The problems are on the energy efficiency worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Efficiency is the fraction of the total input energy that becomes useful output: efficiency = useful output ÷ total input, often shown as a percentage. These worked answers correct the 20 problems of the Physics 9-10 worksheet on energy efficiency, from name the store to motor and load data: each solution shows the formula, the substitution and the final answer with its unit.
Each answer gives the equation, the substitution and the result with its unit in bold. Use them to check your method as well as your numbers. Where g is needed, the value 9.8 N/kg is used, and efficiency is shown as a percentage.
Level: Physics grades 9-10 (GCSE · IGCSE · O Level · Class 9-10) · Curriculum: GCSE P1, HS-PS3-4
Problem 1 Name the store
1. Elastic potential.2. Thermal.
3. Chemical.
4. Gravitational potential.
5. Kinetic.
Problem 2 Hydroelectric chain
1. Gravitational potential (lake) → kinetic (falling water) → kinetic (turbine) → electrical (lamps).2. Thermal energy (from friction and electrical resistance) and sound are wasted in the pipes, turbine and generator.
Problem 3 Fan motor
1. 400 − 300 = 100 J.2. 300 ÷ 400 × 100 = 75 %.
Problem 4 Laptop charger
1. 60 − 9 = 51 J.2. 51 ÷ 60 = 0.85, so 85 %.
Problem 5 Percentages and decimals
1. 40 %.2. 0.85.
3. It would mean more energy comes out than goes in, which breaks the conservation of energy. Some input is always wasted.
Problem 6 Pendulum swing
1. The energy is proportional to height: 7.0 ÷ 8.0 = 0.875 (87.5 %).2. 12.5 % goes to the thermal store of the air and string (friction at the pivot and air resistance), plus a little sound.
Problem 7 Workshop machine
1. 500 − 350 − 100 = 50 J.2. 350 ÷ 500 × 100 = 70 %.
3. The sound arrow carries 50 J and the thermal arrow 100 J, so it is half as wide.
Problem 8 Gas boiler
1. 0.85 × 40 = 34 MJ.2. 40 − 34 = 6 MJ.
Problem 9 Finding the input
1. Input = useful output ÷ efficiency.2. 1800 ÷ 0.60 = 3000 J.
Problem 10 Which charger is better?
1. A: 9.6 ÷ 12.0 = 80 %. B: 12.3 ÷ 15.0 = 82 %.2. A: 12.0 − 9.6 = 2.4 J. B: 15.0 − 12.3 = 2.7 J.
3. B is more efficient, even though it wastes slightly more energy in total, because it takes in more energy. Efficiency compares the useful part with the input.
Problem 11 Scooter motor power
1. 480 ÷ 600 × 100 = 80 %.2. 600 − 480 = 120 W.
3. E = P t = 120 × 900 = 108 000 J (108 kJ).
Problem 12 Lift motor
1. GPE = m g h = 600 × 9.8 × 20 = 117 600 J.2. 117 600 ÷ 168 000 = 0.70, so 70 %.
Problem 13 Comparing converters
1. Wasted = 75 % of 1000 = 750 J.2. Input = 500 ÷ 0.35 = 1430 J (1428.6 J).
3. The hydroelectric station, at 90 %.
Problem 14 Insulating a room
1. 1500 × 0.60 = 900 W.2. E = 1500 × 3600 = 5 400 000 J (5.4 MJ).
3. After: 900 × 3600 = 3 240 000 J; saving = 2 160 000 J (2.16 MJ).
Problem 15 Chain oil
1. Oil reduces friction between the links, so less of the input energy is transferred to the thermal store and sound, and more goes to the kinetic energy of the bicycle.2. Before: more thermal energy in the chain and air. After: less thermal energy and sound in the surroundings; the energy is still conserved.
Problem 16 Two-stage power station
1. 0.90 × 2.0 = 1.8 MJ.2. 0.95 × 1.8 = 1.71 MJ.
3. 0.90 × 0.95 = 85.5 % (1.71 ÷ 2.0).
4. 0.94 × 1.71 = 1.6074, so 1.61 MJ; overall = 1.6074 ÷ 2.0 = 80.4 %.
Problem 17 Diesel generator
1. 5000 − 1500 − 3000 = 500 kJ.2. 1500 ÷ 5000 × 100 = 30 %.
3. t = E ÷ P = 1 500 000 ÷ 5000 = 300 s.
4. Any two: lubricate moving parts, use the hot exhaust to heat water (recover thermal energy), insulate the engine, service the engine regularly, fit a more efficient engine.
Problem 18 Replacing a bulb
1. 5 % of 60 = 3.0 W.2. Input = 3.0 ÷ 0.30 = 10 W.
3. Time = 150 h. Saving = (60 − 10) W × 150 h = 7500 Wh = 7.5 kWh.
4. 7.5 × 3.6 = 27 MJ.
Problem 19 Spot the errors
1. Impossible: output cannot exceed input. Either 120 J is wrong or the input is larger. Efficiency is at most 100 %.2. Energy is not used up. It is transferred to the thermal store of the water and surroundings: energy is conserved.
3. In a heater the useful output is thermal energy, so almost all the input becomes useful. A very high efficiency (nearly 100 %) is good, not terrible.
Problem 20 Motor and load data
1. GPE = m × 9.8 × 0.80: 1.568 J, 3.136 J, 4.704 J, 6.272 J.2. 1.568 ÷ 5.2 = 30 %; 3.136 ÷ 8.9 = 35 %; 4.704 ÷ 12.4 = 38 %; 6.272 ÷ 15.7 = 40 %.
3. Efficiency increases with load. The motor wastes a roughly fixed amount of energy on friction and warming the coils, so a larger useful output makes up a bigger share of the input.
