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AP Environmental Science Energy Resources and Consumption Guide

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Official unit weighting

Energy Resources and Consumption in the AP Environmental Science blueprint

College Board assigns Unit 6, Energy Resources and Consumption, 10–15% of AP Environmental Science multiple-choice questions. This range describes the multiple-choice section, not a percentage of the total exam score.

AP Environmental Science · Unit 6

Energy Resources and Consumption

Systems, evidence, calculations, and solutions
  • May 2027
  • Bluebook
  • Clean-room teaching
  • Read the evidence surface.
  • Trace the environmental mechanism.
  • Check units, scale, and claim scope.
APES-U6
Unit lesson

Unit 6: Energy Resources and Consumption

6.1 Renewable and Nonrenewable Resources

When studying Energy Resources and Consumption, renewable resources replenish on human time scales when use does not exceed renewal; nonrenewable stocks form too slowly to replace current extraction.

6.2 Global Energy Consumption

6.3 Fuel Types and Uses

6.4 Distribution of Natural Energy Resources

6.5 Fossil Fuels

6.6 Nuclear Power

6.7 Energy from Biomass

6.8 Solar Energy

6.9 Hydroelectric Power

6.10 Geothermal Energy

6.11 Hydrogen Fuel Cell

6.12 Wind Energy

6.13 Energy Conservation

Energy conservation reduces energy use through behavior, while efficiency delivers the same service with less energy input.

Vocabulary anchors

Renewable and Nonrenewable Resources; Global Energy Consumption; Fuel Types and Uses; Distribution of Natural Energy Resources; Fossil Fuels; Nuclear Power; Energy from Biomass; Solar Energy; Hydroelectric Power; Geothermal Energy; Hydrogen Fuel Cell; Wind Energy are the official topic anchors used throughout this unit.

Evidence lab: Electricity options for an island grid

Electricity options for an island grid Annual values are normalized per installed megawatt. Source | Capacity factor (%) | Lifecycle emissions (kg CO2e/MWh) | Fuel imported? | Dispatchable? diesel | 72 | 720 | yes | yes wind | 38 | 12 | no | no solar | 24 | 35 | no | no geothermal | 88 | 45 | no | yes

Which source combines the highest capacity factor with dispatchability?

Geothermal — It has an 88% capacity factor and is marked dispatchable.

It is dispatchable but has a lower capacity factor. It is not dispatchable. It has the lowest capacity factor.

Filters a table on two conditions.

Why is diesel classified as nonrenewable?

Its fuel forms far more slowly than it is consumed. — Renewability compares replenishment time with use rate.

Dispatchability does not make a resource renewable. Transportation mode does not determine formation rate. Diesel stores chemical energy.

Applies a resource renewal-time definition.

How many MWh does 1 MW of wind produce annually at 38% capacity factor?

About 3,329 MWh — 1 MW×8,760 h×0.38=3,328.8 MWh.

This assumes 100% output. This uses the solar capacity factor.

Capacity-factor calculation with hours per year.

Why can replacing all diesel nameplate capacity with equal wind capacity reduce reliability?

Wind's lower variable output does not necessarily coincide with demand, so storage, transmission, or firm generation is needed. — Equal nameplate capacity is not equal delivered energy or timing.

The table shows the opposite. The table says no fuel import. Its factor is 72%.

Separates nameplate capacity, annual output, and dispatch timing.

Which portfolio best reduces imports while maintaining firm supply?

Use geothermal as firm low-emission generation and combine wind or solar with storage and demand management. — The portfolio joins dispatchable local supply with variable renewables and balancing.

That retains fuel imports and high emissions. That ignores nighttime and weather variability. Color is not a performance criterion.

Integrates capacity factor, dispatchability, imports, emissions, and balancing.

High-level waste is listed. Withdrawal is high. Uranium is finite.

The table lists sulfur dioxide. No universal neutralization occurs.

Traces precursor to secondary pollutant.

3.255 billion kg

The million multiplier is lost.

The table shows much less. They are distinct hazards.

A drought limits cooling water.

Rooftop solar plus storage

It floods 2,400 hectares. The values differ greatly.

Direct table comparison.

8.64 million kg

That ignores flooding and mortality.

Evidence lab: Surplus wind used for hydrogen

Surplus wind used for hydrogen A wind farm has 100,000 MWh of annual electricity that would otherwise be curtailed. Stage | Efficiency (%) electrolysis | 70 compression and storage | 88 fuel cell | 55

Which stage loses the largest fraction of entering energy?

Fuel cell — Its 55% efficiency corresponds to a 45% stage loss.

Its loss is 30%. Its loss is 12%. Each retains a stated fraction.

Converts stage efficiency to loss fraction.

Why is hydrogen an energy carrier rather than a primary energy source here?

Electricity is used to produce it, and energy is recovered later. — Hydrogen stores transferred energy rather than supplying the original wind input.

It stores chemical energy. The causal direction is reversed. They convert chemical energy through reactions.

Distinguishes primary input from storage carrier.

How much electricity is returned by the fuel cell after all stages?

33,880 MWh — 100,000×0.70×0.88×0.55=33,880 MWh.

This applies only the final efficiency. The efficiencies are added. This is the total loss, not returned energy.

Three-stage multiplicative efficiency calculation.

When can the low round-trip efficiency still be useful?

When surplus electricity would be curtailed and long-duration storage or transport has high system value. — Opportunity cost and storage duration can matter more than efficiency alone.

Using scarce electricity raises opportunity cost. No real conversion requires that. Scaling does not eliminate thermodynamic losses.

Adds counterfactual curtailment and service value to efficiency.

Which comparison is required against batteries?

Round-trip efficiency, duration, power and energy cost, leakage, materials, safety, and seasonal need. — Storage technologies provide different services and constraints.

Color does not determine system service. Those are key variables to measure. Turbine count does not establish storage performance.

Multi-dimensional storage comparison across time scales.

Evidence lab: River-basin energy choice

River-basin energy choice A 60-MW dam would flood 420 ha and block a migratory fish route. Rooftop solar could supply 54 MW at midday but averages 15 MW across the year; batteries would supply 18 MW for four hours.

How much energy can the stated battery deliver when fully discharged?

72 MWh — 18 MW ×4 h=72 MWh.

Continue in the A+ teaching layer for the remaining evidence labs, figure, and final audit.

How it is assessed

How the AP Environmental Science assesses Energy Resources and Consumption

ItemWeight / countWhat it means
Official topic denominator13
Multiple choice80 questions in 90 minutes
Free response3 questions in 70 minutes
Calculation credit
AdministrationMay 2027
Worked examples

Annual wind generation

Read each question as it appears on the exam paper, choose your answer, then tap it to see the feedback.

Practice 1: Calculate annual wind generation

Q. Which setup correctly converts nameplate power to annual energy using capacity factor? Read the exam question below, choose one option, then tap it to check.
SECTION I · MULTIPLE CHOICE · QUESTION 1

A 75-MW wind facility operates at a 36% capacity factor.

How much electricity does the facility generate in an 8,760-hour year?

Select the best answer.

A. 75 MW × 0.36 × 8,760 h = 236,520 MWh

✅ Correct! 😊 Capacity factor converts nameplate power to average power, and multiplying by time produces energy. The surviving unit is MWh.

B. 75 MW × 8,760 h = 657,000 MWh

❌ Not quite 😭 This assumes continuous nameplate output and omits the 36% capacity factor. It overstates annual generation.

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C. 75 MW × 0.36 = 27 MWh

❌ Not quite 😭 The multiplication gives 27 MW of average power, not energy. Hours must be included before the answer can be reported in MWh.

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D. 75 MW × 0.36 × 87,600 h = 2,365,200 MWh

❌ Not quite 😭 This uses ten times the stated annual hours. The structure is otherwise dimensional, but the substituted time makes the result too large.

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Full explanation
  • Step 1Convert 36% to the decimal capacity factor 0.36.
  • Step 2Average power is 75 MW × 0.36 = 27 MW.
  • Step 3Annual energy is 27 MW × 8,760 h = 236,520 MWh.
Answer. A. 75 MW × 0.36 × 8,760 h = 236,520 MWh
Model answer. A — “236,520 MWh”
Scoring guideline. Calculation credit requires distinguishing power from energy, applying capacity factor as a decimal, multiplying by operating time, and reporting MWh.

Practice 2: Evaluate an island electricity portfolio

Q. Which response uses the data to reduce imports while preserving firm supply? Read the exam question below, choose one option, then tap it to check.
SECTION II · FREE RESPONSE · QUESTION 210 points

Per installed megawatt, diesel has a 72% capacity factor, imports fuel, and is dispatchable; wind has 38%, no imported fuel, and is not dispatchable; solar has 24%, no imported fuel, and is not dispatchable; geothermal has 88%, no imported fuel, and is dispatchable.

Identify the source that best combines high capacity factor with dispatchability, calculate annual generation from 1 MW of wind, explain one reliability limitation of equal nameplate replacement, and propose a portfolio that reduces fuel imports.

Show your work.

A. Replace diesel with equal wind nameplate capacity and assume equal annual output and timing.

❌ Not quite 😭 Equal nameplate capacity does not guarantee equal annual generation or availability when demand occurs. Wind has a lower capacity factor and is not dispatchable.

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B. Use geothermal for firm generation and combine wind or solar with storage and demand management.

✅ Correct! 😊 Geothermal supplies dispatchable local power at the highest listed capacity factor. Storage and demand management help balance the variable sources.

C. Retain only diesel because dispatchability makes its fuel renewable.

❌ Not quite 😭 Dispatchability describes control over output, not resource renewal. Diesel still depends on imported fuel that forms far more slowly than it is consumed.

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D. Use only solar because annual average generation proves hourly reliability.

❌ Not quite 😭 An annual average does not show alignment with nighttime or peak demand. Solar is listed as nondispatchable, so balancing resources remain necessary.

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Full explanation
  • Step 1Filter for dispatchability and high capacity factor: geothermal qualifies at 88% and requires no imported fuel.
  • Step 2One MW of wind produces 1 MW × 8,760 h × 0.38 = 3,328.8 MWh/year.
  • Step 3Wind's variable output may not coincide with demand, so equal nameplate replacement does not preserve service.
  • Step 4Combine firm geothermal with wind or solar, storage, and demand management to reduce imports and balance timing.
Answer. B. Use geothermal for firm generation and combine wind or solar with storage and demand management.
Model solution. Geothermal best combines dispatchability with the highest listed capacity factor, 88%, and it does not require imported fuel. One installed MW of wind generates 1 MW × 8,760 h × 0.38 = 3,328.8 MWh per year. Equal wind and diesel nameplate capacities are not equivalent because wind has lower, variable output that may not align with demand. A defensible portfolio would use geothermal as firm local generation and add wind or solar with storage and demand management. This reduces imports while retaining resources that can balance variable supply.
Scoring guideline. Quantitative-data credit requires a valid capacity-factor calculation; reasoning credit requires separating nameplate, annual output, dispatchability, and timing before recommending an equivalent-service portfolio.
Glossary

Key terms for Energy Resources and Consumption

Renewable and Nonrenewable Resources
Global Energy Consumption
Fuel Types and Uses
Distribution of Natural Energy Resources
Fossil Fuels
Nuclear Power
Energy from Biomass
Solar Energy
Hydroelectric Power
Geothermal Energy
Hydrogen Fuel Cell
Wind Energy
FAQ

Energy Resources and Consumption FAQ

Why must power and energy remain separate in calculations?

Power is a rate measured in units such as watts, while energy is power multiplied by time, such as kilowatt-hours. Confusing them produces a dimensional error even when arithmetic looks plausible.

What does capacity factor tell me?

Capacity factor compares actual generation over an interval with generation at continuous nameplate output. It captures availability and operation, not conversion efficiency, market share, or environmental quality.

How should lifecycle emissions be compared across sources?

Use the same functional unit, system boundary, and time horizon. Include relevant extraction, construction, operation, fuel processing, and retirement stages before interpreting the reported grams per kilowatt-hour.

Why is nuclear energy considered low-carbon but not impact-free?

Operational greenhouse emissions are low, yet mining, enrichment, plant construction, thermal discharge, accident risk, and long-lived radioactive waste remain relevant environmental and governance considerations globally.

What makes an energy portfolio more than a list of technologies?

A defensible portfolio matches hourly demand, reliability, transmission, storage, cost, emissions, land and water impacts, and community constraints. Annual generation totals alone do not prove feasible operation.

Study strategy

How to study Energy Resources and Consumption

Create an energy ledger with nameplate power, capacity factor, operating hours, generation, emissions intensity, and total emissions. Carry units through every conversion so megawatts, megawatt-hours, grams per kilowatt-hour, and metric tons cannot collapse together.

Compare technologies using equivalent service. Match the same annual energy and reliability target before discussing lifecycle emissions, extraction, waste, land, water, transmission, storage, and community impacts; otherwise the numbers answer different questions.

Practice portfolio reasoning on an hourly timeline. Mark variable supply, demand peaks, storage power and duration, transmission constraints, and dispatchable resources. Then explain why annual averages are necessary evidence but insufficient proof of reliability.

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