中文（繁體）
中文（简体）
English
S6 Physics: Energy and Use of Energy – Intermediate (10 Questions)
S6 Physics: Energy and Use of Energy – Intermediate (10 Questions)
This paper moves into generation and building systems: plant efficiency and waste heat, transmission voltage against resistive loss, transformer turns ratio, appliance current and equivalent resistance, fission energy, pumped storage, solar water heating, wall and window losses, air conditioner COP with internal gains, and a final item on three transmission and efficiency misconceptions.
Quick answer Plant efficiency is 38 percent with 620 MW of waste heat; 400 kV loses 0.31 MW against 125 MW at 20 kV; the turns ratio is 20 to 1 with 200 primary turns; the dryer draws 5.45 A through 40.3 ohms; 1 kg of uranium gives 8.2 times 10 to the 13 J, about 22.8 GWh; pumping needs 15.7 MW or 12.6 MW of electricity and stores 230 400 cubic metres; the heater takes 6.7 hours; the wall loses 630 W and the window 780 W, 55.3 percent of the total; the air conditioner draws 3.0 and 3.43 kW; and three misconceptions are corrected.
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 and Use of Energy: Sources, Efficiency and Household Electricity |
| Level | S6 |
02 Questions (10 items)
- FoundationA coal plant takes in 1 000 MW of chemical power and delivers 380 MW of electrical power. What is its efficiency, and how much power leaves as waste heat?
- FoundationA transmission line has a total resistance of 5 ohms and carries 100 MW. What current and what loss occur at 400 kV, and what happens at 20 kV?
- FoundationA transformer steps 11 kV up to 220 kV. What is the turns ratio, and how many turns does the primary need if the secondary has 4 000 turns?
- FoundationA 1 200 W hair dryer runs on 220 V. What current does it draw, and what is its equivalent resistance?
- IntermediateEach fission of uranium-235 releases about 200 MeV. How much energy does 1 kg of pure uranium-235 release, and how many GWh is that if all of it became electricity?
- IntermediateA pumped-storage plant lifts water 200 m at 8 cubic metres per second. What is the minimum pumping power, what electrical power is needed at 80 percent efficiency, and how much water is stored in 8 hours?
- IntermediateA solar water heater has 2.0 square metres of collector, an irradiance of 700 W per square metre and 50 percent collection efficiency. How long does it take to heat 100 kg of water from 10 to 50 degrees Celsius, given a specific heat capacity of 4 200 J per kg per kelvin?
- ChallengeA house has 120 square metres of wall with U equal to 0.35 and 20 square metres of window with U equal to 2.6, at a temperature difference of 15 K. What heat flow passes through each, and what percentage of the total is the window?
- ChallengeAn air conditioner has a COP of 2.8 and must remove 8.4 kW of heat from a room. What electrical power does the compressor need, and what if lamps and occupants add another 1.2 kW of heat?
- Error redoA student writes that high-voltage transmission makes the current larger and so wastes more, that an air conditioner at 100 percent efficiency is best, and that a coal plant reaches 90 percent. Correct all three and give the correct relations.
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | The efficiency is 380 divided by 1 000, which is 38 percent, so 1 000 minus 380, namely 620 MW, leaves as waste heat | Heat-engine efficiency is capped by the Carnot limit, so anything above 50 percent is already very good. |
| 2 | At 400 kV the current is 250 A and the loss is 250 squared times 5, about 0.31 MW or 0.31 percent; at 20 kV the current is 5 000 A and the loss is 125 MW, so the power never arrives | For the same power, raising the voltage twentyfold cuts the current twentyfold and the loss by a factor of four hundred. |
| 3 | The ratio is 220 divided by 11, which is 20 to 1, so the primary needs 4 000 divided by 20, namely 200 turns | For an ideal transformer the voltage ratio equals the turns ratio, and since power is unchanged the current changes in the opposite way. |
| 4 | The current is 1 200 divided by 220, about 5.45 A, and the resistance is 220 divided by 5.45, about 40.3 ohms | A hair dryer is not a pure resistor because it contains a fan motor, so the resistance found is only an equivalent value. |
| 5 | Each fission gives 200 times 1.602 times 10 to the minus 19, about 3.20 times 10 to the minus 11 J, and 1 kg holds about 2.56 times 10 to the 24 nuclei, so the total is about 8.2 times 10 to the 13 J, roughly 22.8 GWh | Convert mass to a number of nuclei through the mole, not by multiplying mass directly by the energy per fission. |
| 6 | The ideal power is 1 000 times 9.81 times 200 times 8, about 15.7 MW, and at 80 percent efficiency the input is 15.7 divided by 0.8, about 12.6 MW, while the volume is 8 times 8 times 3 600, namely 230 400 cubic metres | An efficiency below one means the input is larger than the ideal value, so the ideal power is divided by the efficiency, not multiplied. |
| 7 | The collected power is 2.0 times 700 times 0.5, which is 700 W, and the heat needed is 100 times 4 200 times 40, about 1.68 times 10 to the 7 J, so the time is that divided by 700, about 24 000 s or 6.7 hours | Find the heat collected per second first, then divide the total heat by it rather than treating the temperature rise as a time. |
| 8 | The wall loses 0.35 times 120 times 15, which is 630 W, and the window 2.6 times 20 times 15, which is 780 W, so the total is 1 410 W and the window takes 780 over 1 410, about 55.3 percent | The window is small in area but large in U-value, so it often dominates the loss, which is why double glazing pays. |
| 9 | The power is 8.4 divided by 2.8, which is 3.0 kW, and with the internal source the load becomes 9.6 kW, so the power is 9.6 divided by 2.8, about 3.43 kW | Everything that releases heat indoors must be removed, not just the cooling load originally quoted. |
| 10 | For the same power a higher voltage gives a smaller current since power is voltage times current, and the loss falls with the square of current; a larger COP is better and values above 3 are normal, so 100 percent is the worst case; and heat engines are limited in practice to roughly 35 to 45 percent | Each slip applies a rule of thumb or a linear relation where it does not hold. |
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.
Find a tutor freeTutor Registration →
Questions and answers are written originally by the TutorZone team and are not taken from any school, publisher or examination authority paper. Every value is checked by computer.