Option D: Energy and the Environment

What you'll learn

How Earth’s temperature depends on energy balance, greenhouse gases and radiation laws.
How to compare renewable and non-renewable energy sources, in the UK and worldwide.
How to calculate power from solar panels, wind turbines, hydroelectric schemes and insulation.
Why fission, fusion and fuel cells matter in low-carbon energy discussions.

Core quantities you need first

Energy is measured in joules (J). Power is the rate of energy transfer, measured in watts (W), where 1 W means 1 J s⁻¹.

Definition

Power, intensity and efficiency

Power is $P=ΔEΔtP=\frac{\Delta E}{\Delta t}$ . Intensity is power per unit area, $I=PAI=\frac{P}{A}$ , measured in W m⁻². Efficiency is the useful output energy or power divided by the total input energy or power.

For this option, most calculations are really about tracking where energy comes from, where it goes, and how quickly it is transferred.

Earth’s energy balance and temperature rise

Earth receives electromagnetic radiation from the Sun. On a long-term average, Earth’s temperature is steady only if the power absorbed from the Sun equals the power radiated back into space.

If more carbon dioxide, CO₂, is added to the atmosphere, more outgoing infrared radiation is absorbed and re-emitted by greenhouse gases. This reduces the rate at which energy escapes to space, so Earth warms until a new equilibrium is reached.

Schematic of Earth’s energy balance showing incoming solar radiation, reflection, infrared re-radiation and the greenhouse effect

Key Idea

Thermal equilibrium

Thermal equilibrium means incoming power equals outgoing power. Increasing greenhouse gas concentration does not “create” energy; it changes the outgoing radiation rate, so the equilibrium temperature changes.

Global energy demand matters because burning fossil fuels to meet that demand releases CO₂. Climate science uses many validated data sources: satellite radiation measurements, temperature records, ice-core CO₂ data and physical models checked against past climate behaviour.

Solar radiation, Wien’s law and Stefan-Boltzmann law

The Sun behaves approximately like a black body, which is an ideal absorber and emitter of thermal radiation. Its surface temperature is about 5800 K, so it emits strongly in the visible and near-infrared parts of the electromagnetic spectrum.

Solar radiation travels through space as electromagnetic waves. In the atmosphere and at Earth’s surface, some shorter-wavelength radiation is absorbed and re-emitted at longer wavelengths, including infrared. Earth itself, being much cooler than the Sun, re-radiates mainly infrared.

The peak wavelength is found using Wien’s law:

λmax⁡T=constant\lambda_{\max}T=\text{constant}

where the constant is about $2.90 \times 10^{-3}\ \mathrm{m\,K}$ .

The Stefan-Boltzmann law says that the power radiated per unit area by a black body is proportional to $T^4$ :

PA=σT4\frac{P}{A}=\sigma T^4

where $K−4\sigma=5.67 \times 10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$ .

Example

Estimating solar wavelength and surface flux

Use Wien’s law for a surface temperature of 5800 K:
$m\lambda_{\max}=\frac{2.90 \times 10^{-3}\ \mathrm{m\,K}}{5800\ \mathrm{K}}=5.0 \times 10^{-7}\ \mathrm{m}$
Interpret the wavelength: $5.0 \times 10^{-7}\ \mathrm{m}$ is 500 nm, in the visible region, which explains why sunlight is strongly visible.
Use Stefan-Boltzmann law to estimate the power radiated per square metre of the Sun’s surface:
$m−2\frac{P}{A}=(5.67 \times 10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}})(5800\ \mathrm{K})^4=6.4 \times 10^7\ \mathrm{W\,m^{-2}}$

Common Mistake

Using Celsius in radiation laws

Wien’s law and Stefan-Boltzmann law need absolute temperature in kelvin (K), not degrees Celsius.

Sea level rise: density and Archimedes’ principle

The density equation is:

ρ=mV\rho=\frac{m}{V}

where $ρ\rho$ is density in kg m⁻³, $m$ is mass in kg and $V$ is volume in m³.

Archimedes’ principle states that the upthrust on a body in a fluid is equal to the weight of fluid displaced.

For a floating iceberg, the ice already displaces its own weight of seawater. When it melts, it produces roughly the same volume of water as the volume it had displaced, so floating sea ice does not significantly raise sea level in this model.

Ice on land is different: when it melts, extra water enters the ocean.

Example

Estimating sea-level rise from land ice

Convert melted land ice of mass $2.0 \times 10^{15}\ \mathrm{kg}$ into water volume, using water density $m−31000\ \mathrm{kg\,m^{-3}}$ :
$m3V=\frac{m}{\rho}=\frac{2.0 \times 10^{15}\ \mathrm{kg}}{1000\ \mathrm{kg\,m^{-3}}}=2.0 \times 10^{12}\ \mathrm{m^3}$
Spread this volume over ocean area $3.6 \times 10^{14}\ \mathrm{m^2}$ :
$mh=\frac{V}{A}=\frac{2.0 \times 10^{12}\ \mathrm{m^3}}{3.6 \times 10^{14}\ \mathrm{m^2}}=5.6 \times 10^{-3}\ \mathrm{m}$
Convert the result: $5.6 \times 10^{-3}\ \mathrm{m}$ is 5.6 mm, so this amount of land ice would raise mean sea level by about 5.6 mm.

Common Mistake

Melting icebergs

Do not say “all melting ice raises sea level”. Land ice raises sea level; floating icebergs are already displacing seawater.

Renewable and non-renewable energy sources

Definition

Renewable and non-renewable energy

A renewable energy source is replenished on a human timescale, such as solar, wind, tidal, wave, hydroelectric, biomass and geothermal energy. A non-renewable source is depleted when used, such as coal, oil, natural gas and nuclear fuels like uranium.

In the UK, offshore wind has developed strongly because the UK has windy coastal waters. Solar is useful but variable with weather, season and day-night cycles. Large hydroelectric power is limited by geography, but pumped storage is valuable for rapid response. Tidal power is predictable, but barrages can be expensive and environmentally disruptive.

Internationally, energy choices depend strongly on geography, infrastructure, politics and cost. Hydroelectricity is important in mountainous or high-rainfall regions. Solar power is especially attractive in sunny regions. Coal, oil and gas remain widely used because of established infrastructure and high energy density, but they produce greenhouse gases.

Solar power and photovoltaic cells

The Sun’s main energy production mechanism is the main branch of the proton-proton chain, where hydrogen nuclei ultimately fuse into helium. A simplified overall reaction is:

24He+2e++2νe+energy4\,^{1}_{1}\mathrm{H}\rightarrow\,^{4}_{2}\mathrm{He}+2e^+ +2\nu_e+\text{energy}

A photovoltaic cell, or PV cell, is a semiconductor device that converts light energy directly into electrical energy.

For a point source radiating equally in all directions, intensity follows the inverse square law:

I=P4πr2I=\frac{P}{4\pi r^2}

Example

Estimating photovoltaic output

Estimate the solar intensity at Earth using solar power $3.85 \times 10^{26}\ \mathrm{W}$ and Earth-Sun distance $1.50 \times 10^{11}\ \mathrm{m}$ :
$m−2I=\frac{3.85 \times 10^{26}\ \mathrm{W}}{4\pi(1.50 \times 10^{11}\ \mathrm{m})^2}=1.36 \times 10^3\ \mathrm{W\,m^{-2}}$
Find the input power to a panel of area $m21.6\ \mathrm{m^2}$ :
$WP_{\text{in}}=IA=(1.36 \times 10^3\ \mathrm{W\,m^{-2}})(1.6\ \mathrm{m^2})=2.18 \times 10^3\ \mathrm{W}$
Apply an efficiency of 20%:
$WP_{\text{out}}=0.20P_{\text{in}}=0.20(2.18 \times 10^3\ \mathrm{W})=4.4 \times 10^2\ \mathrm{W}$

In practice, ground-level output is reduced by clouds, atmospheric absorption, panel angle, dirt, temperature and electrical losses.

Wind power

The power available from a flowing fluid is:

Pavailable=12Aρv3P_{\text{available}}=\frac{1}{2}A\rho v^3

where $A$ is the swept area of the turbine blades, $ρ\rho$ is air density and $v$ is wind speed.

Efficiency depends on blade design, generator losses, turbulence, wind speed variation, siting, maintenance and the Betz limit: no wind turbine can extract all the kinetic energy from the air.

Example

Calculating wind turbine power

Calculate swept area for blade radius 30 m:
$m2A=\pi r^2=\pi(30\ \mathrm{m})^2=2.83 \times 10^3\ \mathrm{m^2}$
Calculate available wind power for air density $m−31.2\ \mathrm{kg\,m^{-3}}$ and wind speed $s−112\ \mathrm{m\,s^{-1}}$ :
$WP_{\text{available}}=\frac{1}{2}(2.83 \times 10^3\ \mathrm{m^2})(1.2\ \mathrm{kg\,m^{-3}})(12\ \mathrm{m\,s^{-1}})^3=2.9 \times 10^6\ \mathrm{W}$
Apply a turbine efficiency of 35%:
$WP_{\text{out}}=0.35(2.9 \times 10^6\ \mathrm{W})=1.0 \times 10^6\ \mathrm{W}$

Tip

Wind speed matters a lot

Wind power depends on $v^3$ , so doubling wind speed increases available power by a factor of eight, before efficiency losses.

Tidal, hydroelectric and pumped storage schemes

Tidal barrages, hydroelectric dams and pumped storage all use gravitational potential energy:

E_p=mgh

Water loses gravitational potential energy, gains kinetic energy, turns turbines and drives generators.

A tidal barrage traps water at high tide and releases it through turbines when there is a height difference. A hydroelectric scheme uses water stored at height. Pumped storage uses spare electrical energy to pump water uphill, then releases it later at peak demand, so it is an energy store rather than a net energy source.

Example

Calculating hydroelectric power

Use mass flow rate from volume flow rate $s−1Q=40\ \mathrm{m^3\,s^{-1}}$ :
$s−1\frac{\Delta m}{\Delta t}=\rho Q=(1000\ \mathrm{kg\,m^{-3}})(40\ \mathrm{m^3\,s^{-1}})=4.0 \times 10^4\ \mathrm{kg\,s^{-1}}$
Calculate useful output power for height 80 m and efficiency 90%:
$WP=\eta\frac{\Delta m}{\Delta t}gh =0.90(4.0 \times 10^4\ \mathrm{kg\,s^{-1}})(9.81\ \mathrm{m\,s^{-2}})(80\ \mathrm{m}) =2.8 \times 10^7\ \mathrm{W}$
Compare with wind: this 28 MW output is roughly the same as 28 wind turbines each producing 1.0 MW at that moment.

Nuclear fission and fusion

In nuclear fission, a heavy nucleus such as uranium-235 absorbs a neutron and splits into smaller nuclei, releasing energy and more neutrons. These neutrons can continue a chain reaction.

Enrichment means increasing the proportion of fissile uranium-235 in nuclear fuel. Natural uranium contains only a small percentage of uranium-235.

Breeding means converting fertile material into fissile material. For example, uranium-238 can absorb neutrons and eventually form plutonium-239, which can be used as reactor fuel.

Fusion combines light nuclei, such as isotopes of hydrogen, into heavier nuclei. Sustained fusion power is difficult because the plasma must be extremely hot, dense enough, and confined long enough.

Definition

Fusion triple product

The fusion triple product is $nTτnT\tau$ , where $n$ is particle number density, $T$ is plasma temperature and $τ\tau$ is confinement time. A high enough triple product is needed for useful sustained fusion.

Fusion challenges include plasma instability, huge input energy, neutron damage to materials, tritium supply and achieving net electrical output reliably.

Fuel cells

A fuel cell converts chemical energy directly into electrical energy while fuel and oxidant are supplied continuously. In a hydrogen fuel cell, hydrogen is supplied at the anode and oxygen at the cathode. The overall reaction is:

2H2+O2→2H2O2H_2+O_2\rightarrow 2H_2O

Electrons pass through the external circuit, providing useful electrical power.

Fuel cells produce no CO₂ at the point of use if hydrogen is the fuel. Their overall greenhouse gas benefit depends on how the hydrogen is produced: hydrogen from renewable electrolysis is much lower carbon than hydrogen produced from natural gas without carbon capture.

Thermal conduction and insulation

Thermal conduction is energy transfer through a material due to a temperature difference. For a flat slab:

ΔQΔt=−AKΔθΔx\frac{\Delta Q}{\Delta t}=-AK\frac{\Delta \theta}{\Delta x}

Here $ΔQΔt\frac{\Delta Q}{\Delta t}$ is rate of energy transfer, $A$ is area, $K$ is thermal conductivity, $Δθ\Delta \theta$ is temperature difference and $Δx\Delta x$ is thickness. The minus sign shows that heat flows from higher temperature to lower temperature.

Cross-section of heat loss through a wall with layers, temperature drop and U-value relationship

Example

Calculating heat loss through one layer

Use the magnitude of the conduction equation because the question asks for the rate of heat loss:
$∣ΔQΔt∣=AKΔθΔx\left|\frac{\Delta Q}{\Delta t}\right|=AK\frac{\Delta \theta}{\Delta x}$
Substitute $m2A=12\ \mathrm{m^2}$ , $K−1K=0.040\ \mathrm{W\,m^{-1}\,K^{-1}}$ , $K\Delta \theta=18\ \mathrm{K}$ and $m\Delta x=0.20\ \mathrm{m}$ :
$W\left|\frac{\Delta Q}{\Delta t}\right|=(12\ \mathrm{m^2})(0.040\ \mathrm{W\,m^{-1}\,K^{-1}})\frac{18\ \mathrm{K}}{0.20\ \mathrm{m}}=43\ \mathrm{W}$
Interpret the result: the insulation still loses energy, but only at about 43 J every second.

For buildings, heat loss through parallel surfaces is often written using a U-value:

ΔQΔt=UAΔθ\frac{\Delta Q}{\Delta t}=UA\Delta \theta

A lower U-value means better insulation. For layers in contact, ignoring surface resistances:

1U=x1K1+x2K2+x3K3+⋯\frac{1}{U}=\frac{x_1}{K_1}+\frac{x_2}{K_2}+\frac{x_3}{K_3}+\cdots

Example

Calculating heat loss through layered insulation

Add the layer terms for plaster, insulation and brick:
$W−1\frac{1}{U}=\frac{0.012\ \mathrm{m}}{0.50\ \mathrm{W\,m^{-1}\,K^{-1}}} +\frac{0.080\ \mathrm{m}}{0.040\ \mathrm{W\,m^{-1}\,K^{-1}}} +\frac{0.100\ \mathrm{m}}{0.77\ \mathrm{W\,m^{-1}\,K^{-1}}} =2.15\ \mathrm{m^2\,K\,W^{-1}}$
Invert to find the U-value:
$K−1U=\frac{1}{2.15\ \mathrm{m^2\,K\,W^{-1}}}=0.465\ \mathrm{W\,m^{-2}\,K^{-1}}$
Calculate heat loss through area $m250\ \mathrm{m^2}$ with temperature difference 20 K:
$W\frac{\Delta Q}{\Delta t}=UA\Delta \theta =(0.465\ \mathrm{W\,m^{-2}\,K^{-1}})(50\ \mathrm{m^2})(20\ \mathrm{K}) =4.7 \times 10^2\ \mathrm{W}$

Tip

Investigating insulation practically

To measure thermal conductivity, use steady-state heating, measure heater power, temperatures, area and thickness, then plot heat transfer rate per area against temperature gradient. The gradient gives $K$ .

Common Mistake

Adding U-values for layers

For layers in contact through the same area, do not add U-values directly. Add the terms $x / K$ , then invert to find $U$ .

Exam technique

In the exam

Start every energy question by identifying the energy store or power flow: solar intensity, kinetic energy of wind, gravitational potential energy of water, or heat conduction.
Check units carefully: temperatures for radiation laws must be in K, areas in m², densities in kg m⁻³ and powers in W.
For evaluation questions, compare more than “renewable versus non-renewable”: discuss intermittency, reliability, lifecycle CO₂, geography, cost, environmental impact and storage.

Self review

Check yourself

Why does melting land ice raise sea level, while melting floating icebergs does not in the simple Archimedes model?
If wind speed doubles, what happens to the available wind power before efficiency losses?
In a layered wall, how can you tell which layer has the greatest effect on reducing heat loss?

What you'll learn

How Earth’s temperature depends on energy balance, greenhouse gases and radiation laws.
How to compare renewable and non-renewable energy sources, in the UK and worldwide.
How to calculate power from solar panels, wind turbines, hydroelectric schemes and insulation.
Why fission, fusion and fuel cells matter in low-carbon energy discussions.

Core quantities you need first

Energy is measured in joules (J). Power is the rate of energy transfer, measured in watts (W), where 1 W means 1 J s⁻¹.

Definition

Power, intensity and efficiency

For this option, most calculations are really about tracking where energy comes from, where it goes, and how quickly it is transferred.

Earth’s energy balance and temperature rise

Earth receives electromagnetic radiation from the Sun. On a long-term average, Earth’s temperature is steady only if the power absorbed from the Sun equals the power radiated back into space.

Schematic of Earth’s energy balance showing incoming solar radiation, reflection, infrared re-radiation and the greenhouse effect

Key Idea

Thermal equilibrium

Solar radiation, Wien’s law and Stefan-Boltzmann law

The peak wavelength is found using Wien’s law:

λmax⁡T=constant\lambda_{\max}T=\text{constant}

where the constant is about $2.90 \times 10^{-3}\ \mathrm{m\,K}$ .

The Stefan-Boltzmann law says that the power radiated per unit area by a black body is proportional to $T^4$ :

PA=σT4\frac{P}{A}=\sigma T^4

where $K−4\sigma=5.67 \times 10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$ .

Example

Estimating solar wavelength and surface flux

Use Wien’s law for a surface temperature of 5800 K:
$m\lambda_{\max}=\frac{2.90 \times 10^{-3}\ \mathrm{m\,K}}{5800\ \mathrm{K}}=5.0 \times 10^{-7}\ \mathrm{m}$
Interpret the wavelength: $5.0 \times 10^{-7}\ \mathrm{m}$ is 500 nm, in the visible region, which explains why sunlight is strongly visible.
Use Stefan-Boltzmann law to estimate the power radiated per square metre of the Sun’s surface:
$m−2\frac{P}{A}=(5.67 \times 10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}})(5800\ \mathrm{K})^4=6.4 \times 10^7\ \mathrm{W\,m^{-2}}$

Common Mistake

Using Celsius in radiation laws

Wien’s law and Stefan-Boltzmann law need absolute temperature in kelvin (K), not degrees Celsius.

Sea level rise: density and Archimedes’ principle

The density equation is:

ρ=mV\rho=\frac{m}{V}

where $ρ\rho$ is density in kg m⁻³, $m$ is mass in kg and $V$ is volume in m³.

Archimedes’ principle states that the upthrust on a body in a fluid is equal to the weight of fluid displaced.

Ice on land is different: when it melts, extra water enters the ocean.

Example

Estimating sea-level rise from land ice

Convert melted land ice of mass $2.0 \times 10^{15}\ \mathrm{kg}$ into water volume, using water density $m−31000\ \mathrm{kg\,m^{-3}}$ :
$m3V=\frac{m}{\rho}=\frac{2.0 \times 10^{15}\ \mathrm{kg}}{1000\ \mathrm{kg\,m^{-3}}}=2.0 \times 10^{12}\ \mathrm{m^3}$
Spread this volume over ocean area $3.6 \times 10^{14}\ \mathrm{m^2}$ :
$mh=\frac{V}{A}=\frac{2.0 \times 10^{12}\ \mathrm{m^3}}{3.6 \times 10^{14}\ \mathrm{m^2}}=5.6 \times 10^{-3}\ \mathrm{m}$
Convert the result: $5.6 \times 10^{-3}\ \mathrm{m}$ is 5.6 mm, so this amount of land ice would raise mean sea level by about 5.6 mm.

Common Mistake

Melting icebergs

Do not say “all melting ice raises sea level”. Land ice raises sea level; floating icebergs are already displacing seawater.

Renewable and non-renewable energy sources

Definition

Renewable and non-renewable energy

Solar power and photovoltaic cells

The Sun’s main energy production mechanism is the main branch of the proton-proton chain, where hydrogen nuclei ultimately fuse into helium. A simplified overall reaction is:

24He+2e++2νe+energy4\,^{1}_{1}\mathrm{H}\rightarrow\,^{4}_{2}\mathrm{He}+2e^+ +2\nu_e+\text{energy}

A photovoltaic cell, or PV cell, is a semiconductor device that converts light energy directly into electrical energy.

For a point source radiating equally in all directions, intensity follows the inverse square law:

I=P4πr2I=\frac{P}{4\pi r^2}

Example

Estimating photovoltaic output

Estimate the solar intensity at Earth using solar power $3.85 \times 10^{26}\ \mathrm{W}$ and Earth-Sun distance $1.50 \times 10^{11}\ \mathrm{m}$ :
$m−2I=\frac{3.85 \times 10^{26}\ \mathrm{W}}{4\pi(1.50 \times 10^{11}\ \mathrm{m})^2}=1.36 \times 10^3\ \mathrm{W\,m^{-2}}$
Find the input power to a panel of area $m21.6\ \mathrm{m^2}$ :
$WP_{\text{in}}=IA=(1.36 \times 10^3\ \mathrm{W\,m^{-2}})(1.6\ \mathrm{m^2})=2.18 \times 10^3\ \mathrm{W}$
Apply an efficiency of 20%:
$WP_{\text{out}}=0.20P_{\text{in}}=0.20(2.18 \times 10^3\ \mathrm{W})=4.4 \times 10^2\ \mathrm{W}$

In practice, ground-level output is reduced by clouds, atmospheric absorption, panel angle, dirt, temperature and electrical losses.

Wind power

The power available from a flowing fluid is:

Pavailable=12Aρv3P_{\text{available}}=\frac{1}{2}A\rho v^3

where $A$ is the swept area of the turbine blades, $ρ\rho$ is air density and $v$ is wind speed.

Efficiency depends on blade design, generator losses, turbulence, wind speed variation, siting, maintenance and the Betz limit: no wind turbine can extract all the kinetic energy from the air.

Example

Calculating wind turbine power

Calculate swept area for blade radius 30 m:
$m2A=\pi r^2=\pi(30\ \mathrm{m})^2=2.83 \times 10^3\ \mathrm{m^2}$
Calculate available wind power for air density $m−31.2\ \mathrm{kg\,m^{-3}}$ and wind speed $s−112\ \mathrm{m\,s^{-1}}$ :
$WP_{\text{available}}=\frac{1}{2}(2.83 \times 10^3\ \mathrm{m^2})(1.2\ \mathrm{kg\,m^{-3}})(12\ \mathrm{m\,s^{-1}})^3=2.9 \times 10^6\ \mathrm{W}$
Apply a turbine efficiency of 35%:
$WP_{\text{out}}=0.35(2.9 \times 10^6\ \mathrm{W})=1.0 \times 10^6\ \mathrm{W}$

Tip

Wind speed matters a lot

Wind power depends on $v^3$ , so doubling wind speed increases available power by a factor of eight, before efficiency losses.

Tidal, hydroelectric and pumped storage schemes

Tidal barrages, hydroelectric dams and pumped storage all use gravitational potential energy:

E_p=mgh

Water loses gravitational potential energy, gains kinetic energy, turns turbines and drives generators.

Example

Calculating hydroelectric power

Use mass flow rate from volume flow rate $s−1Q=40\ \mathrm{m^3\,s^{-1}}$ :
$s−1\frac{\Delta m}{\Delta t}=\rho Q=(1000\ \mathrm{kg\,m^{-3}})(40\ \mathrm{m^3\,s^{-1}})=4.0 \times 10^4\ \mathrm{kg\,s^{-1}}$
Calculate useful output power for height 80 m and efficiency 90%:
$WP=\eta\frac{\Delta m}{\Delta t}gh =0.90(4.0 \times 10^4\ \mathrm{kg\,s^{-1}})(9.81\ \mathrm{m\,s^{-2}})(80\ \mathrm{m}) =2.8 \times 10^7\ \mathrm{W}$
Compare with wind: this 28 MW output is roughly the same as 28 wind turbines each producing 1.0 MW at that moment.

Nuclear fission and fusion

In nuclear fission, a heavy nucleus such as uranium-235 absorbs a neutron and splits into smaller nuclei, releasing energy and more neutrons. These neutrons can continue a chain reaction.

Enrichment means increasing the proportion of fissile uranium-235 in nuclear fuel. Natural uranium contains only a small percentage of uranium-235.

Breeding means converting fertile material into fissile material. For example, uranium-238 can absorb neutrons and eventually form plutonium-239, which can be used as reactor fuel.

Fusion combines light nuclei, such as isotopes of hydrogen, into heavier nuclei. Sustained fusion power is difficult because the plasma must be extremely hot, dense enough, and confined long enough.

Definition

Fusion triple product

Fusion challenges include plasma instability, huge input energy, neutron damage to materials, tritium supply and achieving net electrical output reliably.

Fuel cells

2H2+O2→2H2O2H_2+O_2\rightarrow 2H_2O

Electrons pass through the external circuit, providing useful electrical power.

Thermal conduction and insulation

Thermal conduction is energy transfer through a material due to a temperature difference. For a flat slab:

ΔQΔt=−AKΔθΔx\frac{\Delta Q}{\Delta t}=-AK\frac{\Delta \theta}{\Delta x}

Cross-section of heat loss through a wall with layers, temperature drop and U-value relationship

Example

Calculating heat loss through one layer

Use the magnitude of the conduction equation because the question asks for the rate of heat loss:
$∣ΔQΔt∣=AKΔθΔx\left|\frac{\Delta Q}{\Delta t}\right|=AK\frac{\Delta \theta}{\Delta x}$
Substitute $m2A=12\ \mathrm{m^2}$ , $K−1K=0.040\ \mathrm{W\,m^{-1}\,K^{-1}}$ , $K\Delta \theta=18\ \mathrm{K}$ and $m\Delta x=0.20\ \mathrm{m}$ :
$W\left|\frac{\Delta Q}{\Delta t}\right|=(12\ \mathrm{m^2})(0.040\ \mathrm{W\,m^{-1}\,K^{-1}})\frac{18\ \mathrm{K}}{0.20\ \mathrm{m}}=43\ \mathrm{W}$
Interpret the result: the insulation still loses energy, but only at about 43 J every second.

For buildings, heat loss through parallel surfaces is often written using a U-value:

ΔQΔt=UAΔθ\frac{\Delta Q}{\Delta t}=UA\Delta \theta

A lower U-value means better insulation. For layers in contact, ignoring surface resistances:

1U=x1K1+x2K2+x3K3+⋯\frac{1}{U}=\frac{x_1}{K_1}+\frac{x_2}{K_2}+\frac{x_3}{K_3}+\cdots

Example

Calculating heat loss through layered insulation

Add the layer terms for plaster, insulation and brick:
$W−1\frac{1}{U}=\frac{0.012\ \mathrm{m}}{0.50\ \mathrm{W\,m^{-1}\,K^{-1}}} +\frac{0.080\ \mathrm{m}}{0.040\ \mathrm{W\,m^{-1}\,K^{-1}}} +\frac{0.100\ \mathrm{m}}{0.77\ \mathrm{W\,m^{-1}\,K^{-1}}} =2.15\ \mathrm{m^2\,K\,W^{-1}}$
Invert to find the U-value:
$K−1U=\frac{1}{2.15\ \mathrm{m^2\,K\,W^{-1}}}=0.465\ \mathrm{W\,m^{-2}\,K^{-1}}$
Calculate heat loss through area $m250\ \mathrm{m^2}$ with temperature difference 20 K:
$W\frac{\Delta Q}{\Delta t}=UA\Delta \theta =(0.465\ \mathrm{W\,m^{-2}\,K^{-1}})(50\ \mathrm{m^2})(20\ \mathrm{K}) =4.7 \times 10^2\ \mathrm{W}$

Tip

Investigating insulation practically

Common Mistake

Adding U-values for layers

For layers in contact through the same area, do not add U-values directly. Add the terms $x / K$ , then invert to find $U$ .

Exam technique

In the exam

Start every energy question by identifying the energy store or power flow: solar intensity, kinetic energy of wind, gravitational potential energy of water, or heat conduction.
Check units carefully: temperatures for radiation laws must be in K, areas in m², densities in kg m⁻³ and powers in W.
For evaluation questions, compare more than “renewable versus non-renewable”: discuss intermittency, reliability, lifecycle CO₂, geography, cost, environmental impact and storage.

Self review

Check yourself

Why does melting land ice raise sea level, while melting floating icebergs does not in the simple Archimedes model?
If wind speed doubles, what happens to the available wind power before efficiency losses?
In a layered wall, how can you tell which layer has the greatest effect on reducing heat loss?

What you'll learn

Core quantities you need first

Earth’s energy balance and temperature rise

Solar radiation, Wien’s law and Stefan-Boltzmann law

Sea level rise: density and Archimedes’ principle

Renewable and non-renewable energy sources

Solar power and photovoltaic cells

Wind power

Tidal, hydroelectric and pumped storage schemes

Nuclear fission and fusion

Fuel cells

Thermal conduction and insulation

Review Option D: Energy and the Environment by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Option D: Energy and the Environment

What you'll learn

Core quantities you need first

Earth’s energy balance and temperature rise

Solar radiation, Wien’s law and Stefan-Boltzmann law

Sea level rise: density and Archimedes’ principle

Renewable and non-renewable energy sources

Solar power and photovoltaic cells

Wind power

Tidal, hydroelectric and pumped storage schemes

Nuclear fission and fusion

Fuel cells

Thermal conduction and insulation

Review Option D: Energy and the Environment by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Option D: Energy and the Environment

What you'll learn

Core quantities you need first

Earth’s energy balance and temperature rise

Solar radiation, Wien’s law and Stefan-Boltzmann law

Sea level rise: density and Archimedes’ principle

Renewable and non-renewable energy sources

Solar power and photovoltaic cells

Wind power

Tidal, hydroelectric and pumped storage schemes

Nuclear fission and fusion

Fuel cells

Thermal conduction and insulation

Review Option D: Energy and the Environment by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Option D: Energy and the Environment Revision Guide

Option D: Energy and the Environment

What you'll learn

Core quantities you need first

Earth’s energy balance and temperature rise

Solar radiation, Wien’s law and Stefan-Boltzmann law

Sea level rise: density and Archimedes’ principle

Renewable and non-renewable energy sources

Solar power and photovoltaic cells

Wind power

Tidal, hydroelectric and pumped storage schemes

Nuclear fission and fusion

Fuel cells

Thermal conduction and insulation

Review Option D: Energy and the Environment by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Option D: Energy and the Environment Revision Guide