Energy transfers, efficiency and power: Sankey diagrams, the watt and the kilowatt-hour
By the end of this lesson you will be able to
- Identify the energy stores and transfer pathways in a device or process, naming two energy forms for every transfer.
- Construct a Sankey diagram to scale from data, and interpret one to find missing energies.
- Calculate efficiency from energies or powers, as a decimal and a percentage, and rearrange it to find an input or an output.
- Calculate power with P = E ÷ t and P = VI, and energy in joules and kilowatt-hours, including the cost of electricity.
- Explain why dissipated energy is described as wasted, and evaluate ways of reducing it.
- Design an investigation into the efficiency of an electric motor to the standard of the on-screen examination.
You met this before
- Energy is stored (kinetic, gravitational potential, elastic, thermal, chemical, nuclear) and transferred mechanically, electrically, by heating or by radiation. It is measured in joules (J).
- Kinetic energy Ek = ½mv².
- Gravitational potential energy Ep = mgΔh, with g = 9.8 N/kg on Earth.
- Conservation of energy: energy cannot be created or destroyed, only transferred; the total after a change equals the total before.
- Electrical power P = VI (watts are volts × amps). A kilowatt is 1 000 W, and the energy bill counts kilowatt-hours.
- Work done W = Fd: pushing a force through a distance transfers energy.
A 0.50 kg ball is dropped from 2.0 m. Calculate its gravitational potential energy at the top and its speed just before it lands, ignoring air resistance.
Reveal
Ep = mgΔh
Ep = 0.50 kg × 9.8 N/kg × 2.0 m
Ep = 9.8 J
All of it becomes kinetic energy: ½mv² = 9.8 J
v² = 2 × 9.8 J ÷ 0.50 kg
v² = 39.2 m²/s²
v = 6.3 m/s
Keep this link between stores in mind: this whole unit is about where energy comes from, where it goes, and what the waste does to the planet.
Revise it in full: Year 1 · Energy · Lesson 1 — What is energy? Stores, transfers and the bar chart · Year 1 · Energy · Lesson 3 — Conservation of energy: the law nothing has ever broken · Year 3 · Circuits · Lesson 4 — Circuit calculations and power: watts, kilowatt-hours and the wire that gets hot
The puzzle
Follow 100 joules of chemical energy in a lump of coal. The power station turns about 37 J of it into electrical energy. The transmission lines lose about 8 % of that on the way to your home, leaving 34 J.
An old filament bulb then turns only 5 % of those 34 J into light: less than 2 J of light from 100 J of coal. Everything else warms cooling water, the air, the wires and the room.
Swap the bulb for an LED that turns 40 % of its input into light and the same 100 J of coal gives nearly 14 J of light — about eight times as much. Or the same light for an eighth of the coal, and an eighth of the carbon dioxide.
Efficiency is not a dull number on a label. It decides how much fuel a country burns. This lesson gives you the tools to follow energy through any device and to put numbers on the waste.
An 8 W LED lamp and a 60 W filament lamp give about the same amount of light. Predict how much electrical energy each takes in 5 hours, and where the difference goes.
Compare your answer
LED: E = Pt
E = 8 W × 5 × 3 600 s
E = 144 000 J = 0.04 kWh
Filament: E = 60 W × 18 000 s
E = 1 080 000 J = 0.30 kWh
Both give about the same light energy, so the extra 0.26 kWh taken by the filament lamp all ends up as thermal energy in the lamp and the room. Energy is not destroyed — it is dissipated where nobody wanted it.
The science — structured notes
1.1 Stores, transfers and the law that never breaks
Energy is stored in a system and transferred when something happens. The on-screen examination usually names energy “forms”: chemical potential, gravitational potential, kinetic, elastic, thermal (heat), electrical, light, sound and nuclear. Examiners expect two forms for every transfer — “chemical potential → thermal”, not just “heat”.
The rule behind every calculation in this unit is conservation of energy: in any change the total energy after equals the total energy before. For a device we write it as energy in = useful energy out + wasted energy out.
| Device or process | Energy in | Useful out | Wasted (dissipated) out |
|---|---|---|---|
| Coal or biomass power station | chemical potential (fuel) | electrical | thermal in cooling water and flue gases |
| Hydroelectric station | gravitational potential (water) | electrical | thermal and sound (friction, turbulence) |
| Wind turbine | kinetic (moving air) | electrical | thermal and sound in the gearbox and generator; air still moving |
| Solar (photovoltaic) panel | light (radiation) | electrical | thermal (the panel warms up); reflected light |
| Electric motor lifting a load | electrical | gravitational potential of the load | thermal (resistance, friction) and sound |
| LED lamp | electrical | light | thermal |
“When a car brakes to a stop, its kinetic energy is used up.”
Reveal answer
False. Energy is never used up. Friction between the brake pads and discs transfers the kinetic energy to thermal energy in the brakes, which then spreads into the air. The energy still exists but is dissipated and no longer useful. A hybrid car’s regenerative brakes turn part of it into electrical energy that recharges the battery (chemical potential energy) instead.
1.2 Sankey diagrams: energy drawn to scale
A Sankey diagram is a flow diagram in which the width of each arrow is proportional to the energy (or power) it carries. The input enters on the left. The useful output carries straight on. Each wasted output bends away, usually downwards.
Because energy is conserved, the widths of the outputs add up exactly to the width of the input. That gives you a check on every diagram, and a way to find a missing value.
A 30 kg child climbs to the top of a slide 2.0 m high. A Sankey diagram for the ride shows 588 J of gravitational potential energy in, 80 J of thermal energy and 8 J of sound energy out. (1) Find the kinetic energy at the bottom. (2) Calculate the child’s speed there, to 2 significant figures. (3) Calculate the efficiency of the slide.
(1) Energy in = useful out + wasted out
Ek = 588 J − 80 J − 8 J
Ek = 500 J
(2) Ek = ½mv², so v = √(2Ek ÷ m)
v = √(2 × 500 J ÷ 30 kg)
v = √33.3 m²/s²
v = 5.77 m/s ≈ 5.8 m/s
(3) efficiency = useful out ÷ total in
efficiency = 500 J ÷ 588 J
efficiency = 0.850 = 85 %
Check the input first: Ep = 30 × 9.8 × 2.0 = 588 J. The November 2021 paper asked exactly this kind of question with its own numbers, and the rounding to 2 s.f. was a mark on its own.
Human muscles are about 25 % efficient. In one second a cyclist’s muscles transfer 1 000 J of chemical energy from food; 250 J does work on the pedals and 750 J warms the body. Draw the Sankey diagram to scale on squared paper.
Choose a scale that fits: 1 small square = 50 J
Input arrow: 1 000 J ÷ 50 J = 20 squares wide
Useful (work on the pedals): 250 J ÷ 50 J = 5 squares, carrying straight on
Wasted (thermal in the body): 750 J ÷ 50 J = 15 squares, bending downwards
Check: 5 + 15 = 20 squares
Label every arrow with the energy form and the value. The 750 J/s of heat is why a cyclist sweats: evaporating sweat carries the dissipated energy away. The diagram shows at a glance that the body is a better heater than an engine.
“50 MJ of the 100 MJ input is transferred as thermal energy to the cooling water and dissipated into the surroundings.”
A student’s Sankey diagram for an electric kettle shows an input arrow 20 mm wide labelled 2 000 J, a useful arrow 17 mm wide labelled “thermal energy in the water 1 700 J” and a waste arrow 5 mm wide labelled “thermal energy to the air 300 J”. What is wrong?
Reveal answer
The labels add up (1 700 + 300 = 2 000 J) but the widths do not: 17 mm + 5 mm = 22 mm, more than the 20 mm input. At the scale of 1 mm per 100 J, the waste arrow must be 3 mm wide. A Sankey diagram drawn to scale must conserve widths as well as numbers.
1.3 Efficiency: how much of the input is useful
Efficiency is the fraction of the input energy that is transferred usefully. It can be calculated from energies or from powers, because both are measured over the same time:
efficiency = useful energy outtotal energy in
efficiency = useful power outtotal power in
Multiply by 100 for a percentage. Efficiency has no unit, and it can never be more than 1 (100 %), because a device cannot give out more energy than it takes in. A calculated efficiency above 100 % tells you a mistake has been made, usually a unit left in grams or centimetres.
A motor lifts a 200 g mass through 80.0 cm. The voltmeter reads 6.0 V, the ammeter 0.45 A, and the lift takes 2.4 s. Calculate the efficiency of the motor.
Convert first: m = 0.200 kg, Δh = 0.800 m
Useful energy out: Ep = mgΔh
Ep = 0.200 kg × 9.8 N/kg × 0.800 m
Ep = 1.57 J
Total energy in: E = VIt
E = 6.0 V × 0.45 A × 2.4 s
E = 6.48 J
efficiency = 1.57 J ÷ 6.48 J
efficiency = 0.242 = 24 %
The other 4.91 J became thermal energy in the motor’s coils (their resistance), thermal energy from friction in the bearings, and sound. Using 200 g and 80.0 cm without converting gives an “efficiency” of about 2 400 000 %, an impossible answer that should make you check.
A gas-fired power station must deliver 700 MW of electrical power and is 35 % efficient. Calculate the power it takes in from the fuel, and the power it wastes.
Reveal answer
total power in = useful power out ÷ efficiency
total power in = 700 MW ÷ 0.35
total power in = 2 000 MW
wasted power = 2 000 MW − 700 MW
wasted power = 1 300 MW
Rearranging efficiency is a common Year 5 step: divide the output by the efficiency to find the input. That 1 300 MW leaves as thermal energy in cooling water and flue gases.
A filament bulb takes in 100 J and gives out 5 J of light. An LED takes in 0.050 W and gives out 0.020 W of light. Compare their efficiencies.
Reveal answer
Filament: 5 J ÷ 100 J = 0.05 = 5 %
LED: 0.020 W ÷ 0.050 W = 0.40 = 40 %
The LED is eight times as efficient as the filament bulb (40 % against 5 %), so for the same light it needs one-eighth of the electrical energy. “Compare” needs both values and one sentence that relates them; listing two numbers is not a comparison.
1.4 Power: how fast energy is transferred
Power is the rate of transferring energy, or the rate of doing work. One watt is one joule per second.
P = Et
E = Pt
P = VI (electrical power)
Power stations are rated in megawatts (MW, 10⁶ W) and gigawatts (GW, 10⁹ W). A kettle is about 2 kW, a phone charger about 10–20 W, a school laptop about 50 W.
Electricity is sold by the kilowatt-hour (kWh): the energy transferred by a 1 kW device in 1 hour. Because 1 kW × 3 600 s = 3 600 000 J, 1 kWh = 3.6 MJ. Use kW and hours for kWh, and W and seconds for joules; never mix them.
A pump lifts 2.0 m³ of water (2 000 kg) from a borehole 12 m deep in 1.0 hour. Its electric motor takes 150 W from a solar panel. Calculate (1) the useful power output, (2) the efficiency of the pump.
(1) Useful energy: Ep = mgΔh
Ep = 2 000 kg × 9.8 N/kg × 12 m
Ep = 235 200 J
Useful power: P = E ÷ t
P = 235 200 J ÷ 3 600 s
P = 65 W (65.3 W)
(2) efficiency = 65.3 W ÷ 150 W
efficiency = 0.436 ≈ 44 %
Thousands of boreholes on farms and cattle posts across southern Africa now run on solar pumps like this one instead of diesel. You will size the panels in the Stretch question at the end of the lesson.
A 2.2 kW kettle is used for 4.0 minutes a day for 30 days. Electricity costs 1.50 pula per kWh (a round figure for this example). Calculate the energy in kWh and in joules, and the cost.
time = 4.0 min × 30 = 120 min
time = 2.0 h
E = Pt
E = 2.2 kW × 2.0 h
E = 4.4 kWh
In joules: 4.4 × 3.6 MJ = 15.8 MJ
cost = 4.4 kWh × 1.50 pula per kWh
cost = 6.60 pula
A 5.0 kW electric car charger runs for 6.0 hours. Calculate the energy transferred in kWh and in MJ. If the car uses 0.15 kWh per km, how far can it then travel?
Reveal answer
E = Pt
E = 5.0 kW × 6.0 h = 30 kWh
E = 30 × 3.6 MJ = 108 MJ
distance = 30 kWh ÷ 0.15 kWh per km
distance = 200 km
(Charging itself is about 90 % efficient, so a little less than this reaches the battery: an efficiency question hiding inside a power question.)
1.5 Chains of devices: efficiencies multiply
Energy often passes through several devices in turn: a power station, transmission lines and a lamp. Each keeps only its own fraction of what it receives, so the overall efficiency is the product of the individual efficiencies, not their sum or their average.
The transmission efficiency in the chain above falls from 92 % to 80 % because long, overloaded lines are used. By what factor does the light from 100 J of coal change, and is it a bigger or smaller effect than changing the bulb?
Reveal answer
New overall (LED) = 0.37 × 0.80 × 0.40
New overall = 0.118, down from 0.136
factor = 0.80 ÷ 0.92 = 0.87
The light falls by about 13 %. Changing a filament bulb to an LED multiplies the light by 0.40 ÷ 0.05 = 8. The weakest link in a chain gives the biggest gain when it is improved — which is why lamps and motors were targeted before power lines.
1.6 Why wasted energy is wasted — and what to do about it
Wasted energy is not destroyed. It is dissipated: spread out into the surroundings, usually as thermal energy, raising the temperature of a large mass of air or water by a tiny amount. Spread-out energy cannot be gathered back to do useful work. That is the answer the May 2021 markscheme wanted when it asked why the heat from braking is “wasted”.
Ways of reducing waste all attack a specific transfer: lubrication and bearings cut friction in motors; thicker cables and high transmission voltages cut the heating of power lines; insulation keeps thermal energy where it is useful; streamlining cuts the energy spent pushing air aside; LEDs produce less heat for the same light; regenerative braking returns kinetic energy to a battery.
Some waste can be reused: combined heat and power stations pipe their “waste” hot water to heat buildings, lifting the overall use of the fuel’s energy from under 40 % to about 80 %.
A hybrid car’s regenerative brakes transfer some of its kinetic energy to its battery. Explain why this lets the car use less fuel over a journey with many stops.
Reveal answer
With ordinary brakes all the kinetic energy becomes thermal energy in the brakes and is dissipated [1].
Regenerative braking uses the motor as a generator, transferring part of the kinetic energy to electrical energy and then to chemical potential energy in the battery [1]. On the next acceleration the battery’s energy is transferred back into kinetic energy by the motor, so the engine has to supply less and burns less fuel [1]. (“The battery is recharged” alone repeats the question and did not score in May 2021.)
A combined heat and power station burns gas at a rate of 100 MW. It generates 38 MW of electricity and pipes 42 MW of hot water to nearby buildings; the rest is lost. Calculate the “electrical efficiency” and the “overall use of energy”, and explain why an engineer might quote either figure — and what each hides.
Reveal answer
electrical efficiency = 38 MW ÷ 100 MW = 38 %
overall use = (38 + 42) MW ÷ 100 MW = 80 %
The overall figure counts the hot water as useful, which it is only if the buildings need heat: in summer, or in a hot climate, it may be dumped and the station is effectively 38 % efficient.
The electrical figure hides the value of the heat. And a joule of electricity is more useful than a joule of warm water, because it can run any device; so “80 %” and “38 %” are not measuring the same kind of usefulness.
Try it — the efficiency of a bounce
A bouncing ball is the simplest energy chain you can test at home. At the top of the drop all its energy is gravitational potential energy, mgh₁. At the top of the first bounce it is back to gravitational potential energy, mgh₂. The fraction it kept, h₂ ÷ h₁, is the efficiency of the bounce; the rest has become thermal energy and sound.
You set the drop height (the independent variable) and can change the ball and the floor.
Common mistakes to avoid
- One energy form for a transfer. “Heat” is not a transfer; “chemical potential → thermal” is.
- “Energy is lost” or “used up”. It is transferred to the surroundings and dissipated; the total is conserved.
- Sankey widths that do not add up. Draw to a stated scale and check that the outputs sum to the input.
- Efficiency with a unit, or over 100 %. It has no unit and cannot exceed 1; above 100 % means a unit error (g, cm).
- Upside-down efficiency. Useful out ÷ total in, never in ÷ out.
- Mixing kW with seconds. kW × h gives kWh; W × s gives J; 1 kWh = 3.6 MJ.
- Averaging efficiencies in a chain. Multiply them.
- Repeating the question. “The battery is recharged” does not explain why less fuel is used; say where the energy goes next.
Study smart: how this is examined
Study routine. Today: draw the power-station Sankey diagram from memory, to scale, and write the efficiency. In two days: redo the motor and borehole EXAMPLES with the book closed, converting every unit first. In a week: run the bounce lab as a full investigation, then answer the Think task and the probe questions without looking at the answers.
How we found out — from horsepower to the conservation of energy
In the 1770s and 1780s the Scottish engineer James Watt was selling improved steam engines to mine owners who pumped water with horses. To make the comparison, he measured what a mill horse could do and defined one horsepower as lifting 33 000 pounds by one foot in a minute — about 746 W in modern units. The SI unit of power, the watt, was named after him in 1882.
Watt’s separate condenser made engines several times more efficient in fuel than the older Newcomen engines. It was the first great efficiency gain of the industrial age, and it was sold on the coal it saved.
What was the “power” that engines and horses had in common? In 1842 the German doctor Julius Robert Mayer argued that heat and work were interchangeable forms of one indestructible quantity. He had noticed that sailors’ venous blood looked redder in the tropics, where the body needs to make less heat.
In Manchester, James Prescott Joule measured it: through the 1840s he let falling weights turn a paddle wheel in water and showed that a fixed amount of work always produced the same rise in temperature. In 1847 Hermann von Helmholtz gave the principle its general mathematical form. By the 1850s “conservation of energy” was the foundation of physics.
The diagram named in this lesson came later. In 1898 the Irish engineer Captain Matthew Sankey published a drawing of the energy flows in a steam engine in which the width of each stream was proportional to the energy it carried. The idea made waste visible at a glance, and engineers have used it ever since.
Watt defined one horsepower as 33 000 foot-pounds per minute. Using 1 foot = 0.3048 m and a weight of 1 pound = 4.448 N, show that one horsepower is about 746 W.
Reveal answer
work per minute = 33 000 × 0.3048 m × 4.448 N
work per minute = 44 740 J
P = E ÷ t
P = 44 740 J ÷ 60 s
P = 746 W
A “work in a time” definition of power is exactly the P = E ÷ t of this lesson, written in the units of a Scottish mine in the 1780s.
Where it is used now
- Power stations in southern Africa. Botswana’s Morupule coal stations and South Africa’s large coal fleet turn roughly a third of the coal’s chemical energy into electricity. Engineers chase every point of efficiency, because a gain of 1 % across a station saves tens of thousands of tonnes of coal a year.
- Energy labels. Fridges, air conditioners and lamps are sold with labels that rank their efficiency or their yearly kWh, so buyers can compare running costs as well as prices.
- Solar water pumping. Farms and cattle posts use solar pumps at boreholes; installers size the panels from exactly the borehole calculation of Section 1.4, using the depth, the daily water needed and the pump’s efficiency.
- Clean cooking. A three-stone fire passes only about a tenth of the wood’s energy to the pot. Improved cookstoves raise that fraction, using less wood and producing less smoke — Lesson 3 returns to this.
- Electric vehicles. An electric motor turns about 90 % of its electrical input into kinetic energy; a petrol engine turns about 25–30 % of the fuel’s energy into motion. Regenerative braking recovers some of the rest.
- Data centres. Operators report a “power usage effectiveness”: total power taken by the building ÷ power reaching the computers. The closer it is to 1, the less goes on cooling.
Science now — and the scientist you need to be
What is happening in 2026
Efficiency is at the centre of energy policy. The International Energy Agency describes energy efficiency as the “first fuel”, and the countries that met at COP28 in Dubai in 2023 agreed to aim to double the global rate of energy-efficiency improvement by 2030.
LED lighting has spread worldwide over the past decade and now makes up most of the lamps sold. Heat pumps, which move thermal energy from outside air into a building instead of generating it, are spreading in colder countries. They deliver roughly three units of heat for each unit of electricity, so they are described by a “coefficient of performance”, not an efficiency above 100 %.
A new pressure comes from artificial intelligence. In 2025 the IEA estimated that data centres used about 1.5 % of the world’s electricity in 2024 and that their demand could roughly double by 2030, driven mainly by AI. Engineers are responding with more efficient chips, liquid cooling and the reuse of server heat.
Careers here include energy engineering, building services, power-station and grid operation, and energy auditing — people who measure where the joules go and find the waste.
The mindshift
What you need to be able to do
Name two energy forms for every transfer. Draw and read Sankey diagrams to scale. Convert units before substituting. Calculate efficiency both ways — from energy and from power — and rearrange it. Move confidently between joules, watts, kilowatt-hours and pula. And when an answer comes out above 100 %, stop and find the unit you forgot to convert.
See an example
Kettle 2.0 kW for 30 min a week:
E = 2.0 kW × 0.50 h = 1.0 kWh
Fridge 120 W, compressor running about a third of the time for 168 h:
E = 0.12 kW × 56 h = 6.7 kWh
Phone charger 15 W for 14 h: 0.21 kWh. The fridge dominates, even though its power is far lower than the kettle’s, because it runs for so long. An assistant that multiplies 2.0 kW by 30 (minutes) gives 60 kWh — the unit trap again.
Key terms
| Conservation of energy | Energy cannot be created or destroyed, only transferred; total energy in = useful energy out + wasted energy out. |
| Dissipated energy | Energy spread into the surroundings, usually as thermal energy, where it can no longer do useful work. |
| Sankey diagram | An energy-flow diagram in which the width of each arrow is proportional to the energy (or power) it carries. |
| Efficiency | Useful energy (or power) out ÷ total energy (or power) in; no unit; at most 1 (100 %). |
| Power | The rate of transferring energy or doing work: P = E ÷ t, in watts (1 W = 1 J/s). |
| Electrical power | P = VI; energy transferred electrically E = VIt. |
| Kilowatt-hour (kWh) | The energy transferred by 1 kW in 1 hour; 1 kWh = 3.6 MJ. The unit on an electricity bill. |
| Overall efficiency | For devices in a chain, the product of the individual efficiencies. |
| Regenerative braking | Braking that transfers kinetic energy to electrical energy and then to chemical energy stored in a battery. |
| Combined heat and power | A power station that also uses its waste thermal energy to heat buildings. |
Think like a scientist
Criterion B A school has: a small 6 V electric motor with a pulley and string; slotted masses of 50 g; a clamp stand; a metre rule; a variable power supply; an ammeter and a voltmeter; a stopwatch (0.01 s); connecting leads; a balance; a foam mat.
A student thinks that a motor lifting a heavier load might be more efficient, because more of the electrical energy goes into the load. She wants to investigate how the mass lifted affects the efficiency of the motor.
(a) Write a research question for the investigation. [2]
(b) Formulate a hypothesis with a testable prediction and scientific reasoning introduced by “because”. [3]
(c) Identify the independent variable and the dependent variable. Explain why the dependent variable cannot be read from a single instrument, and state every quantity that must be measured to calculate it. [4]
(d) Identify two control variables, each with the value held and why it must be held. [4]
(e) Describe how the circuit is set up to measure the electrical power supplied to the motor. [2]
(f) Write a numbered method that another student could follow, including five masses, three trials, what is recorded, how the efficiency is calculated, and one safety point specific to this experiment. [8]
Model answer
(a) How does the mass lifted by a 6 V electric motor, from 50 g to 250 g, affect the efficiency of the motor when it lifts the mass through 0.800 m? [1 for linking the mass lifted to the efficiency; 1 for a range and a fixed height.]
(b) If the mass lifted is increased from 50 g to 250 g, then the efficiency of the motor will increase [1].
This is because part of the electrical energy supplied is wasted in a way that does not depend much on the load — heating in the coils and friction in the bearings of a motor that turns in any case — so a larger useful output (mgΔh) makes that fixed waste a smaller fraction of the total [1].
At very large loads the motor slows and the current rises, heating the coils more (P = I²R), so the efficiency may level off or fall [1].
(c) IV: the mass lifted in kg, set with 50 g slotted masses and checked on the balance [1]. DV: the efficiency of the motor, as a percentage [1].
No instrument reads efficiency: it is a ratio calculated from the useful energy out and the electrical energy in [1]. Measure the mass m (balance), the height lifted Δh (metre rule), the voltage V across the motor (voltmeter), the current I through it (ammeter) and the time t of the lift (stopwatch); useful energy = mgΔh and energy in = VIt [1].
(d) [2 each, value + reason] The height lifted, 0.800 m between two marks on the metre rule, because the useful energy mgΔh depends on it and it would change the time of each lift.
The supply voltage, 6.0 V, set on the power supply and checked on the voltmeter, because a higher voltage increases the power and the speed of the motor, which would change the efficiency independently of the mass. (Also accepted: the same motor, pulley and string throughout.)
(e) The ammeter is connected in series with the motor, in the same loop as the supply [1]; the voltmeter is connected in parallel across the motor’s two terminals [1]. (November 2022 Physics Q4(d) gave marks for a motor, ammeter and voltmeter only, the ammeter in series and the voltmeter in parallel.)
(f)
- [1] Clamp the motor at the top of the stand with the pulley over the bench edge; fix the metre rule vertically and mark heights 0.100 m and 0.900 m above the floor (0.800 m apart).
- [1] Connect the motor, ammeter (series) and voltmeter (parallel) to the supply set at 6.0 V.
- [1] Hang 50 g (0.050 kg, checked on the balance) on the string with its base at the lower mark.
- [1] Switch on and start the stopwatch together; stop it as the base of the mass passes the upper mark; record V and I during the lift, and t.
- [1] Repeat twice more for three trials; repeat any trial whose time differs by more than 0.2 s from the others.
- [1] Repeat for 100, 150, 200 and 250 g — five masses, three trials each.
- [1] Record: mass/kg, V/V, I/A, t/s (trials and mean). Calculate the useful energy mgΔh and the energy in VIt; efficiency is (mgΔh ÷ VIt) × 100 %. Plot efficiency against mass.
- [1] Safety specific: put the foam mat under the hanging masses in case the string slips or the knot fails, and keep feet clear; switch off at once if the motor stalls, because a stalled motor draws a large current and its coils overheat.
[Level grid: 7–8 complete and repeatable with all measured quantities and the processing; 5–6 one element vague; 3–4 sequence right but quantities, repeats or processing missing; 1–2 outline only.]
Probe deeper — exam-style questions
Attempt every question before opening the answer. Marks in brackets show what a full answer is worth.
An electric kettle takes in 2 000 J and transfers 1 700 J to the water. Calculate its efficiency and state what happens to the rest of the energy.
Reveal answer
efficiency = 1 700 J ÷ 2 000 J
efficiency = 0.85 = 85 % [1]
The other 300 J is transferred as thermal energy to the kettle body and the surrounding air (and a little sound), and dissipated [1].
A 9.0 W LED lamp emits 3.6 W of light. Calculate its efficiency and the energy it wastes in 1.0 hour.
Reveal answer
efficiency = 3.6 W ÷ 9.0 W
efficiency = 0.40 = 40 % [1]
wasted power = 9.0 W − 3.6 W = 5.4 W [1]
wasted energy = 5.4 W × 3 600 s
wasted energy = 19 440 J ≈ 19 kJ [1]
A crane lifts a 500 kg load through 20 m in 25 s. Calculate the useful power output. If the crane’s motor is 60 % efficient, calculate its input power in kW.
Reveal answer
Ep = mgΔh
Ep = 500 kg × 9.8 N/kg × 20 m = 98 000 J [1]
P = 98 000 J ÷ 25 s = 3 920 W [1]
input = 3 920 W ÷ 0.60
input = 6 530 W ≈ 6.5 kW [1]
State why the thermal energy produced in a car’s brakes is described as wasted energy.
Reveal answer
It is dissipated into the surroundings and cannot be used by the car to do useful work [1]. (“Because it is lost” repeats the stem and scores 0 — May 2021 Physics Q3(c).)
A 150 W fridge compressor runs for one-third of the time. Calculate the energy it transfers in 30 days in kWh, and the cost at 1.50 pula per kWh.
Reveal answer
running time = 30 × 24 h ÷ 3 = 240 h [1]
E = 0.150 kW × 240 h = 36 kWh [1]
cost = 36 kWh × 1.50 pula per kWh = 54 pula [1]
Complete the energy-flow diagram for regenerative braking in a hybrid car: ______ energy → electrical energy → ______ energy in the battery, with some energy also transferred as ______ energy.
Reveal answer
Kinetic → electrical → chemical potential [1]; also thermal (heat) energy [1]. (May 2021 Physics Q3(d) awarded one mark for both boxes correct.)
7. A solar borehole pump uses 1.6 m² of panels in sunlight of 900 W/m². The panels are 20 % efficient and the pump is 45 % efficient. Calculate the mass of water it lifts from a depth of 40 m in one hour. Then state how many hours of such sunshine are needed to supply a herd that drinks 8 000 kg of water a day.
Reveal answer
electrical power = 1.6 m² × 900 W/m² × 0.20 = 288 W [1]
useful power = 0.45 × 288 W = 129.6 W
energy in 1 h = 129.6 W × 3 600 s = 466 560 J [1]
m = E ÷ (gΔh)
m = 466 560 J ÷ (9.8 N/kg × 40 m)
m = 1 190 kg ≈ 1 200 kg per hour [1]
hours = 8 000 kg ÷ 1 190 kg per hour = 6.7 h [1]
About 7 hours of strong sun — possible on a clear Kalahari day, but a storage tank is needed for cloudy days. The overall efficiency from sunlight to lifted water is only 0.20 × 0.45 = 9 %.
8. To heat 1.0 kg of water from 20 °C to 100 °C needs 336 kJ.
Compare the primary energy (energy in the fuel at the start of the chain) needed by (1) an electric kettle 90 % efficient, powered by a coal station 38 % efficient through lines 92 % efficient, and (2) a gas stove that passes 40 % of the gas’s energy to the pot. Then state one circumstance in which the kettle would be the better choice for the climate.
Reveal answer
(1) overall = 0.38 × 0.92 × 0.90 = 0.315 [1]
primary energy = 336 kJ ÷ 0.315 = 1 070 kJ [1]
(2) primary energy = 336 kJ ÷ 0.40 = 840 kJ [1]
The gas stove needs about 20 % less primary energy for the same heating. But if the electricity comes from solar, wind or hydro, the kettle’s chain starts with no fuel burned at all and releases far less carbon dioxide, so the kettle becomes the better choice for the climate [1].
Summary
- Energy is conserved: energy in = useful out + wasted out. Name two energy forms for every transfer.
- Sankey diagrams show energy flows with arrow widths proportional to energy; the output widths add up to the input width.
- efficiency = useful out ÷ total in, from energies or powers; no unit; never above 1 (100 %). Rearrange to find an input: input = output ÷ efficiency.
- Power is the rate of energy transfer: P = E ÷ t and P = VI. One watt is one joule per second.
- 1 kWh = 3.6 MJ: kW × hours gives kWh; W × seconds gives J.
- In a chain of devices the overall efficiency is the product of the individual efficiencies.
- Wasted energy is dissipated into the surroundings and cannot do useful work; lubrication, insulation, LEDs and regenerative braking reduce it, and combined heat and power reuses it.
Next time → Weather, climate and the greenhouse effect: how the Earth balances the energy it receives from the Sun, why carbon dioxide traps infrared radiation, and how burning fuels is changing the climate.