Open with a striking number: a typical US coal-fired power plant is ~35% efficient — for every 100 units of chemical energy in the coal, 65 units leave the smokestack and cooling tower as waste heat. Ask: 'If I told you we could nearly double that number without building anything new — no solar panels, no wind farms — what would you guess we're doing?' Take 2–3 student guesses, then reveal: capturing the waste heat (cogeneration) and not needing the energy in the first place (efficiency + conservation). Frame today: the cheapest, cleanest kilowatt-hour is the one we never had to generate. Tie to a real case — NYU's cogeneration plant in Manhattan supplies ~90% of the campus's electricity AND heats/cools 22 buildings from the same natural gas, running at ~80% overall efficiency vs the ~33% grid average.

1. 7m

   Conservation vs. Efficiency vs. Curtailment

   Content

   Energy conservation is the umbrella term for reducing energy use. Inside that umbrella live two very different strategies. Energy efficiency means delivering the same service with less energy — swapping a 60 W incandescent bulb for a 9 W LED still gives you ~800 lumens of light, but uses 85% less electricity. Curtailment means cutting back the service itself — turning the bulb off, or setting the thermostat to 62 °F in winter instead of 70 °F. Both reduce kWh, but efficiency preserves quality of life while curtailment sacrifices some of it. An ENERGY STAR refrigerator is efficiency; unplugging the fridge is curtailment. A worked example: a household uses 900 kWh/month. Replacing all incandescents with LEDs cuts 90 kWh/month (efficiency, no lifestyle change). Also raising the AC setpoint from 72 °F to 78 °F cuts another 60 kWh (curtailment — you feel warmer). Both count as conservation; only the first is efficiency.

   Delivery

   Emphasize that on the AP exam, 'propose one conservation measure and one efficiency measure' is a classic FRQ prompt — students lose points by giving two efficiency answers. Cold-call: 'Is carpooling efficiency or curtailment?' (curtailment of solo driving / behavioral). 'Is a hybrid car efficiency or curtailment?' (efficiency — same trip, less fuel). Pre-empt the misconception that conservation always means sacrifice — most US conservation gains since 1975 came from efficiency standards (CAFE, appliance standards), not from people using less. Targets SP 1.

2. 7m

   Cogeneration and the Second Law

   Content

   The second law of thermodynamics guarantees that any heat engine — coal plant, gas turbine, car engine — must reject some heat to a cold reservoir. A conventional condensing coal plant converts ~35% of fuel energy to electricity; the other ~65% leaves as warm water and stack gas. Cogeneration, also called combined heat and power (CHP), captures that rejected heat and pipes it as steam or hot water to nearby buildings or industrial processes. The electrical efficiency drops slightly (say from 35% to 30%) because the steam is extracted at higher temperature, but the CAPTURED heat — often 50–55% of fuel energy — becomes useful output. Total useful energy: 30% electricity + 50% heat = 80% overall efficiency, more than double a standard plant. Worked example: burn 100 MJ of natural gas. Conventional plant → 35 MJ electricity, 65 MJ dumped. CHP plant → 30 MJ electricity + 50 MJ district heat = 80 MJ useful. The catch: the heat customer must be within a few kilometers of the plant, because hot water loses heat over distance. That's why CHP is common at hospitals, universities, and industrial parks, not at remote power stations.

   Delivery

   Anchor the flow diagram in the deck — fuel in on the left, two useful outputs on the right (electricity + heat), a much smaller waste-heat arrow. Ask: 'Why isn't every US power plant a CHP plant?' (they were sited far from cities to keep smoke away; the heat can't travel far). This directly targets the misconception that cogeneration is exotic — it's standard practice in Denmark (~60% of electricity is CHP) and growing in US industry. Targets SP 2 and SP 6.

3. 6m

   Where the Savings Live: Buildings, Appliances, Phantom Loads

   Content

   In the US, buildings consume ~40% of primary energy, with space heating, cooling, and water heating dominating. The single highest-leverage efficiency measure in most homes is the envelope — insulation and air sealing. Adding attic insulation from R-19 to R-49 typically cuts heating energy 15–25% with a payback of 3–7 years. Sealing air leaks around windows, doors, and duct penetrations often saves another 10–15%. Then come appliances: an ENERGY STAR refrigerator uses ~400 kWh/year vs. ~1,500 kWh for a 1990 model. LED lighting cuts lighting energy ~85% vs. incandescent. Finally, phantom loads — devices drawing power in standby (game consoles, cable boxes, chargers) — account for 5–10% of residential electricity, roughly $100/year for an average US household. An energy audit ranks these by cost per kWh saved. Consistently, the cheapest 'kilowatt-hour' costs 2–3¢ to save through efficiency, versus 10–15¢ to generate from a new gas plant and 15–25¢ from new nuclear — hence 'negawatts are cheaper than megawatts' (Amory Lovins).

   Delivery

   Ask students to rank, from memory, which uses more energy in a US home: lighting, refrigerator, or heating/cooling. (Heating/cooling ~50%, water heating ~18%, refrigerator ~4%, lighting ~4%.) This challenges the misconception that turning lights off is the biggest lever — it isn't; the envelope is. Also address the misconception that individual actions are negligible: US appliance standards since 1987 save ~5% of national electricity annually — that's aggregated individual purchases. Targets SP 1 and SP 7.

1. 10m

   Cogeneration Efficiency Calculation (AP FRQ-style)

   Targets SP 2 (interpret CHP flow diagram) and SP 6 (Mathematical Routines). Students work individually for 5 minutes, then pair-share for 3 minutes to check answers; 2 minutes to review as a class. Walk around and check unit tracking — the classic error is forgetting that MJ_electricity and MJ_heat are both useful output. Student handout — Cogeneration at Cornell Cornell University operates a natural-gas cogeneration plant on its Ithaca campus. In one hour of operation, the plant burns natural gas containing 500 GJ of chemical energy. - Electricity produced: 150 GJ - Heat captured as steam delivered to campus buildings: 250 GJ - Waste heat rejected to Cayuga Lake and stack: 100 GJ A nearby conventional natural-gas power plant, using the same fuel input of 500 GJ/hr, produces 225 GJ of electricity and vents 275 GJ as waste heat (no heat is captured). Part 1 — Calculations (show work with units): 1. Calculate the electrical efficiency of the Cornell CHP plant. Electrical efficiency = ______ % 2. Calculate the overall (electrical + thermal) efficiency of the Cornell CHP plant. Overall efficiency = ______ % 3. Calculate the efficiency of the conventional plant. Efficiency = ______ % 4. If the campus needs both 150 GJ of electricity AND 250 GJ of heat each hour, how much natural gas (GJ of fuel) would be required if the campus instead bought grid electricity (at 45% efficiency) and burned separate natural gas in campus boilers (at 85% efficiency) for heat? Fuel needed = ______ GJ Part 2 — Explain (SP 1): 5. In 2–3 sentences, explain why the Cornell CHP plant's overall efficiency exceeds the conventional plant's, even though its electrical efficiency is lower. Reference the second law of thermodynamics. Answer in complete sentences — one-word answers will not receive credit.

   Materials

   • Calculator
   • Student handout (below)

   Example outputs

   • 1) 150/500 = 30% electrical efficiency. 2) (150+250)/500 = 400/500 = 80% overall efficiency. 3) 225/500 = 45% efficiency. 4) Electricity: 150/0.45 = 333 GJ fuel. Heat: 250/0.85 = 294 GJ fuel. Total = 627 GJ — vs. 500 GJ at Cornell CHP, a 20% fuel savings.
   • 5) The second law forces every heat engine to reject some heat. The conventional plant vents that heat to the environment (275 GJ wasted); the CHP plant sacrifices a little electrical output (30% vs 45%) but captures 250 GJ of that rejected heat as useful steam, so more of the fuel's energy leaves the plant as a useful product rather than as pollution.
2. 15m

   Lifetime Lighting Analysis + Conservation RecommendationLab

   Targets SP 4 (measure/verify wattages), SP 5 (interpret the bar chart in the deck), SP 6 (lifetime cost math), and SP 7 (justify a solution). Groups of 3. First 4 minutes: each group plugs each bulb into the Kill-A-Watt via the power strip and records actual measured wattage — this is the SP 4 verification step (spec sheets vs measurement). Next 8 minutes: complete the calculation table and write the recommendation. Last 3 minutes: two groups share their recommendation aloud. If the Kill-A-Watt reads slightly different from the printed wattage (e.g., LED reads 8.7 W instead of 9 W), students should use their measured value — this is the point. Student handout — Which bulb should the school district buy? The district is replacing 5,000 hallway light bulbs that each burn 12 hours per day, 365 days per year. Electricity costs $0.15/kWh. Complete the table using your measured wattages. Part 1 — Measure and calculate (per bulb, over 25,000 hours of use): - Incandescent — measured wattage: ______ W - Rated life: 1,000 hours · Purchase price: $1.00 - kWh used over 25,000 hr = ______ - Electricity cost = ______ - Bulbs needed (25,000 / rated life) = ______ - Bulb purchase cost = ______ - Total lifetime cost = ______ - CFL — measured wattage: ______ W - Rated life: 10,000 hours · Purchase price: $2.00 - kWh used = ______ - Electricity cost = ______ - Bulbs needed = ______ - Bulb purchase cost = ______ - Total lifetime cost = ______ - LED — measured wattage: ______ W - Rated life: 25,000 hours · Purchase price: $5.00 - kWh used = ______ - Electricity cost = ______ - Bulbs needed = ______ - Bulb purchase cost = ______ - Total lifetime cost = ______ Part 2 — Scale up (5,000 bulbs, district-wide): 6. District-wide electricity savings switching from incandescent to LED (kWh over 25,000 hr): ______ 7. District-wide dollar savings: ______ Part 3 — Justify (SP 7, ~4 sentences): 8. Write a recommendation to the school board. Identify whether your recommendation is an efficiency measure or a curtailment measure, cite one specific number from your table, and address one likely objection (upfront cost, disposal, light quality, etc.). Every dollar figure must show its unit; every kWh must show its unit. Answers without units lose credit.

   Materials

   • Kill-A-Watt meter (or similar plug-in wattmeter)
   • One incandescent bulb (60 W, if available — or use published spec)
   • One CFL bulb (14 W)
   • One LED bulb (9 W)
   • Power strip
   • Calculator
   • Student handout (below)

   Example outputs

   • Incandescent: 60 W × 25,000 hr = 1,500 kWh; electricity = 1,500 × $0.15 = $225; bulbs needed = 25; bulb cost = $25; total = $250. CFL: 14 W × 25,000 = 350 kWh; electricity = $52.50; 2.5 bulbs × $2 = $5; total = $57.50. LED: 9 W × 25,000 = 225 kWh; electricity = $33.75; 1 bulb × $5 = $5; total = $38.75.
   • District: incandescent → LED saves (1,500 − 225) × 5,000 = 6,375,000 kWh and (250 − 38.75) × 5,000 = $1,056,250 over 25,000 hr. Recommendation: 'Switch all 5,000 hallway bulbs to LED. This is an efficiency measure — same light output (~800 lumens) with 85% less electricity. Over 25,000 hours the district saves $1.06 million and 6.4 million kWh. The higher upfront cost ($5 vs $1) is repaid in under 4 months of electricity savings per bulb.'

   No-equipment fallback

   If no Kill-A-Watt is available, provide printed manufacturer wattages (Incandescent 60 W, CFL 14 W, LED 9 W) and note that students are using spec-sheet values rather than measured — briefly discuss why measurement matters (LEDs sometimes draw less than rated; incandescents sometimes draw more when line voltage is high).

1. Which of the following is an example of energy efficiency rather than energy curtailment? (A) Setting the office thermostat to 65 °F in winter instead of 70 °F (B) Replacing a 1998 refrigerator with an ENERGY STAR model that uses 70% less electricity (C) Carpooling to work with three coworkers instead of driving alone (D) Turning off lights when leaving a room

   multiple choice(B). Efficiency delivers the same service (refrigeration) with less energy through a technology change. A, C, and D all reduce the service level or amount used — those are curtailment/behavioral conservation. Targets SP 1.

2. A natural gas power plant burns fuel containing 1,000 MJ of chemical energy each second. In conventional operation it produces 400 MJ/s of electricity and rejects the rest as waste heat. If retrofitted for cogeneration, it produces 350 MJ/s of electricity and captures 500 MJ/s of the previously-wasted heat as usable steam for a neighboring paper mill. Calculate (a) the overall efficiency before the retrofit and (b) the overall efficiency after the retrofit. Show work with units.

   calculation(a) Before: 400 MJ/s useful / 1,000 MJ/s fuel = 0.40 = 40% overall efficiency. (b) After: (350 MJ/s electricity + 500 MJ/s heat) / 1,000 MJ/s fuel = 850/1000 = 0.85 = 85% overall efficiency. The retrofit sacrifices 50 MJ/s of electricity but captures 500 MJ/s of formerly-wasted heat, more than doubling useful energy output per unit of fuel burned. Targets SP 6.

3. Explain why the second law of thermodynamics limits any single-purpose electric power plant to well below 100% efficiency, and describe how cogeneration works around this limit without violating the second law.

   short answerThe second law states that no heat engine can convert all input heat into work; some energy must be rejected to a cold reservoir. That's why coal and gas plants max out around 35–45% electrical efficiency. Cogeneration does NOT violate the second law — the plant still rejects heat exactly as required. The trick is that the rejected heat, instead of being dumped into a river or the atmosphere, is captured at a useful temperature (steam or hot water ~100–150 °C) and piped to buildings or industry. The heat is still 'waste' from the electricity-generating perspective, but it becomes a useful product because a nearby customer needs low-grade heat. Total useful output can reach 80–90%. Targets SP 1.

4. A homeowner claims that switching from incandescent to LED bulbs is meaningless because 'one house can't affect climate change.' Using both a specific calculation and an aggregation argument, write a 3–4 sentence rebuttal.

   short answerSample: A single 60 W incandescent replaced by a 9 W LED burning 3 hr/day saves (60 − 9) W × 3 hr × 365 days = 55.8 kWh/year — about 40 kg of CO₂/year on the average US grid. That's small alone. But there are ~130 million US households; if each replaces just 10 bulbs the annual savings exceed 70 billion kWh, roughly the output of 20 large coal plants. Individual efficiency actions are negligible in isolation but decisive in aggregate — which is why federal appliance standards (aggregating individual purchases) have cut ~5% of US electricity demand since 1987. Targets SP 6 and SP 7.

• ☐Calculators (1 per student)
• ☐Kill-A-Watt or equivalent plug-in wattmeter (1 per group of 3)
• ☐60 W incandescent bulb + socket lamp (1 per group; if unavailable use spec-sheet value)
• ☐14 W CFL bulb (1 per group)
• ☐9 W LED bulb (1 per group)
• ☐Power strip (1 per group)
• ☐Printed cogeneration handout (1 per student)
• ☐Printed lifetime lighting handout (1 per student)
• ☐Printed formative assessment (1 per student)