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Eduqas A-level Physics (A420QS) · Component 3 Option D: Energy and the Environment
Mini-Lesson · Option D

Option D: Energy and the Environment

This mini-lesson covers Eduqas Component 3, Option D: the Earth's thermal equilibrium and the effect of CO₂; Wien's law and Stefan's law in a solar context; Archimedes and sea level rise; solar, wind, tidal, hydroelectric and pumped storage power; nuclear fission and fusion; fuel cells; and thermal conduction and U-values.

⚠️ This is an OPTION. Eduqas Component 3 Section B offers four options — A Alternating Currents, B Medical Physics, C The Physics of Sports, D Energy and the Environment. You study exactly ONE of them. Only work through this mini-lesson if Option D: Energy and the Environment is the one your school teaches. Everyone also needs the Component 3 core (Sections 1–10), which has its own mini-lesson.

Press Start when you are ready.

(a)(i)–(iii) The Earth's energy balance

Thermal equilibrium, the greenhouse effect and CO₂

The Earth is in thermal equilibrium when the power it absorbs from the Sun equals the power it re-radiates to space. If the outgoing power falls below the incoming power, the Earth warms until balance is restored at a higher temperature.

The Sun is very hot (~5800 K), so by Wien's law its radiation peaks in the visible — which passes freely through the atmosphere. The Earth is far cooler (~288 K), so its radiation peaks in the infrared. Greenhouse gases — CO₂, water vapour, methane — are largely transparent to visible light but strongly absorb infrared. Radiation gets in easily and out with difficulty, so the surface settles at a higher temperature. Rising CO₂ levels tighten that infrared "blanket".

λmaxT = 2.90 × 10⁻³ m K  ·  P = σAT⁴Wien's law and the Stefan–Boltzmann law. Together they set both where a body radiates and how much.

Archimedes and sea level (a)(iv): a floating iceberg already displaces its own weight of water — so when it melts, the meltwater exactly fills the volume it was displacing and the sea level does not change. Ice sitting on land (Greenland, Antarctica) is displacing nothing; when it melts it adds entirely new water to the ocean and the sea level rises. (Thermal expansion of the warming water raises it further.)

Quick check

Melting ice

?Which melting ice raises sea level, and why?
Quick check

Why CO₂ matters

?Why does adding CO₂ to the atmosphere warm the Earth?
(b)(i) Solar power

The Sun's output and photovoltaics

The Sun's energy comes from the proton–proton chain: hydrogen nuclei fuse, ultimately forming helium-4, and the small loss of mass appears as energy (E = mc²).

I = P / A  ·  I = P / 4πd²Intensity (W m⁻²) obeys an inverse square law for a point source: the same power spread over the surface of an ever-larger sphere.

Photovoltaic cells convert solar radiation directly into electrical energy, typically at 15–25% efficiency. Useful electrical power = intensity × panel area × efficiency.

Calculate

Your turn — solar intensity at the Earth

1The Sun radiates 3.85 × 10²⁶ W and the Earth is 1.50 × 10¹¹ m away. Calculate the intensity of solar radiation above the atmosphere, to the nearest 10 W m⁻².
W m⁻²
Hint: I = P / 4πd² = 3.85 × 10²⁶ ÷ (4π × (1.50 × 10¹¹)²) = 3.85 × 10²⁶ ÷ 2.83 × 10²³.
Calculate

Your turn — a solar array

2A photovoltaic array of area 15 m² and efficiency 20% receives radiation of intensity 800 W m⁻². Calculate its electrical power output.
W
Hint: Power in = 800 × 15 = 12 000 W. Useful power out = 0.20 × 12 000.
(b)(ii)–(iii) Wind, tidal and hydro

Power from moving water and air

P = ½ρAv³The power available in a flowing fluid — air through the swept area of a turbine, or water through a channel. Note the .

Where the cube comes from: the kinetic energy of the fluid is ½mv², and the mass arriving per second is ρAv — so the power is ½(ρAv)v² = ½ρAv³. Doubling the wind speed gives eight times the power, which is why turbine siting matters so enormously.

Efficiency limits: a turbine can never take all the kinetic energy — the air must keep moving to get out of the way (the theoretical maximum, the Betz limit, is about 59%; real turbines manage roughly 40%). Other factors: blade design, drag and generator losses, and the fact that turbines must shut down in very high winds.

Tidal barrages, hydroelectric and pumped storage all work by the same conversion: gravitational potential energy → kinetic energy → electrical energy.

P = (efficiency) × (mass per second) × g × hPumped storage is not a source of energy at all — it is a store. Surplus electricity pumps water uphill at night; at peak demand it runs back down through the turbines within seconds.
Calculate

Your turn — wind power

3A wind turbine has blades of radius 30 m. The wind speed is 10 m s⁻¹ and the density of air is 1.2 kg m⁻³. Calculate the total power available in the wind passing through the swept area, in MW.
MW
Hint: A = πr² = π × 30² = 2827 m². P = ½ρAv³ = 0.5 × 1.2 × 2827 × 10³ = 1.70 × 10⁶ W.
Calculate

Your turn — hydroelectric power

4Water flows through a hydroelectric station at 500 kg s⁻¹, falling through a height of 40 m. The plant is 90% efficient. Calculate the electrical power output, in kW. (g = 9.81 m s⁻²)
kW
Hint: Power available = (mass per second) × g × h = 500 × 9.81 × 40 = 196 200 W. Multiply by 0.90, then divide by 1000.
Quick check

A windier day

?The wind speed at a turbine site doubles. What happens to the power available in the wind?
Sort it

Renewable, non-renewable, or insulation?

Tap an item, then tap the box it belongs in.

♻️ Renewable source

🛢️ Non-renewable source

🏠 Reduces heat loss

(b)(iv)–(c) Nuclear and fuel cells

Fission, fusion and fuel cells

Fission — enrichment and breeding. Natural uranium is over 99% U-238, which does not readily fission with slow neutrons; only about 0.7% is the fissile U-235. Enrichment raises the U-235 fraction to a few percent to sustain a chain reaction. Breeding converts otherwise useless U-238 into fissile plutonium-239 by neutron capture — turning the bulk of the uranium into fuel, at the cost of producing weapons-usable material.

Fusion — why it is so hard. Two positive nuclei must be pushed close enough for the strong nuclear force to grip, against enormous Coulomb repulsion. That means temperatures over 10⁸ K — at which the fuel is a plasma that no material container can touch, so it must be confined magnetically (a tokamak) or inertially (laser compression). Success requires the triple product — the product of the plasma density, its temperature and the confinement time — to exceed a threshold value. Achieving any one of the three is easy; achieving all three at once is the entire problem.

Fuel cells combine hydrogen and oxygen electrochemically to produce electricity directly, with water as the only product at the point of use — no CO₂, no particulates, and higher efficiency than a heat engine (which is limited by thermodynamics). The catch: the hydrogen has to be made, and if it is made by reforming natural gas, the CO₂ has simply been moved rather than removed. Green hydrogen (electrolysis powered by renewables) fixes that, at a cost.

Quick check

Why is fusion so difficult?

?What must be achieved simultaneously for sustained, net-energy-producing fusion?
(d)–(e) Conduction and insulation

Thermal conduction and U-values

Q / t = −KA (Δθ / Δx)The thermal conduction equation. K = thermal conductivity (W m⁻¹ K⁻¹), A = area, Δθ/Δx = the temperature gradient. The minus sign says heat flows down the temperature gradient.

To cut the heat loss you can reduce K (a better insulator), reduce A, or increase the thickness Δx. A trapped layer of air is an excellent insulator (very low K) — which is exactly what cavity wall insulation, double glazing and a wool jumper all exploit.

rate of energy transfer = UAΔθThe U-value (W m⁻² K⁻¹) rolls the conductivity and the thickness of a whole wall — including several materials in contact — into one number. A lower U-value means better insulation.

Materials in contact: for layers in series, the same power flows through each layer, and the temperature drop across each is proportional to its thermal resistance. The U-value of the composite is found from 1/U = sum of (Δx/K) for each layer, plus the surface resistances.

Calculate

Your turn — conduction through a wall

5A brick wall of area 10 m² and thickness 0.10 m has a thermal conductivity of 0.60 W m⁻¹ K⁻¹. The temperature difference across it is 15 K. Calculate the rate of heat loss.
W
Hint: Q/t = KAΔθ/Δx = (0.60 × 10 × 15) ÷ 0.10 = 90 ÷ 0.10.
Calculate

Your turn — U-value

6A window of area 12 m² has a U-value of 1.6 W m⁻² K⁻¹. The temperature difference across it is 18 K. Calculate the rate of heat loss.
W
Hint: Rate = UAΔθ = 1.6 × 12 × 18.
Quick check

What does a low U-value mean?

?Wall X has a U-value of 0.3 W m⁻² K⁻¹; wall Y has a U-value of 1.8 W m⁻² K⁻¹. Which is the better insulator?
Quick check

The fuel cell question

?What is the main environmental benefit of a hydrogen fuel cell — and the main caveat?
Match it

Match each equation to what it gives you

Tap an item on the left, then its partner on the right.

Equation
What it gives
Recap

Option D — the big ideas

Energy balance: in = out at equilibrium; Sun peaks visible, Earth peaks infrared; CO₂ absorbs the infrared

Sea level: floating ice → no change (Archimedes); land ice → rise (plus thermal expansion)

Solar: proton–proton chain; I = P/4πd²; PV output = I × A × efficiency

Wind: P = ½ρAv³ — power goes as v³, so doubling the wind gives 8× the power

Water: tidal, hydro and pumped storage all convert E(p) → E(k) → electrical; pumped storage is a store, not a source

Nuclear: enrichment and breeding (fission); the fusion triple product — density × temperature × confinement time

Fuel cells: water is the only product at the point of use — but the hydrogen has to be made

Insulation: Q/t = KAΔθ/Δx; rate = UAΔθ; low U-value = good insulator

Press Finish to see your score.

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