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S6 Physics: Energy Use – Error Correction Drill (10 Questions)
S6 Physics: Energy Use – Error Correction Drill (10 Questions)
This paper targets the mistakes that recur in the previous three: converting kWh into joules, billing per kilowatt-hour, comparing efficacy at equal light output, effective sunlight hours, plant efficiency and waste heat, transmission loss rising with the square of current, transformer turns ratio with power conservation, heat pump performance above one, the cube law for wind power and the Betz limit, ending with three unit and conservation misconceptions.
Quick answer The kettle takes 0.167 kWh or 600 kJ; the 1.5 kW air conditioner uses 12 kWh costing 14.4 dollars; 60 W against 9 W gives 13.3 against 88.9 lumens per watt and a 93.1 kWh yearly saving; a 300 W panel gives 1.2 kWh a day and 438 kWh a year; a 38 percent plant draws 1053 MW and rejects 653 MW; the 400 kV line loses 0.30 MW against 120 MW at 20 kV; 400 to 20 turns gives 12 V and a primary current of 0.10 A; a heat pump at 3.5 uses 2.86 kWh and saves 71.4 percent; the wind carries 1.04 MW and the Betz limit gives 614 kW; one kWh is 3.6 times 10 to the 6 joules, the waste heat is 0.2 kWh, and the turbine ceiling is 59.3 percent.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 questions |
| Suggested time | about 25 minutes |
| Difficulty mix | Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1 |
| Chapter | Chapter: S6 Physics | Energy Use: Resources, Efficiency and Transmission |
| Level | S6 |
02 Questions (10 items)
- FoundationA 2.0 kW kettle takes 5 minutes to boil. How much electrical energy does it use in kWh, and how many kilojoules is that?
- FoundationAn air conditioner rated 1.5 kW runs for 8 hours at 1.2 dollars per kWh. What energy does it use and what does it cost?
- FoundationA 60 W filament lamp and a 9 W LED both give 800 lumens. What is the luminous efficacy of each in lumens per watt, and how many kWh does the LED save over 365 days at 5 hours a day?
- FoundationA 300 W solar panel receives on average 4.0 effective sunlight hours a day. How much energy does it produce daily and in a 365-day year?
- IntermediateA coal plant of 38 percent efficiency delivers 400 MW of electrical power. What are the fuel input power and the waste heat rejected, in megawatts?
- IntermediateA 400 kV line carries 200 MW with a total resistance of 1.2 ohms. What are the current and the line loss, and what is the loss if the same power goes at 20 kV?
- IntermediateAn ideal transformer has 400 turns on the primary and 20 on the secondary, with 240 V on the primary. What is the secondary voltage, and what primary current flows when the secondary delivers 2.0 A?
- ChallengeA heat pump has a coefficient of performance of 3.5 while a resistive heater has 1.0. To deliver 10 kWh of heat, how much electricity does the heat pump use, and what percentage does it save?
- ChallengeWind blows at 12 m/s through a rotor swept area of 1000 square metres where air density is 1.2 kg per cubic metre. What is the power in the wind, and what can a turbine extract at the Betz limit?
- Error redoA student writes that 1 kWh equals 1000 J, that a device of 80 percent efficiency fed with 1 kWh produces 0.8 kWh of waste heat, and that a wind turbine can extract all the kinetic energy of the wind so its efficiency can reach 100 percent. Identify the three errors and correct each.
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 0.167 kWh, which is 600 kJ, since 2.0 times 5 over 60 gives 0.167 kWh and one kWh is 3.6 times 10 to the 6 joules | A kilowatt-hour is power times hours, and converting to joules needs a factor of 3.6 times 10 to the 6, not one thousand. |
| 2 | 12 kWh and 14.4 dollars, because 1.5 times 8 is 12 kWh and 12 times 1.2 is 14.4 dollars | Electricity is billed per kilowatt-hour, so neither the power alone nor the hours alone gives the charge. |
| 3 | The efficacies are 13.3 and 88.9 lumens per watt, and the saving is 51 watts times 5 hours times 365 days, which is 93.1 kWh | To compare running cost, take the power difference at the same light output and multiply by the hours used. |
| 4 | It produces 1.2 kWh a day and 438 kWh a year, since 0.30 times 4.0 is 1.2 and 1.2 times 365 is 438 | Effective sunlight hours are not the same as daylight hours, because only the period near full rated output counts. |
| 5 | The input is 400 divided by 0.38, about 1053 MW, and the waste heat is the difference, about 653 MW | Efficiency is output over input, so the waste heat is the difference between them rather than 38 percent of the output. |
| 6 | At 400 kV the current is 500 A and the loss is 500 squared times 1.2, which is 0.30 MW; at 20 kV the current is 10 000 A and the loss is 120 MW | Loss rises with the square of the current, so raising the voltage twentyfold cuts the current twentyfold and the loss four hundredfold. |
| 7 | The turns ratio is 20 to 1, so the secondary is 240 divided by 20, which is 12 V; power is conserved at 24 W, so the primary current is 24 divided by 240, which is 0.10 A | Voltage steps down with the turns ratio while current steps up by the same ratio, which is what keeps the power constant. |
| 8 | It uses 10 divided by 3.5, about 2.86 kWh, which saves about 71.4 percent compared with the heater | A heat pump moves heat rather than generating it, which is why its coefficient of performance can exceed one. |
| 9 | The wind power is half times 1.2 times 1000 times 12 cubed, about 1.04 MW, and the Betz limit gives about 614 kW | Wind power rises with the cube of the speed, and 59.3 percent is a theoretical ceiling that no turbine can exceed. |
| 10 | One kWh is 3.6 times 10 to the 6 joules; with 1 kWh in, the useful output is 0.8 kWh and the waste heat is the remaining 0.2 kWh, or 7.2 times 10 to the 5 joules; and a turbine can extract at most 16 over 27, about 59.3 percent, never 100 percent | All three slips confuse conversion with conservation, since energy is conserved but the useful output and the waste heat are separate shares that cannot both equal the input. |
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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