception: Thinks CO₂ already absorbs all the infrared it can, so adding more changes nothing.
  Feedback from the catalogue
  At the band’s centre, yes: there the air already stops almost everything. But at the edges of the band light still escapes, and extra gas catches some of it. And even at the centre, extra gas raises the height from which heat escapes to space, to colder air that radiates less. So each doubling still adds about 3.7 W/m², and the effect grows with the logarithm of the amount, it does not stop.
Once the prediction is locked in, the Scene changes
Greenhouse gas
1120 ppm
The Scene then shows: 7.417 W/m² · Radiative forcing of the gas against pre-industrial 280 ppm, 5.35 ln(C/280) (Myhre et al. 1998); NaN with no gas/nodes/sky/outputs/forcing
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
About 7.4 W/m²: the same step for each doubling.

Worked solution
1. $\Delta F = 5.35 \ln\dfrac{1120}{280} = 5.35 \ln 4$ ⁠.
2. $\ln 4 = 2 \ln 2$ ⁠, so $\Delta F = 2 \times 5.35 \ln 2 \approx 2 \times 3.71 = 7.4\ \mathrm{W/m^2}$ ⁠.
3. The second doubling added exactly as much as the first, even though it added twice as much gas.
The centre of the absorption band is already saturated, so extra gas works at its edges. The effect grows with the logarithm of the amount: more slowly than a straight line, but without stopping.
Step 5 of 6
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How much warmer
Practice · Calculate
Question
Greenhouse gas is locked at 560 ppm. Once the climate settles, each 1 W/m² of forcing warms the Earth by about 0.8 °C, feedbacks included. By how many degrees will the Earth warm in the end, for a doubling of carbon dioxide?

Correct answer
3 °C ± 5%
Setup, choices and feedback
The Scene at this step
Greenhouse gas
560 ppmfixed
Sunlight
100 %
Show the energy
On
Accepted within
± 5%
Computed from the Scene at this step’s setting
Equilibrium warming against pre-industrial, a temperature difference in kelvin, for a climate sensitivity of 3 K per doubling: 3 log₂(C/280); NaN with no gas/nodes/sky/outputs/warming
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0 °C ± 0.2 °C
  Misconception: Thinks CO₂ already absorbs all the infrared it can, so adding more changes nothing.
  Feedback from the catalogue
  At the band’s centre, yes: there the air already stops almost everything. But at the edges of the band light still escapes, and extra gas catches some of it. And even at the centre, extra gas raises the height from which heat escapes to space, to colder air that radiates less. So each doubling still adds about 3.7 W/m², and the effect grows with the logarithm of the amount, it does not stop.
Feedback for a right answer
About $3$ ⁠ degrees, for one doubling.

Hint
Take the forcing for a doubling, and multiply by $0.8$ ⁠.

Worked solution
1. For a doubling, $\Delta F = 5.35 \ln 2 \approx 3.7\ \mathrm{W/m^2}$ ⁠.
2. $\Delta T \approx 0.8 \times 3.7 \approx 3.0$ ⁠ degrees.
3. With no feedbacks it would be only about $1.2$ ⁠ degrees. Extra water vapour and melting ice amplify the warming about two and a half times.
This is the climate sensitivity. Most studies put it at about $3$ ⁠ degrees, between $2.5$ ⁠ and $4$ ⁠. Clouds are the largest uncertainty.
Step 6 of 6
Open this step
Feedbacks
Practice · Choose
Question
Warming sets off changes that in turn change it. Which of them amplify the warming? Choose all that are right.

Correct answer
- Correct Warmer air holds more water vapour, itself a greenhouse gas
- Correct Melting ice uncovers dark sea and land, which absorb more sunlight
Setup, choices and feedback
The Scene at this step
Greenhouse gas
840 ppm
Sunlight
100 %
Show the energy
On
Choices
Shown in a random order. More than one choice is correct: every one must be selected.
- Correct Warmer air holds more water vapour, itself a greenhouse gas
- Correct Melting ice uncovers dark sea and land, which absorb more sunlight
- Wrong Warmer ground glows more strongly in infrared
  Feedback
  That is a steadying feedback: warmer ground radiates more heat, which is what finally stops the warming at a new temperature.
- Wrong The Sun warms up when there is more carbon dioxide
  Feedback
  The gases in our air have no effect on the Sun. It gets its energy from fusion in its core.
Feedback for a right answer
Water vapour and ice: both amplify the warming.

Worked solution
1. Water vapour: for each degree the air holds about 7% more vapour, which catches more infrared. Amplifies.
2. Ice: snow and ice reflect most sunlight to space. When they melt, dark sea and land absorb it. Amplifies.
3. The ground’s glow grows as $T^4$ ⁠, so warmer ground loses more heat. That is the feedback that steadies the temperature in the end.
Clouds cut both ways: by day they shade, and by night they act as a blanket. They are the largest uncertainty.

Across the unit · 12 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

Glaciers and ice sheets

Snow that outlasts the summer

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

A budget of snow

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

Solid ice flows

Question

Correct answer

The Scene at this step

Choices

Worked solution

The ground sinks

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Water missing from the sea

Question

Correct answer

The Scene at this step

Hint

Worked solution

Floating ice

Question

Correct answer

The Scene at this step

Choices

Worked solution

Ice ages and Milankovitch cycles

An oval orbit

Question

Correct answer

The Scene at this step

Hint

Worked solution

Closest in winter

Question

Correct answer

The Scene at this step

Choices

Worked solution

The most oval orbit

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Northern summers

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

A weak summer

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

The record

Question

Correct answer

The Scene at this step

Choices

Worked solution