Year 9 Science | Victorian Curriculum 2.0
Conservation of Energy & efficiency
Topic 08 | Physical sciences | Practice

What you will learn

  • state the Law of Conservation of Energy,
  • identify energy inputs, useful outputs, and wasted outputs in real systems,
  • calculate efficiency as a percentage,
  • draw and interpret a Sankey diagram,
  • analyse efficiency of everyday devices (kettle, car engine, power station, LED bulb).
Why does "conserved" not mean "saved"?

The Law of Conservation of Energy says the total amount of energy in a closed system never changes — energy is transformed from one form to another, not created or destroyed. Yet we talk about “wasting” energy. The resolution: “waste” does not mean energy disappears — it means energy ends up in a form we cannot usefully recover, usually heat dispersed into the surroundings. Conservation is a physical law; efficiency is a human judgement about what counts as useful.

Where you'll see this
  • Home: LED lights vs incandescent; fridge efficiency star ratings.
  • Transport: petrol cars waste most fuel energy as heat; EVs are much more efficient.
  • Industry: heat recovery, co-generation plants.
  • Power generation: coal (~40% efficient), gas turbines (~55%), PV solar panels (~20%).
  • Policy: appliance efficiency standards reduce energy use and emissions.
Worked example 0 Real-world example: boiling water in a kettle

An electric kettle takes 250250250 kJ of electrical energy to boil 111 L of water. The water actually gains 210210210 kJ of heat. Find the efficiency and describe the “lost” energy.

  1. Useful output: 210210210 kJ (heat into the water).
  2. Wasted: 250−210=40250 - 210 = 40250−210=40 kJ (heat into the casing, steam escaping, sound).
  3. Efficiency: 210250×100=84%\dfrac{210}{250} \times 100 = 84\%250210​×100=84%.

Key idea: energy is not destroyed — the 404040 kJ ends up as heat in the room and a small amount of sound. It is just not usefully “in the water”.

1. The Law of Conservation of Energy

In any closed system, the total energy before a process equals the total energy after. Energy can change form:

  • kinetic ↔\leftrightarrow↔ gravitational potential (a bouncing ball, a pendulum),
  • chemical →\to→ thermal →\to→ kinetic (a car engine),
  • light →\to→ electrical (solar cell),
  • electrical →\to→ thermal + light (incandescent bulb),
  • electrical →\to→ kinetic + heat + sound (an electric fan).
Conservation of Energy for a device
Ein=Euseful+EwastedE_{\text{in}} = E_{\text{useful}} + E_{\text{wasted}}Ein​=Euseful​+Ewasted​

The energy put in equals the useful energy out plus the energy transformed into forms we cannot use (usually heat and sound).

2. Useful vs wasted energy

For a given device, “useful” depends on the purpose.

DeviceUseful outputCommon wasted forms
Kettleheat in waterheat in casing, sound
Incandescent bulbvisible lightheat (most of input!)
LED bulbvisible lightsmall amount of heat
Car enginekinetic energy of carheat in exhaust, engine, radiator; sound
Speakersoundheat in voice coil
Hairdryer (to dry hair)heat in air stream + kinetic air flowsound; heat in motor

3. Efficiency

Efficiency is the fraction (or percentage) of input energy that becomes useful output.

Efficiency

Percentage form
η=EusefulEin×100%\eta = \dfrac{E_{\text{useful}}}{E_{\text{in}}} \times 100\%η=Ein​Euseful​​×100%

η\etaη is always between 0%0\%0% and 100%100\%100%.

Power form (equivalent)

If the process runs steadily,

η=PusefulPin×100%.\eta = \dfrac{P_{\text{useful}}}{P_{\text{in}}} \times 100\%.η=Pin​Puseful​​×100%.
Worked example 1 Efficiency of an LED bulb

A 101010 W LED bulb produces 666 W of visible light. Find its efficiency.

  1. η=610×100=60%\eta = \dfrac{6}{10} \times 100 = 60\%η=106​×100=60%.

Compare with a 606060 W incandescent bulb that produces only 333 W of visible light: η=5%\eta = 5\%η=5%. The rest is heat.

Worked example 2 Car engine

A petrol car burns fuel with chemical energy 404040 MJ. It moves the car with kinetic energy 101010 MJ; the rest is heat and friction. Find the efficiency.

  1. η=1040×100=25%\eta = \dfrac{10}{40} \times 100 = 25\%η=4010​×100=25%.
  2. 75%75\%75% of fuel energy is wasted — this is the fundamental reason EVs use much less energy per km than petrol cars.

4. Sankey diagrams

A Sankey diagram shows energy flow. The width of each band is proportional to the amount of energy. The input band splits into “useful” and “wasted” branches whose widths add to the input width.

electrical input 100 Jlight 10 J (useful)heat 90 J (wasted)
Sankey diagram for a typical incandescent light bulb: 100 J in, about 10 J useful light out, about 90 J wasted as heat.
Worked example 3 Reading a Sankey diagram

A Sankey diagram for a car shows: 100 J fuel in, 25 J kinetic energy, 40 J exhaust heat, 30 J radiator/engine heat, 5 J sound and friction. Find the efficiency and check the total balances.

  1. Useful: 25 J. Wasted: 40+30+5=7540 + 30 + 5 = 7540+30+5=75 J.
  2. Total: 25+75=10025 + 75 = 10025+75=100 J — energy is conserved.
  3. Efficiency =25/100=25%= 25/100 = 25\%=25/100=25%.

5. Efficiency of power systems

SystemTypical efficiency
Coal-fired power station35-42%
Combined-cycle gas turbine55-62%
Hydroelectric85-90%
Wind turbine (electrical)35-45% of wind energy
Photovoltaic solar panel18-22%
Modern petrol car engine20-30%
Electric motor85-95%
Incandescent bulb2-5%
LED bulb40-60%

Making systems more efficient reduces energy use and, for fossil-fuel systems, emissions per unit of useful output.

Worked example 4 Comparing petrol and EV

A petrol car uses 454545 MJ of fuel to travel 100100100 km with 25%25\%25% engine efficiency. An EV uses 555555 MJ of electricity to travel 300300300 km with 85%85\%85% motor efficiency. Compare useful energy per km.

  1. Petrol: useful = 0.25×45=11.250.25 \times 45 = 11.250.25×45=11.25 MJ for 100 km = 0.11250.11250.1125 MJ/km.
  2. EV: useful = 0.85×55=46.750.85 \times 55 = 46.750.85×55=46.75 MJ for 300 km = 0.1560.1560.156 MJ/km at the wheels. Actually electrical input per km = 55/300=0.18355 / 300 = 0.18355/300=0.183 MJ/km, vs petrol 0.450.450.45 MJ/km. EV uses less than half the input energy per km.

Key idea: EV efficiency wins even before considering how the electricity is generated.

No machine can exceed 100% efficiency

If a device claimed ”110%110\%110% efficient”, it would be creating energy — impossible under the conservation law. A heat pump may seem to bend this rule (it can move more heat energy than the electrical energy it consumes), but it is moving, not creating, heat; its coefficient of performance is not the same as efficiency.


Practice: Year 9

Fluency

Conservation and efficiency

    1. State the Law of Conservation of Energy.
    2. Define useful energy, wasted energy, and efficiency.
    3. Calculate efficiency for each: (a) 200 J in, 150 J useful out; (b) 400 kJ in, 80 kJ wasted.
    4. A kettle uses 250250250 kJ and transfers 210210210 kJ to the water. Find efficiency.
    5. Sketch a Sankey diagram for a hairdryer that is 70%70\%70% efficient (useful = heated air flow; wasted = motor heat + sound).
    6. Give one example each of: chemical -> electrical; electrical -> light; kinetic -> electrical.
Reasoning

Apply the ideas

    1. A 606060 W incandescent bulb produces about 333 W of light. An LED bulb produces 666 W of light from 101010 W input. Compare efficiencies and comment on energy saving over a year.
    2. Explain why a car engine’s exhaust and radiator are both important to consider when calculating efficiency.
    3. A coal-fired station is 40%40\%40% efficient at converting chemical energy to electrical energy. If 100010001000 MJ of coal is burned, how much electricity is produced, and where does the rest go?
    4. A bouncing ball loses height with each bounce. Explain this in terms of energy conservation.
    5. Why does rubbing hands together produce heat, and what energy transformation is this?
Problem solving

Apply conservation

    1. A lift motor uses 555 kWh of electricity to lift a 100010001000 kg load 202020 m. If gravitational PE gained is 1000×9.8×20=196 0001000 \times 9.8 \times 20 = 196\,0001000×9.8×20=196000 J, calculate the efficiency. (Convert 555 kWh to joules first: 5×3.6×106=1.8×1075 \times 3.6 \times 10^{6} = 1.8 \times 10^{7}5×3.6×106=1.8×107 J.)
    2. A solar panel receives 100010001000 J of sunlight per second on its surface. Its electrical output is 180180180 W. Find its efficiency.
    3. A fan heater draws 200020002000 W. Useful heat output is 180018001800 W. Where is the other 200200200 W going? What efficiency is this?
    4. A power station generates 500500500 MW of electricity. 888 MW is lost as heat in transmission lines. Transmission efficiency?

Challenge

Reasoning

Harder reasoning

    1. A coal-fired power station has efficiency 40%40\%40% for electricity generation, and transmission is 92%92\%92% efficient. An appliance using this electricity has efficiency 70%70\%70%. Calculate the overall efficiency from fuel to useful output and comment on the result.
    2. A domestic heat pump is rated with a coefficient of performance (COP) of 444: for every 111 J of electrical work it moves 444 J of heat into the house. Explain why this does not violate the Law of Conservation of Energy.
    3. A student argues that we should use incandescent bulbs in winter because the “wasted” heat warms the house anyway. Evaluate this argument, including at least one counter-point (think about the cost of electricity vs gas heat, and the path the bulb’s photons take).
    4. Draw a Sankey diagram for a complete chain: coal -> power station -> transmission -> LED bulb. Use the efficiencies 40%, 92%, 50% at each stage. What fraction of the original coal energy becomes visible light?
Year 9 Science study companion | Practice