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OxfordAQA International GCSE Physics: Energy transfers, efficiency, Sankey diagrams, resources (9203) – Revision Notes

Revision notes on energy transfers, dissipation, efficiency, Sankey diagrams and energy resources for OxfordAQA International GCSE Physics (9203).

Subject
Physics
Level
IGCSE
Topic
Energy transfers, efficiency, Sankey diagrams and energy resources
Updated

Aligned to OxfordAQA IGCSE Physics (9203), For exams May/June 2018 onwards. Official specification .

Syllabus page (what it covers and how it is assessed): OxfordAQA IGCSE Physics.

Syllabus points this page covers

9203

  • 3.2.2 Energy transfers, conservation and dissipation of energy
  • 3.2.3 Energy resources

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Quick recall for sections 3.2.2 and 3.2.3 of the OxfordAQA International GCSE Physics (9203) specification, for exams May/June 2018 onwards. For full explanations and worked examples, use the study guide. Then test yourself with the practice questions. See also the OxfordAQA IGCSE Physics hub, the printable checklist and the free diagnostics. Earlier notes on work and power: Energy revision notes.

Key definitions

Term Meaning
System An object, or a group of objects
Useful transfer Energy that ends up where it is wanted
Stored energy Energy held in a store (gravitational potential, elastic potential, chemical in a fuel)
Dissipated energy Energy spread into the surroundings, stored in less useful ways; often called wasted
Conservation Energy cannot be created or destroyed, only transferred usefully, stored or dissipated
Efficiency Useful output divided by total input; no unit
Renewable Replaced naturally as fast as it is used
Non-renewable Being used up much faster than nature forms it, so it can run out

Transfers to recognise (3.2.2 a-b)

Situation Transfer
Released catapult Elastic potential to kinetic
Falling object Gravitational potential to kinetic
Object thrown upwards Kinetic to gravitational potential
Pendulum, top of swing Maximum gravitational potential, zero kinetic
Pendulum, bottom of swing Maximum kinetic, minimum gravitational potential
Braking Kinetic to thermal energy in the surroundings

Pendulum: energy moves back and forth between kinetic and gravitational potential. The swing gets lower because air resistance and pivot friction dissipate a little energy each swing, not because energy disappears.

Conservation and dissipation (3.2.2 c-e)

  • Total energy before equals total energy after.
  • Useful output is less than input, because some energy is dissipated.
  • Friction and air resistance dissipate energy by heating the surroundings.
  • Say “dissipated to the surroundings” or “transferred as thermal energy”, never “lost” or “used up”.
  • Reduce dissipation with lubrication, insulation and streamlining.

Efficiency (3.2.2 f)

efficiency = useful energy out / total energy in
efficiency = useful power out  / total power in
multiply by 100 for a percentage
Want Rearrangement
Useful output efficiency x total input
Total input useful output / efficiency
Wasted energy (power) total input - useful output

Method in steps:

  1. Pick out the useful output and the total input, in matching units (both energy or both power).
  2. Divide useful by total.
  3. Give a decimal or a percentage as asked. The answer must be less than 1 (or 100%).
  4. Check: useful plus wasted equals the input.

Reminder: a motor takes in 150 W and gives 93 W useful. Efficiency = 93/150 = 0.62, or 62%. Wasted power = 57 W.

Sankey diagrams (3.2.2 g)

Drawing method:

  1. Choose a scale (for example 1 mm for 2 kJ) and divide each energy by it to get the arrow widths.
  2. Draw the input arrow on the left.
  3. Let the useful output continue straight on. Branch the dissipated outputs away.
  4. Check that output widths add up to the input width.
  5. Label every arrow with its type of energy and value.

Reading method: measure each width, multiply by the scale, then use useful over input for efficiency.

Reminder: a diagram with input 40 mm wide and useful arrow 10 mm wide, at 1 mm for 2 kJ, shows 80 kJ in and 20 kJ useful, so efficiency is 0.25.

Fuels (3.2.3 a-b)

Factor What to say
Ease of storage Solids and liquids are easier to store than gases, which need pressurised tanks
Energy content Energy released per kilogram; high content means less fuel mass for the same energy
Safety Ease of ignition, toxic fumes, explosion risk

Some energy from burning a fuel goes to the surroundings, so the useful energy is less than the energy content. Fuel efficiency = useful energy transferred / energy released by the fuel. Energy released = mass x energy content per kilogram.

Energy sources (3.2.3 c)

Source Type Environmental points
Coal, oil, natural gas Non-renewable Burning gives carbon dioxide (global warming); coal can give sulfur dioxide (acid rain); mining and oil spills damage habitats
Nuclear fuel Non-renewable No carbon dioxide in use; mining; radioactive waste needs long-term storage
Wind, hydroelectric, tidal Renewable No fuel or emissions in use; visual impact, habitat change, construction impact
Biofuels Renewable Replaceable by growing more; land use; carbon dioxide on burning

Renewable technologies (3.2.3 d)

Technology Advantages Drawbacks
Wave No fuel, no carbon dioxide in use, good for coastal countries Depends on sea conditions, costly to build and repair, may disturb marine life and shipping
Solar No fuel, no emissions in use, works at small scale No output at night, less in cloud, large areas needed, panel manufacture and disposal
Geothermal Steady output day and night, no fuel to buy Needs accessible hot rock, costly drilling, some gas release

For a scenario question, pick the source that suits the stated location, then give one advantage and one drawback.

Worked reminders

Pendulum speed. Bob of mass 0.25 kg, released 0.080 m above the lowest point, g = 9.8 N/kg, nothing dissipated.

GPE = 0.25 x 9.8 x 0.080 = 0.196 J = kinetic energy at the bottom
v^2 = 2 x 0.196 / 0.25 = 1.568   ->   v = 1.25 m/s

Dissipation from braking. A 1500 kg van brakes from 12 m/s to rest.

KE = 0.5 x 1500 x 12^2 = 108 000 J, all dissipated by friction as thermal energy

Rearranged efficiency. A device is 40% efficient and gives 270 J useful energy.

input = 270 / 0.40 = 675 J     dissipated = 675 - 270 = 405 J

Fuel efficiency. 2.0 kg of fuel at 30 MJ/kg releases 60 MJ. Water gains 33 MJ, so efficiency = 33/60 = 0.55, or 55%.

Solar power. 12 m2 of panel receives 450 W per square metre, so the input is 5400 W. At 18% efficiency the useful output is 972 W.

Comparing lamps. LED: 2.7/9.0 = 0.30. Filament: 3.0/60 = 0.050. Same job, six times the efficiency.

Must-know distinctions

  • Energy (joules) versus power (watts): power is energy per second.
  • Dissipated versus destroyed: dissipated energy still exists, spread out in the surroundings.
  • Renewable versus low-impact: a renewable source still has an environmental cost.
  • Useful versus total: efficiency always divides by the total input.

Quick self-test

  1. A device takes in 360 J and transfers 54 J usefully. Give its efficiency as a decimal and the energy dissipated.
  2. A pendulum bob of mass 0.20 kg is 0.45 m above its lowest point. Find its gravitational potential energy (g = 9.8 N/kg).
  3. A Sankey diagram, scale 1 mm for 5 J, has an input arrow 30 mm wide and a useful arrow 12 mm wide. Find the input energy, the useful energy, the efficiency and the dissipated energy.
  4. A 30% efficient device gives 36 W of useful power. What power does it take in?
  5. Burning 0.50 kg of a fuel with energy content 40 MJ/kg heats water by 12 MJ. Find the fuel efficiency as a percentage.
  6. State the energy transfer when a pendulum bob swings from the top of its swing to the bottom.
  7. Explain why a swinging pendulum eventually stops.
  8. Give one advantage and one drawback of solar power.
  9. State what “renewable” means.
  10. Give two factors that affect the choice of a fuel.
  11. A lamp takes in 8.0 W and gives out 2.0 W of light. Give its efficiency as a percentage and state where the other energy goes.
  12. A 2.0 kg ball moves at 6.0 m/s and rolls to rest on rough ground. How much energy is dissipated, and how?

Answers

  1. 54/360 = 0.15. Dissipated = 360 - 54 = 306 J.
  2. 0.20 x 9.8 x 0.45 = 0.882 J.
  3. Input 30 x 5 = 150 J. Useful 12 x 5 = 60 J. Efficiency 60/150 = 0.40. Dissipated 150 - 60 = 90 J (an 18 mm arrow).
  4. Input = 36/0.30 = 120 W.
  5. Energy released = 0.50 x 40 = 20 MJ. Efficiency = 12/20 = 0.60, or 60%.
  6. Gravitational potential energy is transferred to kinetic energy.
  7. Air resistance and friction at the pivot dissipate energy by heating the surroundings, so each swing is lower. The energy is not destroyed.
  8. Advantage: no fuel needed and no emissions while running. Drawback: no output at night and reduced output in cloud.
  9. Replaced naturally as fast as it is used, so it does not run out.
  10. Any two of: ease of storage, energy content, safety.
  11. 2.0/8.0 = 0.25, so 25%. The other 6.0 W is dissipated to the surroundings as thermal energy.
  12. Kinetic energy = 0.5 x 2.0 x 6.0^2 = 36 J, dissipated by friction as thermal energy in the ball and ground.

Where marks are usually lost

  • Calling the missing energy “lost” or “used up” when the correct word is dissipated.
  • Dividing input by output.
  • Mixing joules and watts in one efficiency calculation.
  • Giving an efficiency above 1 or 100%.
  • Sankey arrows that do not add up, or no scale or labels.
  • Forgetting to say that friction or air resistance dissipates energy by heating the surroundings.
  • Listing generic points about a renewable source instead of matching them to the location in the question.
  • Saying renewable sources cause no environmental harm.

Official syllabus

OxfordAQA International GCSE Physics (9203) specification, Version 5.2, for exams May/June 2018 onwards, published by OxfordAQA. Sections 3.2.2 and 3.2.3.

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