Heat transfer, energy balance and climate change
Why is the mean temperature of the Earth about $+15\,{}^\circ\text{C}$ and not below freezing, even though the Earth radiates as much energy into space as it absorbs from the Sun?
Prior knowledge
- Heat and work change internal energy, $\Delta U = Q + W$, and energy is conservedchapter Internal energy, heat and conservation of energy
- Heat is the transfer of energy from hotter to colder because of a temperature differencechapter Internal energy, heat and conservation of energy
Core path: follow the steps in order. Tasks marked Optional give extra practice. Optional tasks: 5.
Think about it
The Moon is on average as far from the Sun as the Earth, and the same amount of sunlight hits each square metre of its surface. However, the Moon has no atmosphere, and the mean temperature of its surface is well below freezing. The same difference can be seen on a small scale on Earth: on a clear winter night the frost gets harsher quickly, but on a cloudy night the temperature falls only a little.
The atmosphere does not produce energy itself. How can it still keep the ground warmer than it would be without it?
Optional Review from the previous lesson
Predict
Make your predictions before investigating 2 questions
Prediction 1: why is the ground warmer?
Prediction 2: when carbon dioxide is added
Investigate
Predict the radiation of the atmosphere downwards
Next you study a model of the energy balance of the ground, in which $240\,\text{W/m}^2$ is absorbed by the ground from the Sun and the ground gives out energy only as thermal radiation. Without an atmosphere the surface radiates $240\,\text{W/m}^2$ and is in balance ($-18\,{}^\circ\text{C}$). With an atmosphere the surface is in balance at $+15\,{}^\circ\text{C}$ and radiates $390\,\text{W/m}^2$.
How much power per square metre must the atmosphere radiate down to the ground?
Hint 1: a guiding question
When the ground is in balance, the power coming into it is equal to the power leaving it.
Hint 2: what is the connection
The ground receives $240\,\text{W/m}^2$ from the Sun and an unknown power from the atmosphere; the power leaving is $390\,\text{W/m}^2$.
Show the solution
Given
with the ground in balance, $P_\text{sun} = 240\,\tfrac{\text{W}}{\text{m}^2}$ is absorbed from the Sun and the surface radiates $P_\text{rad} = 390\,\tfrac{\text{W}}{\text{m}^2}$; the surface gives out energy only as thermal radiation
Find
the power radiated down by the atmosphere $P_\text{down}$ in $\tfrac{\text{W}}{\text{m}^2}$
Principle
The temperature of the surface stays constant ($+15\,{}^\circ\text{C}$), so energy neither accumulates in the surface nor decreases: the power coming into the surface is equal to the power leaving it. There are two inputs, the Sun and the thermal radiation radiated down by the atmosphere, and only one output, the radiation of the surface itself.
Formula
In balance the power coming in is equal to the power leaving:
$$P_\text{sun} + P_\text{down} = P_\text{rad}$$
$$P_\text{down} = P_\text{rad} - P_\text{sun}$$
Conversions
No conversions are needed, because all the powers are given in the same unit $\tfrac{\text{W}}{\text{m}^2}$.
Substitution
Answer
$150\,\tfrac{\text{W}}{\text{m}^2}$
Three significant figures. The example values $390$ and $240\,\tfrac{\text{W}}{\text{m}^2}$ are given as whole numbers, so the difference has the same precision in W/m²; no data limits it any further.
Reasonableness
Check the balance: $240 + 150 = 390\,\tfrac{\text{W}}{\text{m}^2}$ comes in and $390\,\tfrac{\text{W}}{\text{m}^2}$ goes out. The result is positive and smaller than the radiation of the surface, because part of the power leaving is already covered by the Sun. Without an atmosphere $P_\text{down} = 0$ and the surface would radiate only $240\,\tfrac{\text{W}}{\text{m}^2}$, which corresponds to the colder situation.
Energy balance of the ground without and with an atmosphere
Values calculated by a model
The table shows the energy balance of the ground as a mean over the whole Earth per square metre. The numbers are calculated with a simple model and are example values, not measurements. In the model the power absorbed from the Sun is absorbed entirely by the ground, and the ground gives out energy only as thermal radiation. In reality part of the sunlight is absorbed by the atmosphere, and the ground also gives out energy by convection and by the evaporation of water. In both cases a total of $240\,\text{W/m}^2$ leaves into space, as much as is absorbed from the Sun.
The column without is an Earth with no atmosphere, and the column with is an Earth with one. The powers are in W/m².
| Ground | without | with |
|---|---|---|
| absorbed from the Sun | 240 | 240 |
| comes down from the atmosphere | 0 | 150 |
| surface radiates | 240 | 390 |
| temperature | $-18\,{}^\circ\text{C}$ | $+15\,{}^\circ\text{C}$ |
When the balance is broken
Values calculated by a model
The carbon dioxide concentration of the atmosphere is raised all at once and then kept constant. There are more greenhouse gases, so they absorb more thermal radiation, and at first less power leaves into space than the Earth absorbs. The extra power warms the oceans, the land and the air. In the model the power leaving into space increases by $1.2\,\text{W/m}^2$ for every kelvin the Earth warms. The table shows the power leaving into space, the excess, that is, the difference between the absorbed power and the power leaving, and the warming $\Delta T$ from the initial situation; the powers are in W/m². The Earth absorbs $240.0\,\text{W/m}^2$ all the time. The numbers are example values; in reality the warming takes decades, because the large heat capacity of the oceans slows it down.
| Stage | Leaves | Excess | $\Delta T$ (K) |
|---|---|---|---|
| before | 240.0 | 0.0 | 0.0 |
| just after | 236.4 | 3.6 | 0.0 |
| later | 237.6 | 2.4 | 1.0 |
| later still | 238.8 | 1.2 | 2.0 |
The concentration stays constant from the row "just after" onwards.
How many kelvins has the Earth warmed in total when the power leaving is again equal to the power absorbed?
Hint 1: a guiding question
By how much must the power leaving still increase from the value in the row "just after" for the excess to be zero?
Hint 2: what is the connection
The power leaving increases by $1.2\,\text{W/m}^2$ for every kelvin of warming. Just after the change the excess is $3.6\,\text{W/m}^2$.
Hint 3: an intermediate step
The warming needed is $\dfrac{3.6\,\text{W/m}^2}{1.2\,\text{W/(m}^2\text{K)}}$.
Show the solution
Given
row "just after" of the table: power leaving $236.4\,\tfrac{\text{W}}{\text{m}^2}$ and excess $3.6\,\tfrac{\text{W}}{\text{m}^2}$; absorbed power $240.0\,\tfrac{\text{W}}{\text{m}^2}$; in the model the power leaving increases by $k = 1.2\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$ for every kelvin of warming
Find
the total warming $\Delta T$ in kelvins when the power leaving is again equal to the power absorbed
Principle
The concentration stays constant, so the absorbed power stays the same $240.0\,\tfrac{\text{W}}{\text{m}^2}$. The Earth warms while the excess is positive, and as it warms it radiates more. Balance is reached when the warming has increased the power leaving by exactly the excess.
Formula
The power leaving must increase by the excess $\Delta P$, and the increase is $k\,\Delta T$:
$$k\,\Delta T = \Delta P \quad\Rightarrow\quad \Delta T = \frac{\Delta P}{k}$$
Conversions
No conversions are needed, because the excess and the coefficient are given in the units $\tfrac{\text{W}}{\text{m}^2}$ and $\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$ and the square metres cancel.
Substitution
Answer
$3.0\,\text{K}$
Two significant figures. The limiting data are the excess $240.0 - 236.4 = 3.6\,\tfrac{\text{W}}{\text{m}^2}$ (the difference has one decimal and two figures) and the coefficient $1.2\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$ (two figures).
Reasonableness
Check the balance: the power leaving is $236.4 + 1.2 \cdot 3.0 = 240.0\,\tfrac{\text{W}}{\text{m}^2}$, equal to the power absorbed. The result fits the table: at two kelvins the excess is still $1.2\,\tfrac{\text{W}}{\text{m}^2}$ and after the third kelvin it is zero. The warming therefore continues for a long time after the concentration has stopped increasing, but it stops at a new balance.
Conclusion: why does added carbon dioxide warm the Earth?
Based on the tables, write briefly why the ground is warmer with an atmosphere and what happens when carbon dioxide is added. Use the words balance, excess and warming.
Model answer
With an atmosphere, the ground receives atmospheric radiation of $150\,\text{W/m}^2$ in addition to the $240\,\text{W/m}^2$ from the Sun, and in balance the surface radiates the same amount, $390\,\text{W/m}^2$. Only a warmer surface can do that, $+15\,{}^\circ\text{C}$ and not $-18\,{}^\circ\text{C}$. The atmosphere does not produce energy, it returns part of the radiation of the surface. When carbon dioxide is added, at first less leaves into space than is absorbed (an excess of $3.6\,\text{W/m}^2$). The excess warms the Earth, and a warming Earth radiates more, so the excess decreases. A new balance is reached when the warming is $3.6 / 1.2 = 3.0\,\text{K}$. The tables are example values of a simple model; the real values are based on satellite and surface measurements and on climate models.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Theory summary
Key idea
Because of a temperature difference, energy transfers by itself, as a net amount, from a warmer object to a colder one. Energy that transfers in this way is called heat, and it transfers in three ways. In conduction, energy passes through matter from one particle to the next, as faster vibrating particles nudge their slower neighbours. Metals conduct heat well, because in them free electrons also carry energy, and they carry most of it; air and wool conduct it poorly. In convection, a warm liquid or gas flows and carries energy with it: warm air rises above a radiator, because it has expanded and is lighter than the surrounding air.
In thermal radiation, an object emits electromagnetic radiation, which needs no medium. This is how the Sun's energy reaches the Earth through empty space, and how the glow of a fireplace warms your face. All objects radiate, and the hotter an object is, the more radiation it emits and the shorter the wavelengths of that radiation are. Thermal radiation therefore leaves both objects, so the colder object also radiates towards the warmer one. The warmer one radiates more, however, so the net transfer of energy is to the colder object.
Heat losses of a building
A large part of the heating energy of a house is used because heat is conducted out through the walls, roof, windows and floor. The heat insulation of a building part is described by its U-value, or thermal transmittance, which tells you how many watts of thermal power pass through one square metre when the difference between the indoor and outdoor temperature is one kelvin. The smaller the U-value, the better the structure insulates. The U-value of a modern wall is about $0.17\,\text{W/(m}^2\text{}\cdot\text{K)}$, but even a good window has about $1.0\,\text{W/(m}^2\text{}\cdot\text{K)}$, so a small window lets through as much heat as many times its area of wall. Good insulation saves energy and reduces the emissions from heating.
Energy balance
The temperature of an object stays constant when the power coming into it is equal to the power leaving it. This is called energy balance. The Earth receives energy as radiation from the Sun, and about $30\,\%$ of it is reflected straight back into space by clouds, ice and light surfaces. The rest is absorbed, and the Earth radiates the same power away as long-wavelength thermal radiation. Without an atmosphere, the balance would settle at a mean temperature of about $-18\,{}^\circ\text{C}$.
Greenhouse effect
Water vapour, carbon dioxide and methane in the atmosphere let sunlight mostly through, although about a fifth of it is absorbed in the atmosphere. They absorb the thermal radiation emitted by the ground and radiate part of it back down. The ground therefore receives more energy than from the Sun alone, and the balance settles at about $+15\,{}^\circ\text{C}$. This is called the greenhouse effect, and it answers the question at the start.
Climate change
Human activity, mainly the use of fossil fuels, has raised the carbon dioxide concentration of the atmosphere from about 280 parts per million to over 420 parts per million. When greenhouse gases are added, for a while less energy escapes into space than arrives from the Sun. Today the difference is about $1\,\text{W}$ for every square metre of the Earth. The extra energy warms the oceans, the land and the air until a new balance settles at a higher temperature. This is called climate change. A power difference that looks small is huge over the whole area of the Earth, and most of it is stored in the oceans, whose large heat capacity slows the warming down but also keeps it going for a long time. Energy choices such as insulation, heat pumps and zero-emission power plants are therefore also ethical choices that affect future generations.
Thermal power and U-value
The thermal power conducted through a building part is
$$P = U A \Delta T$$
where $P$ is the thermal power (W), $U$ is the U-value of the building part (W/(m²·K)), $A$ is the area (m²) and $\Delta T$ is the difference between the indoor and outdoor temperature (K). The formula holds when the temperatures on both sides of the building part stay constant. In energy balance the power coming into an object and the power leaving it are equal, and the temperature stays constant.
Common misconception
Many people mix up the greenhouse effect and ozone depletion, although they are different things: ozone depletion lets more ultraviolet radiation through, but it hardly warms the Earth. Another mistake is to think that the greenhouse effect is entirely harmful. The natural greenhouse effect keeps the Earth habitable, and the problem is its strengthening because of human emissions. A third mistake is to believe that a woollen sweater produces heat. It only slows down the conduction and convection of heat away from the body, just as the insulation in a wall slows down the escape of heat from a house.
Summary
- Heat transfers by conduction, convection and thermal radiation, and because of a temperature difference the net transfer of energy is from the warmer object to the colder one. All objects emit thermal radiation, the colder one too.
- The thermal power conducted through a building part is $P = UA\Delta T$. A small U-value means good heat insulation.
- In energy balance the power coming in and the power going out are equal, and the temperature stays constant.
- Greenhouse gases absorb the thermal radiation of the ground and radiate part of it back, which raises the equilibrium temperature of the ground from about $-18\,{}^\circ\text{C}$ to about $+15\,{}^\circ\text{C}$.
- An increase in greenhouse gases breaks the balance for a while, and the Earth warms until a new balance settles at a higher temperature.
Optional Return to your predictions
Compare your predictions with the two tables and the model of energy balance. Then return to the question at the start: how can the atmosphere keep the ground warmer, even though it does not produce energy itself? Why does the frost get less harsh on a cloudy night than on a clear one?
Would you change your prediction?
Apply
Worked example
Problem
The walls of a detached house have an area of 120 m² and a U-value of $0.17\,\text{W/(m}^2\text{}\cdot\text{K)}$, and the windows have an area of 20.0 m² and a U-value of $1.0\,\text{W/(m}^2\text{}\cdot\text{K)}$. It is $21\,{}^\circ\text{C}$ indoors and $-15\,{}^\circ\text{C}$ outdoors. How large is the thermal power conducted through the walls and windows, and how much energy does it transfer in one day?
Step 1/7
Given and find. Walls: $A_\text{s} = 120\,\text{m}^2$, $U_\text{s} = 0.17\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$. Windows: $A_\text{i} = 20.0\,\text{m}^2$, $U_\text{i} = 1.0\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$. Temperatures $T_\text{in} = 21\,{}^\circ\text{C}$ and $T_\text{out} = -15\,{}^\circ\text{C}$. You are asked for the thermal power $P$ and the energy $E$ in one day.
Step 2/7
Principle. Heat is conducted through a structure from warm to cold, and when the temperatures stay constant, the power is proportional to the area and the temperature difference: $P = UA\Delta T$. The walls and windows conduct heat in parallel, so their powers are added. The power is constant, so the energy is the power multiplied by the time.
Step 3/7
Rearranged formula. First solve the power and then the energy:
$$\begin{aligned} P &= (U_\text{s} A_\text{s} \\ &\quad + U_\text{i} A_\text{i})\,\Delta T \\[6pt] E &= Pt \end{aligned}$$
Step 4/7
Unit conversions. The temperature difference is the same in kelvins and in degrees Celsius:
The time is $t = 24\,\text{h}$, and the power is converted to kilowatts so that the energy comes out in kilowatt-hours.
Step 5/7
Substitution. Calculate the power and then the energy:
$$\begin{aligned} P &= (0.17 \cdot 120 \\ &\quad + 1.0 \cdot 20.0)\,\tfrac{\text{W}}{\text{K}} \cdot 36\,\text{K} \\ &= (20.4 + 20.0)\,\tfrac{\text{W}}{\text{K}} \cdot 36\,\text{K} \\ &= 1454.4\,\text{W} \\ E &= 1.4544\,\text{kW} \cdot 24\,\text{h} \\ &= 34.9\,\text{kWh} \end{aligned}$$
Step 6/7
Result. The power conducted through the walls and windows is $P \approx 1.5\,\text{kW}$, and in one day it transfers the energy $E \approx 35\,\text{kWh}$. The result is given to two significant figures. The limiting data are the U-values $0.17$ and $1.0$ and the temperature difference $36\,\text{K}$ (a difference of whole degrees), all with two figures; the areas $120\,\text{m}^2$ and $20.0\,\text{m}^2$ are more precise.
Step 7/7
Reasonableness. The area of the windows is only a sixth of the area of the walls, but the same amount of heat goes through them ($20.0\,\text{W/K}$ and $20.4\,\text{W/K}$), because their U-value is almost six times as large. With rounded numbers, $40\,\text{W/K} \cdot 36\,\text{K} \approx 1.4\,\text{kW}$ and $1.4\,\text{kW} \cdot 24\,\text{h} \approx 34\,\text{kWh}$. The heating of the house replaces this power, and the roof, floor and ventilation add to the need.
Optional Find the error
Optional Apply to a new situation
The total area of the walls, door, top and bottom of a refrigerator is 4.0 m², and their mean U-value is $0.30\,\text{W/(m}^2\text{}\cdot\text{K)}$. It is $5\,{}^\circ\text{C}$ in the refrigerator and $21\,{}^\circ\text{C}$ in the kitchen. Heat is conducted through the walls into the refrigerator, and for the temperature of the refrigerator to stay constant, its machinery must move the same energy back to the kitchen.
How much heat is conducted into the refrigerator in one day? Assume that the door is not opened. Give the answer in kilowatt-hours.
Hint 1: a guiding question
In which direction does heat transfer, and what is the temperature difference between the inside and the outside?
Hint 2: what is the connection
The power conducted through the walls is $P = UA\Delta T$, and the energy in one day is $E = Pt$.
Hint 3: an intermediate step
Multiply by the hours and convert to kilowatt-hours.
Show the solution
Given
$A = 4.0\,\text{m}^2$, $U = 0.30\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$, $T_\text{in} = 5\,{}^\circ\text{C}$, $T_\text{kitchen} = 21\,{}^\circ\text{C}$, $t = 1\,\text{day}$; the door is not opened
Find
the energy $E$ conducted into the refrigerator in one day, in kilowatt-hours
Principle
Heat is conducted from the warm kitchen into the cold refrigerator, and when the temperatures stay constant, the power is proportional to the area and the temperature difference: $P = UA\Delta T$. The power is constant, because the door is not opened, so the energy is the power multiplied by the time.
Formula
$P = UA\Delta T$ and $E = Pt$, so
$$E = UA\Delta T\,t$$
Conversions
The temperature difference is the same in kelvins:
The time is $t = 24\,\text{h}$, and the power is converted to kilowatts so that the energy comes out in kilowatt-hours.
Substitution
Answer
$0.46\,\text{kWh}$
Two significant figures. The limiting data are the area $4.0\,\text{m}^2$, the U-value $0.30$ and the difference of whole degrees $\Delta T = 21\,{}^\circ\text{C} - 5\,{}^\circ\text{C} = 16\,\text{K}$, all with two figures, so the calculator result $0.4608\,\text{kWh}$ is rounded.
Reasonableness
With rounded numbers $20\,\text{W} \cdot 24\,\text{h} \approx 0.5\,\text{kWh}$, so the result is correct. The temperature of the refrigerator stays constant only if the machinery moves the same energy out. The power is only about $20\,\text{W}$, because the temperature difference is small and the walls are well insulated. Opening the door would increase the need, because warm air would get in by convection.
Check your understanding
Exit ticket
Answer on your own first and only then open the answer.
- The U-value of a wall is $0.20\,\text{W/(m}^2\text{}\cdot\text{K)}$ and its area is 50 m². It is $20\,{}^\circ\text{C}$ indoors and $-10\,{}^\circ\text{C}$ outdoors. How large is the thermal power transferred through the wall?
Answer
$P$$= UA\Delta T$$= 0.20\,\tfrac{\text{W}}{\text{m}^2\text{K}} \cdot 50\,\text{m}^2 \cdot 30\,\text{K}$$= 300\,\text{W}$
1. Concept
2. How the heat is transferred
Optional 3. Energy balance of the atmosphere
In the model of the first table of the Investigate stage the ground radiates $390\,\text{W/m}^2$. Of this, $100\,\text{W/m}^2$ passes straight through the atmosphere into space, and the atmosphere absorbs the rest. The atmosphere radiates $150\,\text{W/m}^2$ down to the ground. The whole Earth is in balance, so a total of $240\,\text{W/m}^2$ leaves into space.
How much power per square metre does the atmosphere radiate upwards into space? The numbers are example values.
Hint 1: a guiding question
What two parts does the $240\,\text{W/m}^2$ leaving into space consist of?
Hint 2: what is the connection
What leaves into space is the radiation that passed straight through from the ground and the power radiated upwards by the atmosphere: $240\,\text{W/m}^2 = 100\,\text{W/m}^2 + P_\text{up}$.
Hint 3: an intermediate step
Finally check the balance of the atmosphere itself: it absorbs $390\,\text{W/m}^2 - 100\,\text{W/m}^2 = 290\,\text{W/m}^2$, and up and down together it must radiate the same amount.
Show the solution
Given
radiation of the surface $390\,\tfrac{\text{W}}{\text{m}^2}$, of which $P_\text{direct} = 100\,\tfrac{\text{W}}{\text{m}^2}$ passes straight into space; radiated down by the atmosphere $150\,\tfrac{\text{W}}{\text{m}^2}$; the total leaving into space is $P_\text{tot} = 240\,\tfrac{\text{W}}{\text{m}^2}$
Find
the power radiated upwards into space by the atmosphere $P_\text{up}$ in $\tfrac{\text{W}}{\text{m}^2}$
Principle
The whole Earth is in balance, so the power leaving into space is equal to the power absorbed from the Sun, and it consists of two parts: radiation that passes straight through from the ground and radiation sent upwards by the atmosphere. Energy is conserved, and there is no other route out.
Formula
The power leaving into space is the sum of the parts:
$$P_\text{direct} + P_\text{up} = P_\text{tot}$$
$$P_\text{up} = P_\text{tot} - P_\text{direct}$$
Conversions
No conversions are needed, because all the powers are given in the unit $\tfrac{\text{W}}{\text{m}^2}$.
Substitution
Answer
$140\,\tfrac{\text{W}}{\text{m}^2}$
Three significant figures. The example values $240$ and $100\,\tfrac{\text{W}}{\text{m}^2}$ are given as whole numbers, so the difference has the same precision in W/m²; no data limits it any further.
Reasonableness
Check the balance of the atmosphere itself: it absorbs $390 - 100 = 290\,\tfrac{\text{W}}{\text{m}^2}$ and radiates $150 + 140 = 290\,\tfrac{\text{W}}{\text{m}^2}$, so the atmosphere is in balance too. The result is smaller than $240\,\tfrac{\text{W}}{\text{m}^2}$, because part of the power leaving into space comes straight from the surface. The atmosphere radiates slightly more down than up, because the lower air is warmer.
Suggestion: one answer does not show everything you can do, but review the theory summary and do the core exercises.
4. Explain and transfer
Your friend claims: "The Earth radiates as much energy into space as it absorbs from the Sun, so extra carbon dioxide cannot change its temperature. Besides, the greenhouse effect is entirely harmful." Use energy balance and thermal radiation to explain what is right and what is wrong in the claim.
Model answer
The first part of the claim is true in a state of balance: when the Earth radiates into space the same power as it absorbs from the Sun, its temperature stays constant. The balance is not permanent, however. When carbon dioxide is added, the atmosphere absorbs more of the thermal radiation of the ground and radiates part of it back down, and for a while less power leaves into space than the Earth absorbs. The extra power warms the oceans, the land and the air. A warmer Earth radiates more, so the warming continues until the power leaving is again equal to the power absorbed. A new balance therefore settles at a higher temperature. The greenhouse effect is not entirely harmful either: the natural greenhouse effect raises the mean temperature of the ground from about $-18\,{}^\circ\text{C}$ to about $+15\,{}^\circ\text{C}$. The problem is its strengthening because of human emissions.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Review the chapter's concepts
Cards you mark Review later collect in your review deck.
- Conduction
- Energy is transferred through the interactions of the particles of matter without any bulk motion of the matter.
- Convection
- Energy is carried along with a moving liquid or gas.
- Thermal radiation
- Energy is transferred as electromagnetic radiation and needs no medium.
- Thermal insulator
- A material in which energy is transferred slowly by conduction.
- Energy balance
- The energy of a system stays constant when as much energy enters it as leaves it.
- Radiative equilibrium
- The Earth's temperature stays steady when the power received equals the power radiated to space.
- Greenhouse effect
- The atmosphere absorbs part of the Earth's thermal radiation and raises the mean surface temperature.
- Greenhouse gas
- A gas in the atmosphere that absorbs and emits infrared radiation.
- Carbon dioxide
- A long-lived greenhouse gas whose concentration grows with the use of fossil fuels.
- Albedo
- The fraction of the radiation arriving at a surface that is reflected back.
- Ice–albedo feedback
- Less ice means less reflection and more of the Sun's energy absorbed.
- Ocean warming
- The large specific heat capacity of water lets the oceans store a large amount of energy.
- Sea-level rise
- Caused mainly by melting land ice and the thermal expansion of sea water.
- Melting of sea ice
- Floating ice melting does not in itself raise the sea level significantly.
- Climate change
- The long-term warming of the Earth changes the energy balance, glaciers, oceans and other parts of the climate system.
Exercises
Level 1 · basic practice 8 exercises
One formula or one step of reasoning. Core exercises first, the rest is extra practice.
- Exercise 15-1Core
The roof of a log cabin has an area of $45\,\text{m}^2$ and a U-value of $0.24\,\text{W/(m}^2\text{}\cdot\text{K)}$. It is $22\,{}^\circ\text{C}$ in the cabin and $-18\,{}^\circ\text{C}$ outdoors.
How large is the thermal power conducted out through the roof?
Answer
$430\,\text{W}$
Show the solution
Given
$A = 45\,\text{m}^2$, $T_\text{in} = 22\,{}^\circ\text{C}$, $T_\text{out} = -18\,{}^\circ\text{C}$; roof $U = 0.24\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$
Find
the thermal power $P$ conducted through the roof, in watts
Principle
Heat is conducted through the structure from warm to cold, and when the temperatures stay constant, the power is proportional to the area and the temperature difference. The U-value tells you how much power passes through one square metre when the difference is one kelvin.
Formula
$P = UA\Delta T$, where $\Delta T = T_\text{in} - T_\text{out}$; the power is already solved: $P = UA(T_\text{in} - T_\text{out})$
Conversions
The temperature difference is the same in degrees Celsius and in kelvins, so
$\Delta T$$= 22\,{}^\circ\text{C} - (-18\,{}^\circ\text{C})$$= 40\,\text{K}$No other conversions are needed, because $U$ is already in $\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$.
Substitution
$P$$= 0.24\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}} \cdot 45\,\text{m}^2 \cdot 40\,\text{K}$$= 432\,\text{W}$Answer
$430\,\text{W}$
Two significant figures. The limiting data are the U-value $0.24$, the area $45\,\text{m}^2$ and the difference of whole degrees $\Delta T = 40\,\text{K}$, all with two figures, so the calculator result $432\,\text{W}$ is rounded.
Reasonableness
With rounded numbers $0.25 \cdot 45 \cdot 40 \approx 450\,\text{W}$, so $432\,\text{W}$ is the right order of magnitude. Units:
$\tfrac{\text{W}}{\text{m}^2\cdot\text{K}} \cdot \text{m}^2 \cdot \text{K}$$= \text{W}$If it were the same temperature indoors and outdoors, $\Delta T = 0$ and $P = 0$, as it should be. The power corresponds to a small electric radiator, and the walls, windows and floor add to the heating need of the cabin.
- Exercise 15-2Core
A measurement on the front door of an old detached house shows that heat is conducted through the door at a power of $96\,\text{W}$ when it is $21\,{}^\circ\text{C}$ indoors and $-19\,{}^\circ\text{C}$ outdoors. The area of the door is $2.0\,\text{m}^2$.
What is the U-value of the door?
Answer
$1.2\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$
Show the solution
Given
$P = 96\,\text{W}$, $A = 2.0\,\text{m}^2$, $T_\text{in} = 21\,{}^\circ\text{C}$, $T_\text{out} = -19\,{}^\circ\text{C}$
Find
the U-value $U$ of the door, in $\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$
Principle
The door conducts heat out steadily when the temperatures stay constant, so the measured power is proportional to the area and the temperature difference. The constant of proportionality is exactly the U-value, and you get it by dividing the power by the area and the difference.
Formula
$P = UA\Delta T$, so $U = \dfrac{P}{A\Delta T}$
Conversions
The temperature difference is the same in kelvins and in degrees Celsius:
$\Delta T$$= 21\,{}^\circ\text{C} - (-19\,{}^\circ\text{C})$$= 40\,\text{K}$No other conversions are needed, because the power is in watts and the area in square metres.
Substitution
$U$$= \dfrac{96\,\text{W}}{2.0\,\text{m}^2 \cdot 40\,\text{K}}$$= \dfrac{96\,\text{W}}{80\,\text{m}^2\cdot\text{K}}$$= 1.2\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$Answer
$1.2\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$
Two significant figures. The limiting data are the power $96\,\text{W}$, the area $2.0\,\text{m}^2$ and the difference of whole degrees $\Delta T = 40\,\text{K}$, all with two figures.
Reasonableness
Check with the formula: $1.2 \cdot 2.0 \cdot 40 = 96\,\text{W}$, as measured. The U-value of a new front door is about $1.0\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$ and that of a modern wall $0.17\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$, so the value of the old door is of the same order as a new door and clearly worse than a wall, as you would expect.
- Exercise 15-3
The outer wall of a district-heated flat has an area of $28\,\text{m}^2$ and a U-value of $0.25\,\text{W/(m}^2\text{}\cdot\text{K)}$. On a frosty night heat is conducted out through the wall at a power of $210\,\text{W}$.
How large is the difference between the indoor and outdoor temperature?
Answer
$30\,\text{K}$
Show the solution
Given
$A = 28\,\text{m}^2$, $U = 0.25\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$, $P = 210\,\text{W}$
Find
the difference $\Delta T$ between the indoor and outdoor temperature, in kelvins
Principle
The power conducted through the wall is proportional to the temperature difference when the area and the U-value of the wall do not change. So from a known power you can work out how large the temperature difference across the wall is.
Formula
$P = UA\Delta T$, so $\Delta T = \dfrac{P}{UA}$
Conversions
No conversions are needed: the power is in watts, the area in square metres and $U$ in $\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$, so the result comes out directly in kelvins. A temperature difference in kelvins is the same as in degrees Celsius.
Substitution
$\Delta T$$= \dfrac{210\,\text{W}}{0.25\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}} \cdot 28\,\text{m}^2}$$= \dfrac{210\,\text{W}}{7.0\,\tfrac{\text{W}}{\text{K}}}$$= 30\,\text{K}$Answer
$30\,\text{K}$
Two significant figures. The limiting data are the power $210\,\text{W}$, the U-value $0.25$ and the area $28\,\text{m}^2$, all with at least two figures, so the result $30\,\text{K}$ is a two-figure value.
Reasonableness
For example, $21\,{}^\circ\text{C}$ indoors and $-9\,{}^\circ\text{C}$ outdoors gives a difference of $30\,\text{K}$, which is an ordinary winter night. Back-calculation: $0.25 \cdot 28 \cdot 30 = 210\,\text{W}$. Units: $\text{W} / (\text{W/K}) = \text{K}$.
- Exercise 15-4
On a frosty night heat is conducted out of a detached house at a mean power of $3.5\,\text{kW}$.
How many kilowatt-hours of heat must the heating of the house produce during a $14\,\text{h}$ night for the indoor temperature to stay the same?
Answer
$49\,\text{kWh}$
Show the solution
Given
$P = 3.5\,\text{kW}$, $t = 14\,\text{h}$
Find
the energy $E$ the heating must produce, in kilowatt-hours
Principle
The indoor temperature stays the same only if the heating produces the same energy as escapes from the house. The power is constant, so the energy is the power multiplied by the time.
Formula
$E = Pt$
Conversions
No conversions are needed, because a kilowatt multiplied by an hour is directly a kilowatt-hour: $1\,\text{kW} \cdot 1\,\text{h} = 1\,\text{kWh}$. For comparison in joules, $1\,\text{kWh} = 3.6\,\text{MJ}$.
Substitution
$E$$= 3.5\,\text{kW} \cdot 14\,\text{h}$$= 49\,\text{kWh}$Answer
$49\,\text{kWh}$
Two significant figures. The limiting data are the power $3.5\,\text{kW}$ and the time $14\,\text{h}$, each with two figures.
Reasonableness
With rounded numbers $3.5 \cdot 14 \approx 4 \cdot 12 = 48$, so the result is correct. In joules the energy is $49 \cdot 3.6\,\text{MJ} \approx 176\,\text{MJ}$. The consumption over a whole frosty day would be $3.5\,\text{kW} \cdot 24\,\text{h} = 84\,\text{kWh}$, which fits an electrically heated detached house.
- Exercise 15-5
The figure shows the energy flows of the Earth per square metre at the top of the atmosphere. The mean temperature of the Earth stays constant.
How much power per square metre leaves into space as thermal radiation? Read the values you need from the figure.
Answer
$239\,\tfrac{\text{W}}{\text{m}^2}$
Show the solution
Given
read from the figure, $341\,\tfrac{\text{W}}{\text{m}^2}$ comes from the Sun and $102\,\tfrac{\text{W}}{\text{m}^2}$ of it is reflected; the mean temperature of the Earth stays constant
Find
the power per square metre leaving into space as thermal radiation $P_\text{rad}$
Principle
The temperature stays constant, so the Earth is in energy balance: the power it absorbs is equal to the thermal radiation leaving into space. The reflected radiation does not warm the Earth, so the absorbed part is the incoming power minus the reflected power.
Reading the graph
The left arrow of the figure gives the incoming power $P_\text{in} = 341\,\tfrac{\text{W}}{\text{m}^2}$ and the upper right arrow the reflected power $P_\text{refl} = 102\,\tfrac{\text{W}}{\text{m}^2}$. The lower right arrow, whose value is a question mark, is the thermal radiation asked for.
Formula
$P_\text{abs} = P_\text{in} - P_\text{refl}$ and in balance $P_\text{rad} = P_\text{abs}$
Conversions
Not needed: all the powers are in the unit $\tfrac{\text{W}}{\text{m}^2}$.
Substitution
$P_\text{rad}$$= 341\,\tfrac{\text{W}}{\text{m}^2} - 102\,\tfrac{\text{W}}{\text{m}^2}$$= 239\,\tfrac{\text{W}}{\text{m}^2}$Answer
$239\,\tfrac{\text{W}}{\text{m}^2}$
Three significant figures, because the values read, $341$ and $102$, are given in whole watts per square metre and the difference is known with the same precision.
Reasonableness
Check the balance: $102 + 239 = 341\,\tfrac{\text{W}}{\text{m}^2}$, so all the incoming power is accounted for. The reflected fraction is $102 / 341 \approx 0.30$, that is, about $30\,\%$ of the incoming radiation. The result is smaller than the incoming power, as it should be.
- Exercise 15-6
You sit in front of a campfire on a windless evening, and your face feels warm even though the air between the campfire and your face is cool.
How does heat mainly transfer from the campfire to your face?
Answer
By thermal radiation, because the hot surface of the campfire emits radiation that travels through the air to the skin.
Show the solution
Find
how heat is transferred from the campfire to the face when the air between them is cool
Principle
Heat is transferred by conduction, by convection or as thermal radiation. Thermal radiation passes through a medium without warming it much, so it can warm a body even when there is cool air in between. Conduction and convection require heat to travel with the medium or through it.
Model answer
Heat is transferred mainly as thermal radiation: the hot surface of the campfire emits radiation that passes through the air and is absorbed by the skin. Convection does not explain it, because the air heated by the campfire rises upwards and does not flow horizontally to the face. Conduction through the air does not explain it either, because air is a poor conductor of heat and the air in between stays cool. The face does not touch the hot embers either, so direct conduction by contact is ruled out.
- Exercise 15-7
Which statement about the greenhouse effect is correct?
Answer
Greenhouse gases absorb the thermal radiation of the surface and radiate part of it back down, so the surface warms.
Show the solution
Find
which statement about the greenhouse effect is correct
Principle
The greenhouse effect arises from the interaction of the gases of the atmosphere with the thermal radiation of the surface. Greenhouse gases absorb the thermal radiation emitted by the surface and radiate part of it back down, which raises the equilibrium temperature of the surface.
Model answer
The correct statement is the one saying that greenhouse gases absorb the thermal radiation of the surface and radiate part of it back down, so the surface is warmer. Thanks to the effect, the mean temperature of the surface is about $+15\,{}^\circ\text{C}$ and not about $-18\,{}^\circ\text{C}$. Ozone depletion is a different matter: it increases ultraviolet radiation but hardly warms the Earth. Greenhouse gases do not prevent the radiation of the Sun from reaching the surface. The natural greenhouse effect exists even without human emissions, and the emissions strengthen it.
- Exercise 15-8
It is $21\,{}^\circ\text{C}$ inside a house and $-15\,{}^\circ\text{C}$ outside on a frosty night. Which statement about heat transfer through the wall is correct?
Answer
Heat is transferred, as a net amount, from indoors to outdoors, because the temperature difference transfers energy from the warmer to the colder.
Show the solution
Find
which statement about heat transfer through the wall is correct when it is $21\,{}^\circ\text{C}$ indoors and $-15\,{}^\circ\text{C}$ outdoors
Principle
A temperature difference transfers energy, as a net amount, from the warmer body to the colder one. Thermal radiation is emitted by both, however, and the warmer one emits more. The quantity that is transferred is energy, not cold.
Model answer
Heat is transferred as a net amount from indoors to outdoors, because the temperature difference transfers energy from the warmer to the colder. Cold does not move anywhere: the inner surface feels cold because heat is transferred from the skin and the room towards the wall. The transfer is not equal in both directions, because then the indoor temperature would stay the same without heating, which is not true in frost. Heat is conducted also through a solid wall, and insulation only reduces the power that passes through it.
Level 2 · applied practice 7 exercises
Two steps, working with measured data, a unit conversion or rearranging a formula. Core exercises first, the rest is extra practice.
- Exercise 15-9Core
The walls of a summer cabin have an area of $70\,\text{m}^2$ and a U-value of $0.50\,\text{W/(m}^2\text{}\cdot\text{K)}$. The roof has an area of $50.0\,\text{m}^2$ and a U-value of $0.30\,\text{W/(m}^2\text{}\cdot\text{K)}$. Heat conducted through other structures is ignored. On a winter weekend the cabin is heated by electric radiators with a total power of $2.0\,\text{kW}$.
What is the lowest outdoor temperature at which the radiators can keep the cabin at $20\,{}^\circ\text{C}$?
Answer
$-20\,{}^\circ\text{C}$
Show the solution
Given
walls $A_\text{s} = 70\,\text{m}^2$, $U_\text{s} = 0.50\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$; roof $A_\text{y} = 50.0\,\text{m}^2$, $U_\text{y} = 0.30\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$; power of the radiators $P = 2.0\,\text{kW}$; $T_\text{in} = 20\,{}^\circ\text{C}$
Find
the lowest outdoor temperature $T_\text{out}$ at which the radiators are enough
Principle
The walls and the roof conduct heat in parallel, so their powers are added. The indoor temperature stays constant when the power of the radiators is equal to the conducted power. The lowest outdoor temperature corresponds to the situation in which the radiators work at full power.
Formula
$P$$= (U_\text{s} A_\text{s} + U_\text{y} A_\text{y})(T_\text{in} - T_\text{out})$so
$T_\text{out}$$= T_\text{in} - \dfrac{P}{U_\text{s} A_\text{s} + U_\text{y} A_\text{y}}$Conversions
The power is converted to watts: $2.0\,\text{kW} = 2000\,\text{W}$. The temperature difference is the same in kelvins and in degrees Celsius, so the outdoor temperature can be calculated directly in degrees Celsius.
Substitution
$U_\text{s} A_\text{s} + U_\text{y} A_\text{y}$$= 0.50 \cdot 70\,\tfrac{\text{W}}{\text{K}} + 0.30 \cdot 50.0\,\tfrac{\text{W}}{\text{K}}$$= 50\,\tfrac{\text{W}}{\text{K}}$$T_\text{out}$$= 20\,{}^\circ\text{C} - \dfrac{2000\,\text{W}}{50\,\tfrac{\text{W}}{\text{K}}}$$= 20\,{}^\circ\text{C} - 40\,\text{K}$$= -20\,{}^\circ\text{C}$Answer
$-20\,{}^\circ\text{C}$
Two significant figures. The limiting data are the power $2.0\,\text{kW}$ and the conduction coefficient $50\,\tfrac{\text{W}}{\text{K}}$ (the sum $35 + 15.0$ keeps two figures, because the wall term $0.50 \cdot 70$ is a two-figure value), so $\Delta T = 40\,\text{K}$ and $T = 20\,{}^\circ\text{C} - 40\,\text{K} = -20\,{}^\circ\text{C}$ in whole degrees.
Reasonableness
Back-calculation: $50\,\tfrac{\text{W}}{\text{K}} \cdot 40\,\text{K} = 2000\,\text{W}$, which is the power of the radiators. The result is below the indoor temperature, as it should be. In colder weather the loss exceeds $2.0\,\text{kW}$, and the cabin cools until the balance settles at a lower indoor temperature.
- Exercise 15-10Core
The roof of a detached house has an area of $130\,\text{m}^2$ and a U-value of $0.090\,\text{W/(m}^2\text{}\cdot\text{K)}$. The heating season lasts 230 days, and the difference between the indoor and outdoor temperature is on average $21\,\text{K}$ during it. The house uses district heating, and the price of heat is $0.11\,\text{€/kWh}$.
How much does the heat escaping through the roof cost during the heating season?
Answer
$150\,\text{€}$
Show the solution
Given
$A = 130\,\text{m}^2$, $U = 0.090\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$, $\Delta T = 21\,\text{K}$, $t = 230\,\text{days}$, price $h = 0.11\,\tfrac{\text{€}}{\text{kWh}}$
Find
the cost $H$ of the escaping heat in euros over the heating season
Principle
The temperature difference stays constant on average, so the conducted power is constant and the energy is the power multiplied by the time. The cost is the amount of energy multiplied by the unit price.
Formula
$P = UA\Delta T$, $E = Pt$ and $H = hE$, so $H = hUA\Delta T\,t$
Conversions
The price is per kilowatt-hour, so the power is converted to kilowatts and the time to hours:
$t$$= 230 \cdot 24\,\text{h}$$= 5520\,\text{h}$and $P$ in watts is divided by a thousand.
Substitution
$P$$= 0.090\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}} \cdot 130\,\text{m}^2 \cdot 21\,\text{K}$$= 245.7\,\text{W}$$= 0.2457\,\text{kW}$$E$$= 0.2457\,\text{kW} \cdot 5520\,\text{h}$$= 1356.264\,\text{kWh}$$H$$= 0.11\,\tfrac{\text{€}}{\text{kWh}} \cdot 1356.264\,\text{kWh}$$\approx 149.2\,\text{€}$Answer
$150\,\text{€}$
Two significant figures. The limiting data are the U-value $0.090$, the price $0.11\,\tfrac{\text{€}}{\text{kWh}}$ and the temperature difference $21\,\text{K}$, all with two figures, so the calculator result $149.2\,\text{€}$ is rounded.
Reasonableness
With rounded numbers $0.25\,\text{kW} \cdot 5500\,\text{h} \approx 1400\,\text{kWh}$ and $1400 \cdot 0.11 \approx 150\,\text{€}$. A well-insulated roof, despite its large area, loses only about $250\,\text{W}$. District heating for a whole house typically costs over a thousand euros a year, so one building part is only a small share of the bill.
- Exercise 15-11
The old double-glazed windows of a post-war house have a U-value of $2.8\,\text{W/(m}^2\text{}\cdot\text{K)}$. They are replaced with triple-glazed windows with a U-value of $0.80\,\text{W/(m}^2\text{}\cdot\text{K)}$. The total area of the windows is $14\,\text{m}^2$.
How many kilowatt-hours of heat does the replacement save during a frosty week (7 days), when it is $21\,{}^\circ\text{C}$ indoors and $-24\,{}^\circ\text{C}$ outdoors for the whole week?
Answer
$210\,\text{kWh}$
Show the solution
Given
old $U_\text{v} = 2.8\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$, new $U_\text{u} = 0.80\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$, $A = 14\,\text{m}^2$, $T_\text{in} = 21\,{}^\circ\text{C}$, $T_\text{out} = -24\,{}^\circ\text{C}$, $t = 7\,\text{days}$
Find
the energy $E$ saved in a week, in kilowatt-hours
Principle
Through each type of window the power $UA\Delta T$ is conducted, with the same area and the same temperature difference, when the temperatures stay constant. The saving is the difference of the loss powers multiplied by the time.
Formula
$\Delta P = U_\text{v} A\Delta T - U_\text{u} A\Delta T$, so
$E$$= \Delta P\,t$$= (U_\text{v} - U_\text{u})A\Delta T\,t$Conversions
The temperature difference is the same in kelvins:
$\Delta T$$= 21\,{}^\circ\text{C} - (-24\,{}^\circ\text{C})$$= 45\,\text{K}$The time is
$t$$= 7 \cdot 24\,\text{h}$$= 168\,\text{h}$and the power is converted to kilowatts.
Substitution
$\Delta P$$= (2.8 - 0.80)\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}} \cdot 14\,\text{m}^2 \cdot 45\,\text{K}$$= 1260\,\text{W}$$= 1.26\,\text{kW}$$E$$= 1.26\,\text{kW} \cdot 168\,\text{h}$$= 211.68\,\text{kWh}$Answer
$210\,\text{kWh}$
Two significant figures. The limiting data are the difference of the U-values $2.8 - 0.80 = 2.0$ (the difference has the decimals of the less precise number $2.8$, so two figures), the area $14\,\text{m}^2$ and the temperature difference $45\,\text{K}$, all with two figures; the time $7\,\text{days} = 168\,\text{h}$ is exact.
Reasonableness
Check the balance: from the old windows $2.8 \cdot 14 \cdot 45\,\text{W} = 1.76\,\text{kW}$ escaped and from the new ones $0.80 \cdot 14 \cdot 45\,\text{W} = 0.50\,\text{kW}$, a difference of $1.26\,\text{kW}$. In a week the old ones would lose $296\,\text{kWh}$ and the new ones $85\,\text{kWh}$. The saving is greatest in frost, when $\Delta T$ is large.
- Exercise 15-12
Snow and ice reflect sunlight much better than a dark sea surface or the ground.
Suppose, hypothetically, that the reduction of snow and ice would decrease the fraction of the Sun's radiation reflected by the Earth from $30.0\,\%$ to $29.0\,\%$ (the real change is smaller). On average $340\,\text{W/m}^2$ arrives from the Sun.
By how much would the power absorbed by the Earth per square metre increase? Compare the result with the current energy imbalance, which is about $1.0\,\text{W/m}^2$.
Answer
$3.4\,\tfrac{\text{W}}{\text{m}^2}$
Show the solution
Given
$P_\text{in} = 340\,\tfrac{\text{W}}{\text{m}^2}$, $r_1 = 30.0\,\% = 0.300$, $r_2 = 29.0\,\% = 0.290$
Find
the increase in the absorbed power per square metre $\Delta P_\text{abs}$ and its comparison with the current imbalance $1.0\,\tfrac{\text{W}}{\text{m}^2}$
Principle
The incoming power divides into a reflected and an absorbed part, and only the absorbed part warms the Earth. When the reflected fraction decreases and the power of the Sun stays the same, the absorbed power increases.
Formula
$P_\text{abs} = (1 - r)P_\text{in}$, so
$\Delta P_\text{abs}$$= (1 - r_2)P_\text{in} - (1 - r_1)P_\text{in}$$= (r_1 - r_2)P_\text{in}$Conversions
The percentages are converted to decimal numbers: $30.0\,\% = 0.300$ and $29.0\,\% = 0.290$. The power stays in the unit $\tfrac{\text{W}}{\text{m}^2}$.
Substitution
$\Delta P_\text{abs}$$= (0.300 - 0.290) \cdot 340\,\tfrac{\text{W}}{\text{m}^2}$$= 0.010 \cdot 340\,\tfrac{\text{W}}{\text{m}^2}$$= 3.4\,\tfrac{\text{W}}{\text{m}^2}$Answer
$3.4\,\tfrac{\text{W}}{\text{m}^2}$
Two significant figures. There are two limiting data: the difference $0.300 - 0.290 = 0.010$ (two figures, because the percentages are given to a tenth of a percent) and the incoming power $340\,\tfrac{\text{W}}{\text{m}^2}$ (at least two figures). If the percentages had been given only as whole numbers, the difference $30 - 29 = 1$ would be a one-figure value. The result is $\dfrac{3.4}{1.0} = 3.4$ times the current imbalance.
Reasonableness
A separate calculation: the absorbed power increases from $238\,\tfrac{\text{W}}{\text{m}^2}$ to $241.4\,\tfrac{\text{W}}{\text{m}^2}$, a difference of $3.4\,\tfrac{\text{W}}{\text{m}^2}$. The increase is positive, because reflection decreases. A change of one percentage point is small, but large compared with the imbalance, and the warming melts ice, which increases absorption further.
- Exercise 15-13
Two identical cups are filled with $0.25\,\text{kg}$ of hot water at a temperature of $80.0\,{}^\circ\text{C}$. Cup A is open, and cup B has a lid. The cups are in the same room, and the temperatures are measured at five-minute intervals (example data). The heat capacity of the cup itself is ignored, and the specific heat capacity of water is $4190\,\text{J/(kg}\cdot\text{K)}$.
a) Calculate how large a mean thermal power the lid keeps from escaping during the first $10\,\text{min}$.
b) What does the lid prevent in the open cup, and why does the water in cup B still cool?
Enter the result of part a) in the answer box.
Temperature of the water as it cools (example data) $t\,(\text{min})$ $T_\text{A}\,({}^\circ\text{C})$ $T_\text{B}\,({}^\circ\text{C})$ 0 80.0 80.0 5 73.9 76.6 10 69.0 73.0 15 64.8 70.1 20 61.2 67.4 Answer
$7.0\,\text{W}$
Show the solution
Given
$m = 0.25\,\text{kg}$, $c = 4190\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$, initial temperature $80.0\,{}^\circ\text{C}$; from the table at the time $t = 10\,\text{min}$, $T_\text{A} = 69.0\,{}^\circ\text{C}$ (open) and $T_\text{B} = 73.0\,{}^\circ\text{C}$ (lid)
Find
the mean thermal power $P_\text{lid}$, in watts, that the lid prevents from escaping during the first $10\,\text{min}$
Principle
The decrease in the internal energy of the water is equal to the heat transferred to the surroundings when the phase does not change: $Q = cm\Delta T$. The cups and the room are alike, so the extra cooling of the open cup is due to the lack of a lid. Power is heat divided by time.
Formula
$\Delta T_\text{diff} = T_\text{B} - T_\text{A}$,
$Q$$= cm\Delta T_\text{diff}$$P_\text{lid}$$= \dfrac{Q}{t}$$= \dfrac{cm\,(T_\text{B} - T_\text{A})}{t}$Conversions
The time is converted to seconds:
$10\,\text{min}$$= 10 \cdot 60\,\text{s}$$= 600\,\text{s}$A temperature difference is the same in kelvins and in degrees Celsius.
Substitution
$T_\text{B} - T_\text{A}$$= 73.0\,{}^\circ\text{C} - 69.0\,{}^\circ\text{C}$$= 4.0\,\text{K}$$P_\text{lid}$$= \dfrac{4190\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}} \cdot 0.25\,\text{kg} \cdot 4.0\,\text{K}}{600\,\text{s}}$$= \dfrac{4190\,\text{J}}{600\,\text{s}}$$= 6.98\,\text{W}$Answer
$7.0\,\text{W}$
Two significant figures. The limiting datum is the temperature difference $4.0\,\text{K}$, which is a two-figure value.
Reasonableness
The open cup gives out
$\dfrac{4190 \cdot 0.25 \cdot 11.0}{600}\,\text{W}$$\approx 19.2\,\text{W}$and the cup with a lid
$\dfrac{4190 \cdot 0.25 \cdot 7.0}{600}\,\text{W}$$\approx 12.2\,\text{W}$so the difference is $19.2\,\text{W} - 12.2\,\text{W} = 7.0\,\text{W}$, as calculated. The lid thus removes about a third of the loss. The power is small compared with, for example, the losses of a house, which fits a small cup.
Model answer b)
The lid prevents above all evaporation: the evaporating water takes the heat of vaporisation with it, and convection carries the vapour and the warm air away from above the cup. The lid stops this flow, so the water in cup B cools more slowly. It still cools, because heat is conducted through the wall of the cup and the lid into the surrounding air and the cup radiates heat, and the lid does not prevent these. According to the table, the temperature of cup B falls from $80.0\,{}^\circ\text{C}$ to $67.4\,{}^\circ\text{C}$ in $20\,\text{min}$.
- Exercise 15-14
A student calculates how large a thermal power is conducted through a window. The window has an area of $1.6\,\text{m}^2$ and a U-value of $1.0\,\text{W/(m}^2\cdot\text{K)}$. It is $21\,{}^\circ\text{C}$ indoors and $-15\,{}^\circ\text{C}$ outdoors. The student's solution is: “The temperature difference is $21\,{}^\circ\text{C} - 15\,{}^\circ\text{C} = 6\,\text{K}$, so $P = UA\Delta T = 1.0 \cdot 1.6 \cdot 6\,\text{W} = 9.6\,\text{W}$.”
a) Where is the error in the student's solution, and how would you notice already with rounded numbers that the result is too small?
b) Calculate correctly the thermal power conducted through the window.
Enter the result of part b) in the answer box.
Answer
$58\,\text{W}$
Show the solution
Given
$A = 1.6\,\text{m}^2$, $U = 1.0\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$, $T_\text{in} = 21\,{}^\circ\text{C}$, $T_\text{out} = -15\,{}^\circ\text{C}$; the student's result $\Delta T = 6\,\text{K}$ and $P = 9.6\,\text{W}$
Find
the student's error and the correct thermal power $P$ in watts
Principle
The power conducted through a window is $P = UA\Delta T$, and $\Delta T$ is the difference between the indoor and the outdoor temperature. When it is frosty outside, the outdoor temperature is negative, and subtracting it increases the difference, because subtracting $-15$ is the same as adding $15$.
Formula
$\Delta T$$= T_\text{in} - T_\text{out}$$P$$= UA\Delta T$$= UA\,(T_\text{in} - T_\text{out})$Conversions
A temperature difference is the same in degrees Celsius and in kelvins, so no conversions are needed:
$\Delta T$$= 21\,{}^\circ\text{C} - (-15\,{}^\circ\text{C})$$= 36\,\text{K}$The other units are already $\text{m}^2$ and $\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$.
Substitution
$P$$= 1.0\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}} \cdot 1.6\,\text{m}^2 \cdot 36\,\text{K}$$= 57.6\,\text{W}$Answer
$58\,\text{W}$
Two significant figures. The limiting data are $U = 1.0$, $A = 1.6\,\text{m}^2$ and the difference of whole degrees $\Delta T = 36\,\text{K}$.
Reasonableness
The student calculated $21 - 15 = 6$, that is, treated the frost reading of the outdoor temperature as a positive number. The correct difference is $21 + 15 = 36\,\text{K}$. The error can be noticed with rounded numbers: the temperature difference between the room and the frost is about $20 + 15 = 35\,\text{K}$, so the power is about $1.0 \cdot 1.6 \cdot 35 \approx 56\,\text{W}$, not $9.6\,\text{W}$. The student's value is $36 / 6 = 6$ times too small. Back-calculation: $57.6 / (1.0 \cdot 1.6) = 36\,\text{K}$.
Model answer a)
The student calculated the temperature difference incorrectly: they subtracted $15$ from $21$ instead of subtracting $-15$. With rounded numbers the error can be noticed, because the difference between the room and the frost is clearly over $30\,\text{K}$, so the result $9.6\,\text{W}$ is about six times too small.
- Exercise 15-15
Students investigate the insulating ability of a wall panel. The area of the panel is $0.50\,\text{m}^2$. The inner side of the panel is heated with an electric resistor that keeps the temperatures constant, and the power $P$ conducted through the panel is measured at different temperature differences $\Delta T$ (example data).
a) Plot the points in a $(\Delta T, P)$ coordinate system and fit a straight line through the origin to them. Justify why the line should pass through the origin. The line passes approximately through the point $(50.0\,\text{K};\ 8.75\,\text{W})$. Determine the U-value of the panel from the slope of the line.
b) What does the placement of the points on a line tell you about the dependence between the power and the temperature difference? Why did the temperatures have to stay constant during the measurement?
Enter the U-value of part a) in the answer box.
Power conducted through the panel at different temperature differences (example data) $\Delta T\,(\text{K})$ $P\,(\text{W})$ 10.0 1.8 20.0 3.4 30.0 5.4 40.0 6.9 50.0 8.8 Answer
$0.35\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$
Show the solution
Given
$A = 0.50\,\text{m}^2$; the points $(\Delta T, P)$ of the table; the line passes approximately through the points $(0\,\text{K}$; $0\,\text{W})$ and $(50.0\,\text{K}$; $8.75\,\text{W})$
Find
the U-value $U$ of the panel in the unit $\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$
Principle
When the temperatures stay constant, the power conducted through the panel is $P = UA\Delta T$, so $P$ is proportional to the temperature difference $\Delta T$. In the graph $(\Delta T, P)$ the line passes through the origin, because without a temperature difference no heat is transferred ($\Delta T = 0$, so $P = 0$). The slope of the line is $UA$.
Reading the graph
In real series of measurements the data points are usually not exactly on the same line because of measurement uncertainty, so individual points deviate from the line because of the variation in the measurements. The line is drawn between the points, and the slope is read from two points of the line, not from two neighbouring data points: $(0\,\text{K}$; $0\,\text{W})$ and $(50.0\,\text{K}$; $8.75\,\text{W})$.
Formula
$k$$= \dfrac{\Delta P}{\Delta(\Delta T)}$$k = UA$, so $U = \dfrac{k}{A}$
Conversions
Not needed: the power is in watts, the temperature difference in kelvins and the area in square metres.
Substitution
$k$$= \dfrac{8.75\,\text{W} - 0\,\text{W}}{50.0\,\text{K} - 0\,\text{K}}$$= 0.175\,\tfrac{\text{W}}{\text{K}}$$U$$= \dfrac{0.175\,\tfrac{\text{W}}{\text{K}}}{0.50\,\text{m}^2}$$= 0.35\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$Answer
$0.35\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$
Two significant figures, like the area $0.50\,\text{m}^2$.
Reasonableness
Back-calculation: $0.35 \cdot 0.50 \cdot 40.0 = 7.0\,\text{W}$, and the table has the value $6.9\,\text{W}$ at $\Delta T = 40.0\,\text{K}$, so the line passes between the points. The value is larger than that of a modern wall, $0.17\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$, about double, but much smaller than that of a window, $1.0\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$, that is, a plausible value for a somewhat older insulated wall panel.
Model answer b)
The points lie on a line, so the power is proportional to the temperature difference: when the difference doubles, the power doubles, as the formula $P = UA\Delta T$ says. The temperatures had to stay constant, because the formula holds only when both surfaces of the structure are at a uniform temperature and the temperatures do not change during the measurement. Otherwise the panel would store or release energy, and the measured power would not equal $UA\Delta T$.
Level 3 · challenging 5 exercises
Optional: application, two ideas together or a new situation where you reason for yourself.
- Exercise 15-16
$0.60\,\text{kg}$ of hot coffee is packed into a thermos flask for a ski trip. The flask has two nested walls with the air removed from between them, and the surfaces of the walls are shiny.
Explain which ways of heat transfer the vacuum and the shiny surfaces prevent and how. Through which route does heat still escape?
Heat escapes from the flask at a mean power of $1.4\,\text{W}$. By how many kelvins does the coffee cool in $10.0\,\text{h}$? The specific heat capacity of coffee is the same as that of water, $4190\,\text{J/(kg}\cdot\text{K)}$.
Answer
$20\,\text{K}$
Show the solution
Given
$m = 0.60\,\text{kg}$, $P = 1.4\,\text{W}$, $t = 10.0\,\text{h}$; from the table $c = 4190\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$ (coffee like water)
Find
the cooling of the coffee $\Delta T$ in kelvins, and an explanation of which ways of heat transfer the vacuum and the shiny surfaces prevent and through which route heat still escapes
Principle
The coffee loses only as much heat as escapes from the flask, so energy is conserved: the decrease in the internal energy of the coffee is equal to the escaping energy $Pt$. The coffee only cools and its phase does not change, so use $Q = cm\Delta T$.
Formula
$cm\Delta T = Pt$, so $\Delta T = \dfrac{Pt}{cm}$
Conversions
The time is converted to seconds, because the power is in watts, that is, joules per second:
$t$$= 10.0 \cdot 3600\,\text{s}$$= 36\,000\,\text{s}$The mass is already in kilograms.
Substitution
$\Delta T$$= \dfrac{1.4\,\text{W} \cdot 36\,000\,\text{s}}{4190\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}} \cdot 0.60\,\text{kg}}$$= \dfrac{50\,400\,\text{J}}{2514\,\tfrac{\text{J}}{\text{K}}}$$\approx 20.05\,\text{K}$Answer
$20\,\text{K}$
Two significant figures. The limiting data are the mass $0.60\,\text{kg}$ and the power $1.4\,\text{W}$ (both with two figures); the time $10.0\,\text{h}$ and the specific heat capacity $4190\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$ are three-figure values. In a vacuum there are no particles, so heat is neither conducted nor carried by convection from one wall to the other. The shiny surfaces reflect thermal radiation back. Heat escapes mainly by conduction through the neck and the stopper of the flask.
Reasonableness
Coffee at $90\,{}^\circ\text{C}$ in the morning is still about $70\,{}^\circ\text{C}$ in the evening, as in a good thermos flask. Balance: the $4190 \cdot 0.60 \cdot 20.05\,\text{J} \approx 50\,400\,\text{J}$ given out by the coffee is equal to the $1.4 \cdot 36\,000\,\text{J}$ that escaped. Units: $\text{W} \cdot \text{s} = \text{J}$ and $\tfrac{\text{J}}{\text{J}/\text{K}} = \text{K}$.
- Exercise 15-17
Because of the increase in greenhouse gases, the Earth now receives on average $1.0\,\text{W/m}^2$ more power than it radiates into space. The surface area of the Earth is $5.1 \cdot 10^{14}\,\text{m}^2$.
How much extra energy accumulates on the Earth in a year (365 days)?
About $90\,\%$ of it is stored in the oceans. Suppose that it warms evenly the top 700 metres of the oceans, whose mass is $2.6 \cdot 10^{20}\,\text{kg}$ and whose specific heat capacity is $4.0 \cdot 10^{3}\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$.
By how many kelvins does the layer warm in ten years? Why does the large mass of water slow down the warming of the climate?
Answer
$\Delta T = 0.14\,\text{K}$
Show the solution
Given
$I = 1.0\,\tfrac{\text{W}}{\text{m}^2}$, $A = 5.1 \cdot 10^{14}\,\text{m}^2$, $t_\text{v} = 365\,\text{days}$, share of the oceans $f = 0.90$, $m = 2.6 \cdot 10^{20}\,\text{kg}$, $c = 4.0 \cdot 10^{3}\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$, time $10\,\text{years}$
Find
the energy accumulating in a year $E_\text{v}$ and the warming of the ocean layer $\Delta T$ in ten years, and an explanation of why a large mass of water slows down the warming
Principle
The extra power is constant, so the energy is the power times the time. The seawater only warms and its phase does not change, so the energy stored in it is $Q = cm\Delta T$.
Formula
$E_\text{v} = IAt_\text{v}$ and $cm\Delta T = 10fE_\text{v}$, so
$\Delta T$$= \dfrac{10fIAt_\text{v}}{cm}$Conversions
A year in seconds:
$t_\text{v}$$= 365 \cdot 24 \cdot 3600\,\text{s}$$= 3.1536 \cdot 10^{7}\,\text{s}$Ten years is the factor $10$ in the formula, and the percentage is $90\,\% = 0.90$.
Substitution
$E_\text{v}$$= 1.0\,\tfrac{\text{W}}{\text{m}^2} \cdot 5.1 \cdot 10^{14}\,\text{m}^2 \cdot 3.1536 \cdot 10^{7}\,\text{s}$$\approx 1.608 \cdot 10^{22}\,\text{J}$$\Delta T$$= \dfrac{10 \cdot 0.90 \cdot 1.608 \cdot 10^{22}\,\text{J}}{4.0 \cdot 10^{3}\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}} \cdot 2.6 \cdot 10^{20}\,\text{kg}}$$= \dfrac{1.447 \cdot 10^{23}\,\text{J}}{1.04 \cdot 10^{24}\,\tfrac{\text{J}}{\text{K}}}$$\approx 0.139\,\text{K}$Answer
$E_\text{v} = 1.6 \cdot 10^{22}\,\text{J}$
And $\Delta T = 0.14\,\text{K}$, two significant figures. The limiting data are the power $1.0\,\tfrac{\text{W}}{\text{m}^2}$, the area $5.1 \cdot 10^{14}\,\text{m}^2$, the mass $2.6 \cdot 10^{20}\,\text{kg}$ and the specific heat capacity $4.0 \cdot 10^{3}\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$, all with two figures; the calculator result $0.139\,\text{K}$ is rounded. A large mass of water slows down the warming, because a large heat capacity $cm$ needs a lot of energy for a change of one kelvin.
Reasonableness
Order of magnitude: $10^{23}\,\text{J}$ divided by $10^{24}\,\tfrac{\text{J}}{\text{K}}$ is about $0.1\,\text{K}$. The heat capacity is
$4.0 \cdot 10^{3} \cdot 2.6 \cdot 10^{20}\,\tfrac{\text{J}}{\text{K}}$$= 1.04 \cdot 10^{24}\,\tfrac{\text{J}}{\text{K}}$For comparison, the mass of the atmosphere is only about $5 \cdot 10^{18}\,\text{kg}$, and the same energy would warm it by tens of kelvins. The ocean slows down the warming, but the stored energy keeps it going for a long time.
- Exercise 15-18
A detached house is heated by an air-to-water heat pump whose coefficient of performance is on average 2.8 over the heating season.
More insulation wool is blown into the roof, and its U-value decreases from $0.35\,\text{W/(m}^2\text{}\cdot\text{K)}$ to $0.10\,\text{W/(m}^2\text{}\cdot\text{K)}$. The area of the roof is $100\,\text{m}^2$. The heating season lasts $5800\,\text{h}$, and the difference between the indoor and outdoor temperature is on average $19\,\text{K}$ during it.
How many kilowatt-hours of electricity does the extra insulation save over the heating season?
A neighbour thinks insulation does not matter to the owner of a heat pump, because the pump gets its heat free from the outdoor air. Justify why insulation is still worthwhile.
Answer
$980\,\text{kWh}$
Show the solution
Given
coefficient of performance $\varepsilon = 2.8$, $U_1 = 0.35\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$, $U_2 = 0.10\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}}$, $A = 100\,\text{m}^2$, $\Delta T = 19\,\text{K}$, $t = 5800\,\text{h}$
Find
the electrical energy $W$ saved by the extra insulation, in kilowatt-hours, and a justification of why insulation is worthwhile with a heat pump
Principle
Insulation reduces the heat conducted through the roof, and the heat pump has to deliver less heat indoors. The coefficient of performance tells you that the pump delivers heat $\varepsilon$ times the electrical energy it uses, so the electricity is the heat divided by $\varepsilon$.
Formula
$Q = (U_1 - U_2)A\Delta T\,t$ and $W = \dfrac{Q}{\varepsilon}$, so
$W$$= \dfrac{(U_1 - U_2)A\Delta T\,t}{\varepsilon}$Conversions
The time is already in hours and the result is wanted in kilowatt-hours, so the power $\text{W}$ is converted to kilowatts by dividing by a thousand: $\text{Wh} / 1000 = \text{kWh}$.
Substitution
$Q$$= (0.35 - 0.10)\,\tfrac{\text{W}}{\text{m}^2\cdot\text{K}} \cdot 100\,\text{m}^2 \cdot 19\,\text{K} \cdot 5800\,\text{h}$$= 475\,\text{W} \cdot 5800\,\text{h}$$= 2\,755\,000\,\text{Wh}$$= 2755\,\text{kWh}$$W$$= \dfrac{2755\,\text{kWh}}{2.8}$$\approx 983.9\,\text{kWh}$Answer
$980\,\text{kWh}$
Two significant figures. The limiting data are the coefficient of performance $2.8$, the difference of the U-values $0.35 - 0.10 = 0.25$ and the temperature difference $19\,\text{K}$, all with two figures; the area $100\,\text{m}^2$ and the time $5800\,\text{h}$ are values with at least two figures. The calculator result $983.9\,\text{kWh}$ is rounded. The pump does not create heat but moves it using electricity, and the more heat the house needs, the more electricity is used. That is why insulation is worthwhile: a smaller heating need means a smaller electricity bill.
Reasonableness
Back-calculation: $983.9 \cdot 2.8 \approx 2755\,\text{kWh}$. The electricity saving is smaller than the heat saving of $2755\,\text{kWh}$, because the pump delivers $2.8$ kWh of heat from one kilowatt-hour of electricity; in addition to the electricity, heat comes from the outdoor air. With direct electric heating ($\varepsilon = 1$) the saving would be the whole $2755\,\text{kWh}$. In frost the coefficient of performance decreases and the backup resistance heater starts, so the saving is then greatest.
- Exercise 15-19
The ice cover of the Arctic Ocean has shrunk. The table gives the area of the ice at the September minimum (example data). Use a model to estimate what follows from the loss of ice. The open sea that replaced the ice receives the same solar radiation as the ice did, on average $90\,\text{W/m}^2$. Albedo tells you what share of the incoming radiation is reflected: the albedo of ice is $0.60$ and that of open sea $0.06$. The surface area of the Earth is $5.1 \cdot 10^{14}\,\text{m}^2$.
a) Read from the table the decrease in the area of the ice from 1980 to 2020 and convert it into square metres.
b) By how much greater a power does the area now absorb? Then calculate this increase per area of the whole Earth, on average, and compare it with the energy imbalance $1.0\,\text{W/m}^2$.
c) Explain why the loss of ice is a positive feedback (an effect that strengthens the change) and which simplification of the model makes the estimate rough.
Enter the power per area of the whole Earth from part b) in the answer box.
Area of Arctic Ocean ice at the September minimum (example data) Year $A\,(10^{6}\,\text{km}^2)$ 1980 7.5 1990 6.1 2000 6.4 2010 4.9 2020 3.9 Answer
$0.34\,\tfrac{\text{W}}{\text{m}^2}$
Show the solution
Given
from the table $A_{1980} = 7.5 \cdot 10^{6}\,\text{km}^2$ and $A_{2020} = 3.9 \cdot 10^{6}\,\text{km}^2$; radiation power $I = 90\,\tfrac{\text{W}}{\text{m}^2}$, albedos $a_\text{ice} = 0.60$ and $a_\text{sea} = 0.06$, area of the Earth $A_\text{Earth} = 5.1 \cdot 10^{14}\,\text{m}^2$, energy imbalance $1.0\,\tfrac{\text{W}}{\text{m}^2}$
Find
the increase in absorbed power caused by the loss of ice, per area of the Earth, $\Delta P_\text{Earth}$, in the unit $\tfrac{\text{W}}{\text{m}^2}$
Principle
A surface absorbs the share $1 - a$ of the incoming radiation, because the albedo $a$ is reflected. When ice is replaced by open sea, the same area absorbs a larger share of the same radiation, so the absorbed power increases. The increase is divided by the area of the Earth so that it can be compared with the energy imbalance.
Formula
$\Delta A$$= A_{1980} - A_{2020}$$\Delta P$$= I\,\Delta A\,\bigl[(1 - a_\text{sea}) - (1 - a_\text{ice})\bigr]$$= I\,\Delta A\,(a_\text{ice} - a_\text{sea})$so
$\Delta P_\text{Earth}$$= \dfrac{\Delta P}{A_\text{Earth}}$Conversions
The area is converted to square metres: $1\,\text{km}^2 = 10^{6}\,\text{m}^2$, so
$\Delta A$$= 3.6 \cdot 10^{6}\,\text{km}^2$$= 3.6 \cdot 10^{6} \cdot 10^{6}\,\text{m}^2$$= 3.6 \cdot 10^{12}\,\text{m}^2$Substitution
$\Delta A$$= 7.5 \cdot 10^{6}\,\text{km}^2 - 3.9 \cdot 10^{6}\,\text{km}^2$$= 3.6 \cdot 10^{6}\,\text{km}^2$$\Delta P$$= 90\,\tfrac{\text{W}}{\text{m}^2} \cdot 3.6 \cdot 10^{12}\,\text{m}^2 \cdot (0.60 - 0.06)$$= 1.7496 \cdot 10^{14}\,\text{W}$$\Delta P_\text{Earth}$$= \dfrac{1.7496 \cdot 10^{14}\,\text{W}}{5.1 \cdot 10^{14}\,\text{m}^2}$$= 0.343\,\tfrac{\text{W}}{\text{m}^2}$Answer
$0.34\,\tfrac{\text{W}}{\text{m}^2}$
Two significant figures, because the radiation power $90\,\tfrac{\text{W}}{\text{m}^2}$ and the difference of the albedos $0.54$ are two-figure values. It is about a third of the energy imbalance $1.0\,\tfrac{\text{W}}{\text{m}^2}$, because $0.34 / 1.0 \approx 0.3$.
Reasonableness
The change in the area of ice is $3.6 \cdot 10^{12} / 5.1 \cdot 10^{14} \approx 0.7\,\%$ of the area of the Earth. Over it the absorbed power increases by the factor $0.54$ relative to $90\,\tfrac{\text{W}}{\text{m}^2}$, that is, $0.007 \cdot 90 \cdot 0.54 \approx 0.34\,\tfrac{\text{W}}{\text{m}^2}$, as calculated. If the albedos were the same, the increase would be zero, as it should be. The result is positive, because open sea absorbs more.
Model answer c)
The feedback is positive because it strengthens the original change: warming melts ice, so reflection decreases and the absorbed power increases, which warms things even more. The model is rough because it uses all the time the same radiation power $90\,\tfrac{\text{W}}{\text{m}^2}$ and fixed albedos. In reality the radiation of the Sun varies strongly with the season, the albedo of ice changes because of melt ponds, and clouds change the absorbed power. The loss of ice cannot continue indefinitely, because there is only a limited amount of ice. The biggest simplification is that the loss of the September minimum is calculated as if the sea were open all year round at the annual mean of $90\,\tfrac{\text{W}}{\text{m}^2}$, so the estimate is an upper limit.
- Exercise 15-20
The temperature of the Earth is estimated with a simple model. The surface absorbs a power of $239\,\text{W/m}^2$ from the Sun and radiates thermal radiation upwards at a power $S$ per square metre. The atmosphere absorbs the share $0.80$ of this radiation, and the remaining $20\,\%$ passes straight into space. The atmosphere radiates the energy it has absorbed half upwards into space and half downwards to the surface. In the model the atmosphere does not absorb the radiation of the Sun, and both the top of the atmosphere and the surface are in energy balance. The table gives the thermal power radiated by the surface per square metre as a function of its temperature (example data).
a) Write the energy balances (the outgoing power is equal to the incoming power) for the top of the atmosphere and for the surface and calculate the power $S$ radiated by the surface.
b) Deduce the temperature of the surface from the table by interpolation and give it in degrees Celsius.
c) What would the temperature of the surface be without the atmosphere, if it radiated only the power $239\,\text{W/m}^2$ it absorbs? What does the difference tell you about the greenhouse effect, and why is the model still rough?
Enter the temperature of part b) in degrees Celsius in the answer box.
Thermal power radiated by the surface as a function of temperature (example data) $T\,(\text{K})$ $P\,(\text{W/m}^2)$ 255 240 273 315 288 390 303 478 318 580 Answer
$16\,{}^\circ\text{C}$
Show the solution
Given
$P_\text{abs} = 239\,\tfrac{\text{W}}{\text{m}^2}$, the share absorbed by the atmosphere $f = 0.80$, the remaining $1 - f = 0.20$ goes straight into space; the atmosphere radiates half upwards and half downwards; the points $(T, P)$ of the table
Find
the power $S$ radiated by the surface and the temperature $T$ of the surface in degrees Celsius
Principle
In energy balance the incoming and outgoing powers are equal. At the top of the atmosphere the power leaving into space must be $239\,\tfrac{\text{W}}{\text{m}^2}$, because the power absorbed from the Sun is the same. In addition to the Sun, the surface receives the power radiated downwards by the atmosphere, and the greenhouse effect raises the temperature.
Formula
At the top $239\,\tfrac{\text{W}}{\text{m}^2} = (1 - f)\,S + \tfrac{fS}{2}$, so
$S$$= \dfrac{239\,\text{W/m}^2}{1 - f/2}$Check at the surface: $S = 239\,\tfrac{\text{W}}{\text{m}^2} + \tfrac{fS}{2}$. The temperature is interpolated between the points $(T_1, P_1)$ and $(T_2, P_2)$ of the table:
$T$$= T_1 + \dfrac{S - P_1}{P_2 - P_1}\,(T_2 - T_1)$Conversions
A temperature in kelvins is converted to degrees Celsius by subtracting $273.15$: $T_\text{C} = T - 273.15\,\text{K}$.
Substitution
$S$$= \dfrac{239\,\tfrac{\text{W}}{\text{m}^2}}{1 - 0.40}$$= 398.3\,\tfrac{\text{W}}{\text{m}^2}$Check: $239 + 0.40 \cdot 398.3 = 398.3\,\tfrac{\text{W}}{\text{m}^2}$. The value $398.3$ is between the points $288\,\text{K}$ ($390$) and $303\,\text{K}$ ($478$) of the table:
$T$$= 288\,\text{K} + \dfrac{398.3 - 390}{478 - 390} \cdot 15\,\text{K}$$= 289.4\,\text{K}$$T_\text{C}$$= 289.4\,\text{K} - 273.15\,\text{K}$$= 16.3\,{}^\circ\text{C}$Answer
$16\,{}^\circ\text{C}$
(calculated $16.3\,{}^\circ\text{C}$), two significant figures, because linear interpolation on a curved table limits the result to an accuracy of about $\pm 0.5\,\text{K}$.
Reasonableness
A temperature of $16\,{}^\circ\text{C}$ is close to the real mean temperature of the Earth, about $+15\,{}^\circ\text{C}$. The power $398.3\,\tfrac{\text{W}}{\text{m}^2}$ is larger than $239\,\tfrac{\text{W}}{\text{m}^2}$, as it should be, because the surface receives extra energy from the atmosphere. The limiting case $f = 0$ gives $S = 239\,\tfrac{\text{W}}{\text{m}^2}$, which corresponds in the table to about $255\,\text{K}$.
Model answer c)
Without the atmosphere the surface would radiate only the power it absorbs, $239\,\tfrac{\text{W}}{\text{m}^2}$, which corresponds in the table to $255\,\text{K}$, that is, $255 - 273.15 \approx -18\,{}^\circ\text{C}$. With the atmosphere the model gives a temperature of $16.3\,{}^\circ\text{C}$, about $34\,\text{K}$ warmer, so the greenhouse effect raises the surface temperature by this much. The model is still rough: the real atmosphere absorbs the radiation of the Sun, the share $0.80$ varies with wavelength, and clouds, convection and the evaporation of water transfer energy upwards from the surface in other ways than by radiation.