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S6 Physics: Energy and Use of Energy – Foundation (10 Questions)
S6 Physics: Energy and Use of Energy – Foundation (10 Questions)
This paper builds the basics: converting between kWh and joules, electricity bills, lamp efficacy in lumens per watt, solar panel output, the cube law and Betz limit for wind, hydro power from density, head and flow, carbon intensity, heat pump COP, wall U-value losses, and a final item that clears three common misconceptions.
Quick answer The kettle uses 0.167 kWh or 600 kJ; the air conditioner 12 kWh at 14.4 dollars; 60 W gives 13.3 lumens per watt against 88.9 for the LED, saving 51 kWh; the panel gives 72 W; the wind holds 188 kW with a 112 kW limit; hydro gives 785 kW and 667 kW out; 4 500 kWh emits 2 700 kg; the heat pump saves 71.4 percent; the wall loses 129.6 W or 3.1 kWh a day; and three misconceptions are corrected.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 questions |
| Suggested time | about 20 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)
- FoundationAn electric kettle rated at 2 000 W boils water for 5 minutes. How much energy does it use in kWh, and how many joules is that?
- FoundationAn air conditioner of 1 500 W runs for 8 hours. How much energy does it use, and what is the cost at 1.2 dollars per kWh?
- FoundationA 60 W filament lamp and a 9 W LED both give a luminous flux of 800 lm. What is the efficacy of each in lm per W, and how much energy does the LED save over 1 000 hours?
- FoundationA solar panel of area 0.5 square metres has an efficiency of 18 percent under a noon irradiance of 800 W per square metre. What electrical power does it deliver at that moment?
- IntermediateA wind turbine has blades of radius 10 m in wind of 10 m/s where the air density is 1.2 kg per cubic metre. What is the power in the wind, and what is the most that can be extracted at the Betz limit of 16 over 27?
- IntermediateA small hydro plant has a head of 40 m and a flow of 2 cubic metres per second with water density 1 000 kg per cubic metre and g equal to 9.81. What is the power of the falling water, and what is the output at 85 percent efficiency?
- IntermediateA household uses 4 500 kWh in a year and the grid carbon intensity is 0.6 kg of carbon dioxide per kWh. What are the annual indirect emissions in kilograms?
- ChallengeA heat pump has a COP of 3.5, delivering 3.5 kWh of heat for each kWh of electricity. If a 100 percent efficient electric heater supplied the same heat, what percentage of electricity does the heat pump save?
- ChallengeA wall has a thermal transmittance of 1.8 W per square metre per kelvin, an area of 6 square metres and a temperature difference of 12 K. What is the heat flow, and how much energy leaks through it in 24 hours?
- Error redoA student writes that kWh is a unit of power, that an LED always saves energy because its resistance is low, and that doubling wind speed doubles turbine power. Correct all three and state the right relations.
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | The energy is 2 000 W times 5 sixtieths of an hour, about 0.167 kWh, which is 2 000 times 300, namely 600 000 J or 600 kJ | Convert either the power to kilowatts or the time to hours before multiplying, otherwise the unit of the answer is wrong. |
| 2 | The energy is 1.5 kW times 8 h, which is 12 kWh, so the bill is 12 times 1.2, namely 14.4 dollars | Electricity is billed by energy in kWh rather than by power, so the time must be multiplied in first. |
| 3 | The filament lamp gives 800 divided by 60, about 13.3 lm per W, while the LED gives 800 divided by 9, about 88.9 lm per W, so the saving is 51 times 1 000 watt-hours, namely 51 kWh | Compare lamps by efficacy in lumens per watt; a lower wattage alone does not prove a saving. |
| 4 | The output is 0.5 times 800 times 0.18, which is 72 W | The efficiency multiplies the incident power, and the incident power is the area times the irradiance. |
| 5 | The swept area is pi times 100, about 314 square metres, so the wind power is half times 1.2 times 314 times 10 cubed, about 188 kW, and the limit is 188 times 16 over 27, about 112 kW | Wind power rises with the cube of speed while the swept area rises with the square of radius. |
| 6 | The power is 1 000 times 9.81 times 40 times 2, about 785 kW, so the output is 785 times 0.85, about 667 kW | Hydro power is density times g times head times flow; leaving out any factor changes the answer by orders of magnitude. |
| 7 | The emissions are 4 500 times 0.6, which is 2 700 kg of carbon dioxide | The electricity itself carries no emissions; they come from generation, so the grid carbon intensity is the conversion factor. |
| 8 | The heater gives only 1 kWh of heat per kWh of electricity, so the fraction saved is 1 minus 1 divided by 3.5, about 71.4 percent | A heat pump moves heat rather than making it, which is why its COP can exceed one. |
| 9 | The heat flow is 1.8 times 6 times 12, which is 129.6 W, so in 24 hours it loses 129.6 times 86 400, about 1.12 times 10 to the 7 J or 3.1 kWh | The U-value already accounts for thickness and material, so multiplying by area and temperature difference is enough. |
| 10 | kWh is a unit of energy, power times time; saving depends on efficacy in lumens per watt and the flux actually needed, not on resistance; and wind power varies with the cube of speed, so doubling the speed gives eight times the power | All three slips come from confusing power with energy, resistance with efficacy, and a linear law with a cubic one. |
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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