Conservation of Energy & efficiency

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).

Worked example 0 Real-world example: boiling water in a kettle

An electric kettle takes $250$ kJ of electrical energy to boil $1$ L of water. The water actually gains $210$ kJ of heat. Find the efficiency and describe the “lost” energy.

Useful output: $210$ kJ (heat into the water).
Wasted: $250 - 210 = 40$ kJ (heat into the casing, steam escaping, sound).
Efficiency: $\dfrac{210}{250} \times 100 = 84\%$ .

Key idea: energy is not destroyed — the $40$ 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

E_{\text{in}} = E_{\text{useful}} + E_{\text{wasted}}

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.

Device	Useful output	Common wasted forms
Kettle	heat in water	heat in casing, sound
Incandescent bulb	visible light	heat (most of input!)
LED bulb	visible light	small amount of heat
Car engine	kinetic energy of car	heat in exhaust, engine, radiator; sound
Speaker	sound	heat in voice coil
Hairdryer (to dry hair)	heat in air stream + kinetic air flow	sound; heat in motor

3. Efficiency

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

Efficiency

Percentage form

\eta = \dfrac{E_{\text{useful}}}{E_{\text{in}}} \times 100\%

$\eta$ is always between $0\%$ and $100\%$ .

Power form (equivalent)

If the process runs steadily,

\eta = \dfrac{P_{\text{useful}}}{P_{\text{in}}} \times 100\%.

Worked example 1 Efficiency of an LED bulb

A $10$ W LED bulb produces $6$ W of visible light. Find its efficiency.

$\eta = \dfrac{6}{10} \times 100 = 60\%$ .

Compare with a $60$ W incandescent bulb that produces only $3$ W of visible light: $\eta = 5\%$ . The rest is heat.

Worked example 2 Car engine

A petrol car burns fuel with chemical energy $40$ MJ. It moves the car with kinetic energy $10$ MJ; the rest is heat and friction. Find the efficiency.

$\eta = \dfrac{10}{40} \times 100 = 25\%$ .
$75\%$ 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.

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.

Useful: 25 J. Wasted: $40 + 30 + 5 = 75$ J.
Total: $25 + 75 = 100$ J — energy is conserved.
Efficiency $= 25/100 = 25\%$ .

5. Efficiency of power systems

System	Typical efficiency
Coal-fired power station	35-42%
Combined-cycle gas turbine	55-62%
Hydroelectric	85-90%
Wind turbine (electrical)	35-45% of wind energy
Photovoltaic solar panel	18-22%
Modern petrol car engine	20-30%
Electric motor	85-95%
Incandescent bulb	2-5%
LED bulb	40-60%

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 $45$ MJ of fuel to travel $100$ km with $25\%$ engine efficiency. An EV uses $55$ MJ of electricity to travel $300$ km with $85\%$ motor efficiency. Compare useful energy per km.

Petrol: useful = $0.25 \times 45 = 11.25$ MJ for 100 km = $0.1125$ MJ/km.
EV: useful = $0.85 \times 55 = 46.75$ MJ for 300 km = $0.156$ MJ/km at the wheels. Actually electrical input per km = $55 / 300 = 0.183$ MJ/km, vs petrol $0.45$ 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.

Practice: Year 9

Fluency

Conservation and efficiency

State the Law of Conservation of Energy.
Define useful energy, wasted energy, and efficiency.
Calculate efficiency for each: (a) 200 J in, 150 J useful out; (b) 400 kJ in, 80 kJ wasted.
A kettle uses $250$ kJ and transfers $210$ kJ to the water. Find efficiency.
Sketch a Sankey diagram for a hairdryer that is $70\%$ efficient (useful = heated air flow; wasted = motor heat + sound).
Give one example each of: chemical -> electrical; electrical -> light; kinetic -> electrical.

Reasoning

Apply the ideas

A $60$ W incandescent bulb produces about $3$ W of light. An LED bulb produces $6$ W of light from $10$ W input. Compare efficiencies and comment on energy saving over a year.
Explain why a car engine’s exhaust and radiator are both important to consider when calculating efficiency.
A coal-fired station is $40\%$ efficient at converting chemical energy to electrical energy. If $1000$ MJ of coal is burned, how much electricity is produced, and where does the rest go?
A bouncing ball loses height with each bounce. Explain this in terms of energy conservation.
Why does rubbing hands together produce heat, and what energy transformation is this?

Problem solving

Apply conservation

A lift motor uses $5$ kWh of electricity to lift a $1000$ kg load $20$ m. If gravitational PE gained is $1000 \times 9.8 \times 20 = 196\,000$ J, calculate the efficiency. (Convert $5$ kWh to joules first: $5 \times 3.6 \times 10^{6} = 1.8 \times 10^{7}$ J.)
A solar panel receives $1000$ J of sunlight per second on its surface. Its electrical output is $180$ W. Find its efficiency.
A fan heater draws $2000$ W. Useful heat output is $1800$ W. Where is the other $200$ W going? What efficiency is this?
A power station generates $500$ MW of electricity. $8$ MW is lost as heat in transmission lines. Transmission efficiency?

Challenge

Reasoning

Harder reasoning

A coal-fired power station has efficiency $40\%$ for electricity generation, and transmission is $92\%$ efficient. An appliance using this electricity has efficiency $70\%$ . Calculate the overall efficiency from fuel to useful output and comment on the result.
A domestic heat pump is rated with a coefficient of performance (COP) of $4$ : for every $1$ J of electrical work it moves $4$ J of heat into the house. Explain why this does not violate the Law of Conservation of Energy.
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).
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?

Answers

Answer key

Attempt the practice first. When you're ready to check, expand the answers below.

Show the full answer key

Year 9 answers

Fluency

Conservation and efficiency

Energy cannot be created or destroyed, only transformed from one form to another. In a closed system the total energy stays constant.
Useful energy: output that fulfils the device’s purpose. Wasted energy: output in forms we cannot or do not use (usually heat, sometimes sound). Efficiency: the fraction of input energy that becomes useful output.
(a) $\eta = 150/200 \times 100 = 75\%$ . (b) Useful $= 400 - 80 = 320$ kJ. $\eta = 320/400 \times 100 = 80\%$ .
$\eta = 210/250 \times 100 = 84\%$ .
Sankey: input band (say 100 J) splits into useful (70 J heated air flow) and wasted (30 J heat + sound). Widths must add to input width.
Chemical -> electrical: battery. Electrical -> light: LED. Kinetic -> electrical: generator (wind turbine or hydro).

Reasoning

Apply the ideas

Incandescent: $\eta = 3/60 = 5\%$ . LED: $\eta = 6/10 = 60\%$ . LED gives twice the light for one-sixth the electricity, so the yearly saving at 4 h/day is about $(60 - 10) \times 4 \times 365 = 73\,000$ Wh $= 73$ kWh per bulb. Across a house that is a large reduction in bill and emissions.
The engine’s wasted energy leaves the car mainly as hot exhaust gas and heat carried away by the radiator; these are the largest loss paths, so any calculation of efficiency must include them to be realistic.
Electricity $= 0.40 \times 1000 = 400$ MJ. The rest ( $600$ MJ) is mostly waste heat lost at the boiler and cooling towers, plus smaller losses at the turbine and generator.
Each bounce loses some kinetic energy to heat, sound, and air resistance. That energy is not returned, so kinetic energy on the way up is smaller each time, reducing the bounce height. Total energy is still conserved (the lost energy warms the ball, floor, and air).
Kinetic energy -> heat (via friction between skin surfaces). Some sound is also produced.

Problem solving

Apply conservation

$\eta = 196\,000 / 1.8 \times 10^{7} \times 100 \approx 1.09\%$ . That is low because 5 kWh is a huge amount of energy for a 20 m lift — probably this is a single lift of many, or most energy is elsewhere. (Or the numbers represent all the runs over an hour, not one.)
$\eta = 180/1000 \times 100 = 18\%$ .
The other $200$ W is lost as heat in the motor and casing, and some sound. $\eta = 1800/2000 = 90\%$ .
$\eta = (500 - 8)/500 \times 100 = 492/500 \times 100 = 98.4\%$ .

Reasoning

Challenge

Overall $= 0.40 \times 0.92 \times 0.70 = 0.2576$ , or about $25.8\%$ . Only about a quarter of the coal’s chemical energy becomes useful output; most is lost as heat at the power station, with smaller losses in transmission and in the appliance.
A heat pump does not create energy; it moves heat from a colder outside to a warmer inside using electrical work. Total energy into the house = electrical input + heat taken from outside = heat delivered. Conservation holds; COP just describes the ratio of heat delivered to work input.
Counter-points: (i) Electricity is often more expensive per joule than gas heat, so using electric bulbs for heating is wasteful financially. (ii) The light that leaves through windows is not returned; part of the energy leaves the house rather than warming it. (iii) In summer the heat is unwanted and may drive more air-conditioning. (iv) Switching to LEDs + efficient heating saves more energy and emissions overall.
Overall $= 0.40 \times 0.92 \times 0.50 = 0.184$ or $18.4\%$ of the original coal energy becomes visible light. Sankey: coal 100 -> power station (40 to electricity, 60 waste heat) -> transmission (36.8 delivered, 3.2 line losses) -> LED bulb (18.4 light, 18.4 heat).

Prefer paper? Print the answer key as a separate booklet: open print view ->