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S6 Physics: Energy and Use of Energy – Challenge (10 Questions)
S6 Physics: Energy and Use of Energy – Challenge (10 Questions)
This paper is multi-step calculation and decision making: wind farm capacity factor, cost per kilometre for electric and petrol cars, rooftop solar coverage, weighted grid carbon intensity, heat pump against gas boiler emissions, usable battery duration, AC against DC long-distance transmission, street lamp retrofit savings, solar payback, and a final item on the Betz limit and two other misconceptions.
Quick answer Each turbine gives 1.32 MW and the farm 34.7 GWh a year; the electric car costs 0.18 dollars per km against 3.43 for petrol, a ratio of 19; the roof gives 18 kWh a day and covers 146 percent; carbon intensity is 0.57 kg per kWh; the heat pump emits 0.179 against 0.222 kg, 19.5 percent less; the battery gives 8 hours in theory and 7.2 in practice; 500 kV AC loses 16 percent against 3.9 percent for 800 kV DC; street lamps use 150 GWh and save 90 GWh, avoiding 51 500 tonnes; payback is 10.9 years; and the Betz limit is 59.3 percent.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 questions |
| Suggested time | about 30 minutes |
| Difficulty mix | Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1 |
| Chapter | Chapter: S6 Physics | Energy and Use of Energy: Sources, Efficiency and Household Electricity |
| Level | S6 |
02 Questions (10 items)
- FoundationA wind farm has 10 turbines with 30 m blades, a power coefficient of 0.45, wind at 12 m/s and air density 1.2. What does each turbine deliver, and what is the annual output of the farm at a capacity factor of 0.30?
- FoundationAn electric car has a 60 kWh battery and a real range of 400 km, charged at 1.2 dollars per kWh. A petrol car does 7 km per litre at 24 dollars per litre. Compare the energy cost per kilometre and the ratio.
- FoundationA 20 square metre roof is fully covered with 20 percent efficient panels, and the site averages 4.5 kWh per square metre per day. What is the average daily output, and what fraction of a 4 500 kWh annual demand does it cover?
- IntermediateA grid mix is 50 percent coal at 0.90 kg of carbon dioxide per kWh, 30 percent gas at 0.40, 15 percent nuclear at 0.01 and 5 percent renewables at 0.02. What is the average carbon intensity?
- IntermediateA heat pump has a COP of 3.2 on a grid of 0.57 kg of carbon dioxide per kWh, while a gas boiler is 90 percent efficient and burns gas emitting 0.20 kg per kWh. Compare the emissions per kWh of heat.
- IntermediateA battery store is rated 100 MW and 400 MWh with a round-trip efficiency of 90 percent. How long can it discharge at 50 MW in theory, and how long in practice?
- ChallengeFor the same 1 000 MW, a 500 kV AC line has 40 ohms of total resistance and an 800 kV DC line has 25 ohms. Compare the current, the loss in MW and the loss as a percentage.
- ChallengeA city has 150 000 street lamps of 250 W each, lit 11 hours a night. What is the annual energy in GWh, what is saved by switching all to 100 W LEDs, and how many tonnes of carbon dioxide are avoided at 0.57 kg per kWh?
- FoundationA 4 kW rooftop solar system costs 60 000 dollars, generates 4 600 kWh a year and offsets electricity at 1.2 dollars per kWh. What is the simple payback period, and what is the annual saving after that within a 25-year life?
- Error redoA student writes that a wind turbine can extract all the kinetic energy of the wind at 100 percent efficiency, that nuclear plants emit the most carbon dioxide, and that solar panels can reach 100 percent. Correct each and state the Betz limit with its reason.
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | The swept area is about 2 827 square metres, so the wind power is about 2.93 MW and the output about 1.32 MW per turbine, giving 13.2 MW for the farm and about 34.7 GWh a year at a capacity factor of 0.30 | The capacity factor converts rated capacity into real annual output, because a turbine is never at full load all year. |
| 2 | The electric car uses 60 divided by 400, which is 0.15 kWh per km, costing 0.18 dollars per km, while the petrol car costs 24 divided by 7, about 3.43 dollars per km, so the ratio is about 19 | Compare on the same per-kilometre basis, not the price of a kWh against the price of a litre. |
| 3 | The daily output is 20 times 0.20 times 4.5, which is 18 kWh, so the year gives 6 570 kWh and covers 6 570 over 4 500, about 146 percent | When the irradiance is already given in kWh per square metre per day, multiply by area and efficiency without adding a further factor for hours. |
| 4 | The weighted sum is 0.45 plus 0.12 plus 0.0015 plus 0.001, which gives about 0.57 kg of carbon dioxide per kWh | Carbon intensity is a weighted average whose weights are the shares of generation. |
| 5 | The heat pump emits 0.57 divided by 3.2, about 0.179 kg, while the boiler emits 0.20 divided by 0.90, about 0.222 kg, so the heat pump is about 19.5 percent lower | The climate advantage of a heat pump depends on how clean the grid is, and shrinks as the grid gets dirtier. |
| 6 | The theoretical time is 400 divided by 50, which is 8 hours, but only 360 MWh is usable after the 90 percent round trip, so the real time is 7.2 hours | The rated megawatt-hours are usually a direct-current figure, and the round trip loses energy on both charging and discharging. |
| 7 | The AC line carries 2 000 A and loses 160 MW, which is 16 percent, while the DC line carries 1 250 A and loses about 39 MW, which is 3.9 percent | Long-distance DC links avoid one conversion stage and run at a higher voltage, so the resistive loss drops sharply. |
| 8 | The lamps use 150 000 times 0.25 times 11 times 365, about 150 GWh a year, the LEDs need 60 GWh, so 90 GWh is saved, avoiding about 51 500 tonnes of carbon dioxide | Convert large numbers to kW and GWh before applying the carbon intensity, so the powers of ten stay under control. |
| 9 | The annual saving is 4 600 times 1.2, which is 5 520 dollars, so the payback is 60 000 divided by 5 520, about 10.9 years, and each later year saves 5 520 dollars | The simple payback ignores interest and maintenance, so real projects must also allow for panel degradation and servicing. |
| 10 | The Betz limit is 16 over 27, about 59.3 percent, because the air must leave with some speed for flow to continue, and extracting everything would block the wind; nuclear plants emit almost nothing while running, with most emissions from construction; and solar cells are limited by thermodynamics and the spectrum, with single junctions reaching about 20 to 25 percent | All three slips mistake a theoretical ceiling for an achievable 100 percent. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
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