The cheapest, cleanest kilowatt-hour is the one you never use. Before we build a single new power plant, we can slash demand simply by wasting less — better insulation, LED bulbs, efficient engines, capturing waste heat. Energy efficiency and conservation are the least glamorous and most powerful tools in the whole unit, and they're where APES concentrates its math: given a power rating and hours of use, calculate energy, cost, and CO₂; given input and useful output, calculate efficiency. This lesson closes Unit 6 by drilling those calculations with clean dimensional analysis — the single most reliable way to earn FRQ 3 points — and by cataloging the efficiency measures the exam expects.
Both reduce demand, emissions, and cost. Most systems are far from perfectly efficient: a typical incandescent bulb converts only ~5–10% of energy to light (the rest is waste heat); an internal-combustion car engine is only ~20–25% efficient; a coal plant is ~33–40% efficient. Efficiency improvements have huge leverage.
Energy efficiency is quantified as:
efficiency (%) = (useful energy output ÷ energy input) × 100
[DIAGRAM: Energy-flow (Sankey-style) diagram for an incandescent bulb — 100 units energy in; ~10 units out as light (useful), ~90 units lost as heat. Compare to an LED — 100 in, ~90 out as light, ~10 lost as heat. Efficiency = useful out ÷ in.]
The exam's guaranteed calculations. Core relationships:
- Energy = Power × Time → kWh = kW × h
- Cost = Energy × price → $ = kWh × ($/kWh)
- Convert units carefully: 1 kW = 1,000 W.
- CO₂ or fuel: multiply energy by an emission factor (e.g., kg CO₂ per kWh) or divide by an energy-content factor.
The method: write each quantity with its units, arrange conversion factors so units cancel, leaving the target unit. This is the same dimensional analysis you used for aquifers (Lesson 15) and half-lives (Lesson 20).
Efficiency/conservation is the top "reduce energy impact" FRQ answer, and the math (energy → cost → CO₂) is the backbone of FRQ 3. Mastering clean dimensional analysis here pays off across the entire exam.
Classify each as conservation or efficiency: (i) replacing incandescent bulbs with LEDs, (ii) turning off lights when leaving a room.
Solution: (i) Efficiency — technology delivering the same light with less energy. (ii) Conservation — a behavior change that uses less energy.
Interpretation: Technology = efficiency; behavior = conservation.
A power plant takes in 1,000 MW of energy from fuel and produces 380 MW of electricity. Calculate its efficiency.
Strategy: efficiency = useful output ÷ input × 100.
Solution:
efficiency = (380 / 1,000) × 100 = 38%
Answer: 38% (typical of a coal plant; the other 62% is lost mostly as waste heat).
Interpretation: Most fuel energy in a thermal plant is lost as heat — hence the value of cogeneration.
A household uses 900 kWh per month. Electricity costs $0.12 per kWh. (a) Monthly cost? (b) If LED upgrades cut use by 20%, monthly savings?
Strategy: cost = kWh × $/kWh; savings = 20% of cost.
Solution:
(a) 900 kWh × $0.12/kWh = $108/month
(b) 20% × 900 = 180 kWh saved → 180 × $0.12 = $21.60/month saved
Answer: $108/month; save $21.60/month.
Interpretation: Track units ($/kWh × kWh = $). Percentage savings apply to energy, then convert to dollars.
A home uses 12,000 kWh/year. The grid emits 0.5 kg CO₂ per kWh. (a) Annual CO₂? (b) If efficiency cuts use to 9,000 kWh, how much CO₂ is avoided?
Strategy: multiply energy by the emission factor; units cancel.
Solution:
(a) 12,000 kWh/yr × 0.5 kg CO₂/kWh = 6,000 kg CO₂/yr
(b) new: 9,000 kWh × 0.5 = 4,500 kg CO₂/yr
avoided: 6,000 − 4,500 = 1,500 kg CO₂/yr
Answer: 6,000 kg/yr; avoids 1,500 kg CO₂/yr.
Interpretation: kWh × (kg CO₂/kWh) = kg CO₂ — the kWh cancels. This is the archetypal FRQ 3 chain.
1 kW = 1,000 W; kWh = kW × h, not W × h (unless you then divide by 1,000).200/500 × 100 = 40%.1.2 kW × 0.5 h × 30 = 18 kWh.800 × $0.15 = $120.(B) Hybrid/electric vehicles and higher fuel-economy standards.
Conservation = using less through behavior (e.g., turning off lights); efficiency = technology delivering the same service with less energy (e.g., LED bulbs). Reducing demand is cheaper and cleaner because avoided energy needs no fuel, produces no emissions, and costs less than building/operating a new power plant ("the cheapest kWh is the one you don't use").
(a) 0.06 kW × 6 h = 0.36 kWh/day. (b) 0.36 × 365 = 131.4 kWh/yr × $0.10 = $13.14/yr. (c) LED: 0.01 kW × 6 h × 365 = 21.9 kWh/yr × $0.10 = $2.19; savings = $13.14 − $2.19 = $10.95/yr.
FRQ rubric (10 pts):
- (a) 1 pt setup 10,000 kWh × $0.13; 1 pt = $1,300/yr. (2)
- (b) 1 pt setup 10,000 kWh × 0.6 kg/kWh; 1 pt = 6,000 kg CO₂/yr. (2)
- (c) 1 pt new use 10,000 × 0.75 = 7,500 kWh; 1 pt new CO₂ 7,500 × 0.6 = 4,500 kg; 1 pt avoided 6,000 − 4,500 = 1,500 kg CO₂/yr (accept 2,500 kWh × 0.6 = 1,500). (3)
- (d) 1 pt efficiency measure + justification (LEDs, insulation, Energy Star, efficient HVAC); 1 pt conservation measure + justification (thermostat setbacks, turning off devices, reduced use); 1 pt both justifications tie to lower kWh/CO₂/cost. (3)
A town wants to cut electricity use and CO₂ emissions. An average home uses 10,000 kWh/year; the grid emits 0.6 kg CO₂ per kWh; electricity costs $0.13/kWh.
(a) Calculate one home's annual electricity cost. Show work. (2 pts) (b) Calculate one home's annual CO₂ emissions. Show work. (2 pts) (c) An efficiency program cuts each home's use by 25%. Calculate the new annual kWh and the CO₂ avoided per home. Show work. (3 pts) (d) Propose one efficiency measure and one conservation measure the town could promote, and justify each. (3 pts)
MC:
1. (B) Efficiency (technology).
2. (B) Conservation (behavior).
3. (B) 200/500 × 100 = 40%.
4. (C) Mechanical pumps/fans = active solar, not passive.
5. (B) 1.2 kW × 0.5 h × 30 = 18 kWh.
6. (B) Capturing and using waste heat.
7. (C) 800 × $0.15 = $120.
8. (B) Waste heat.
9. (B) Useful output ÷ input × 100.
10. (B) Hybrid/electric vehicles and higher fuel-economy standards.
Conservation = using less through behavior (e.g., turning off lights); efficiency = technology delivering the same service with less energy (e.g., LED bulbs). Reducing demand is cheaper and cleaner because avoided energy needs no fuel, produces no emissions, and costs less than building/operating a new power plant ("the cheapest kWh is the one you don't use").
(a) 0.06 kW × 6 h = 0.36 kWh/day. (b) 0.36 × 365 = 131.4 kWh/yr × $0.10 = $13.14/yr. (c) LED: 0.01 kW × 6 h × 365 = 21.9 kWh/yr × $0.10 = $2.19; savings = $13.14 − $2.19 = $10.95/yr.
FRQ rubric (10 pts):
- (a) 1 pt setup 10,000 kWh × $0.13; 1 pt = $1,300/yr. (2)
- (b) 1 pt setup 10,000 kWh × 0.6 kg/kWh; 1 pt = 6,000 kg CO₂/yr. (2)
- (c) 1 pt new use 10,000 × 0.75 = 7,500 kWh; 1 pt new CO₂ 7,500 × 0.6 = 4,500 kg; 1 pt avoided 6,000 − 4,500 = 1,500 kg CO₂/yr (accept 2,500 kWh × 0.6 = 1,500). (3)
- (d) 1 pt efficiency measure + justification (LEDs, insulation, Energy Star, efficient HVAC); 1 pt conservation measure + justification (thermostat setbacks, turning off devices, reduced use); 1 pt both justifications tie to lower kWh/CO₂/cost. (3)
⭐ Exam strategy: FRQ 3 lives on dimensional analysis. Write every number with units, line up conversion/emission factors so unwanted units cancel, and check that the surviving unit is what the question asked for. kWh × $/kWh = $ and kWh × kg CO₂/kWh = kg CO₂ are the two chains you'll use most.
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