4.1 Energy — revision question pack

7 specification points · notes, questions, answers and worked methods

Checked against AQA 8463 section 4.1. Review basis: the qualification registry sourced from the AQA GCSE Physics (8463) specification; registry verification recorded 17 July 2026.

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4.1.1.1 · Energy stores and systems

Explanation

A system is one object or a group of objects being considered. When a system changes, energy is transferred between its stores or between the system and its surroundings.
Describe the change by naming the store that decreases, the store that increases and the transfer pathway: mechanically, electrically, by heating or by radiation.
Energy stores include kinetic, thermal, gravitational potential, elastic potential, chemical, magnetic, electrostatic and nuclear.
Total energy is conserved, so increases in stores equal the decrease elsewhere.
Energy is not used up; it may become dissipated into less useful thermal stores.

Worked example

Describe the energy changes when a battery-powered motor lifts a load.

1.The battery's chemical energy store decreases.
2.Energy is transferred electrically to the motor and mechanically to the load.
3.The load's gravitational potential energy store increases; some energy is dissipated to thermal stores.

Answer: Energy moves from the battery's chemical store to the load's gravitational store, with some thermal dissipation.

Common mistakes

Don't fall into the trap of saying the load gains ‘electrical energy’ instead of naming its gravitational potential energy store.
Don't fall into the trap of claiming energy is used up rather than transferred or dissipated.
Don't fall into the trap of naming stores but omitting the transfer pathway between them.

Exam tip

For ‘describe the energy changes’, name the decreasing store, transfer pathway and increasing store.

Tier 1 · Easy

A wheeled toy is given a push across a level floor and gradually stops. Describe the main energy-store changes after it is released.
[2 marks]
Total for this question: 2
An energy audit of a flywheel shows that its kinetic store decreases by $260\,\text{J}$ while thermal stores increase by $260\,\text{J}$ . A student says, ‘The flywheel has lost $260\,\text{J}$ of energy.’ Explain why this statement is incorrect.
[2 marks]
Total for this question: 2

Tier 2 · Standard

An electrically powered hotplate transfers $96\,\text{kJ}$ of energy. The pan's thermal energy store increases by $61\,\text{kJ}$ and the food's thermal energy store increases by $27\,\text{kJ}$ . Calculate the energy transferred to other stores and state where it is likely to be stored.
[3 marks]
Total for this question: 3
An energy audit gives these values, all in kilojoules: test A, input $2.40$ , increases in kinetic, thermal and other stores $1.10$ , $0.90$ and $0.40$ ; test B, input $2.70$ , increases $1.30$ , $1.00$ and $0.40$ ; test C, input $3.00$ , increases $1.50$ , $1.10$ and $0.50$ . The recorder copied one input value incorrectly. Identify the inconsistent test and calculate the corrected input.
[2 marks]
Total for this question: 2
Two repelling magnets are held close together, with one fixed and one attached to a stationary trolley. The trolley is released and later compresses a spring until it is momentarily stationary. Describe the sequence of changes in energy stores and name the transfer pathway during each stage. Ignore friction.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A model launcher begins with $240\,\text{J}$ in its elastic potential energy store. After the model has left the launcher, the spring is fully relaxed. The model has $150\,\text{J}$ in its kinetic energy store and $54\,\text{J}$ in its gravitational potential energy store; these and the thermal stores account for the whole system. Calculate the energy in thermal stores and describe the complete redistribution.
[4 marks]
Total for this question: 4
An energy-flow diagram for a battery-powered fan has a $50\,\text{mm}$ input arrow, a $32\,\text{mm}$ useful-output arrow and a $23\,\text{mm}$ dissipated-output arrow. Its caption says, ‘Energy is conserved because $32+23=50$ ; the battery's electrical energy store supplies both outputs.’ The caption is incorrect. Give two corrections, calculate the correct width of the dissipated-output arrow, and describe the stores and pathways correctly.
[5 marks]
Total for this question: 5
A hot block is lowered into water in a well-insulated container. The block's thermal energy store decreases by $720\,\text{J}$ ; the water's thermal store increases by $600\,\text{J}$ and the container and thermometer gain the remainder. First take only the water as the system, then take the container and all its contents as the system. For each boundary, describe the energy transfers. Use the values to show why the second system has no net energy transfer across its boundary.
[5 marks]
Total for this question: 5
An electric delivery van accelerates and later stops. The van's motor recharges the battery as the van slows. During acceleration, the battery's chemical energy store decreases by $640\,\text{kJ}$ , the van's kinetic store increases by $430\,\text{kJ}$ and thermal stores increase by $210\,\text{kJ}$ . During braking, the kinetic store decreases by $430\,\text{kJ}$ and the battery's chemical store increases by $310\,\text{kJ}$ . Calculate the increase in thermal stores during braking. Describe the transfer pathways in both stages and show that the complete journey conserves energy.
[6 marks]
Total for this question: 6
A person turns the handle of a rechargeable torch. The person's chemical energy store decreases by $5.00\,\text{kJ}$ and the torch battery's chemical store increases by $3.20\,\text{kJ}$ . The rest increases thermal stores. Calculate the energy transferred to thermal stores. On an energy-transfer diagram, $1.00\,\text{kJ}$ is represented by $20\,\text{mm}$ . State the width of the input, chemical-store and thermal-store arrows for the charging stage, and show that the arrow widths account for all of the input energy.
[6 marks]
Total for this question: 6

4.1.1.2 · Changes in energy

Explanation

Calculate energy in kinetic, elastic potential and gravitational potential stores using $E_k=\dfrac{1}{2}mv^2$ , $E_e=\dfrac{1}{2}ke^2$ and $E_p=mgh$ .
Use kilograms, metres per second, newtons per metre, metres and the supplied value of $g$ .
The elastic equation applies while the spring obeys the linear force–extension relationship, and $e$ is extension rather than total length.
If a question states that transfers are complete with no dissipation, equate the decrease in one store to the increase in another.
Speed and extension are squared, so doubling either makes the corresponding energy four times larger.

Worked example

A $3.0\,\text{kg}$ object moves at $8.0\,\text{m/s}$ . Calculate its kinetic energy.

1. $E_k=\dfrac{1}{2}mv^2$ .
2. $E_k=0.5\times3.0\times8.0^2$ .

Answer: $96\,\text{J}$

Common mistakes

Don't fall into the trap of using $E_k=mv/2$ and failing to square the speed.
Don't fall into the trap of using the spring's total length instead of its extension in $E_e=\frac12ke^2$ .
Don't fall into the trap of using a mass in grams rather than kilograms.

Exam tip

Write the correct store equation and convert every quantity to SI units before substitution.

Tier 1 · Easy

Calculate the kinetic energy of a $2.0\,\text{kg}$ cart moving at $3.0\,\text{m/s}$ .
[2 marks]
Total for this question: 2
Two identical trolleys move at $2.5\,\text{m/s}$ and $5.0\,\text{m/s}$ . Without calculating either energy, determine the ratio of their kinetic energy stores and explain your answer.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A $35\,\text{kg}$ climber gains $4.2\,\text{m}$ in vertical height. Calculate the increase in the climber's gravitational potential energy store. Use $g=9.8\,\text{N/kg}$ .
[2 marks]
Total for this question: 2
A $0.80\,\text{kg}$ cart is tested at speeds of $1.5\,\text{m/s}$ , $2.0\,\text{m/s}$ and $2.5\,\text{m/s}$ . Use an equation from the Physics Equations Sheet to calculate its kinetic energy at each speed. Explain whether equal increases in speed produce equal increases in kinetic energy.
[4 marks]
Total for this question: 4
A warehouse lift raises a load by $15.0\,\text{m}$ , increasing its gravitational potential energy store by $7.35\,\text{kJ}$ . Determine the mass of the load in kilograms. Use $g=9.8\,\text{N/kg}$ and an equation from the Physics Equations Sheet.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A spring of spring constant $320\,\text{N/m}$ is compressed by $0.15\,\text{m}$ . It launches a $0.45\,\text{kg}$ cart on a level frictionless track. Calculate the cart's launch speed.
[4 marks]
Total for this question: 4
A spring compressed by $6.0\,\text{cm}$ transfers all its stored energy to a $180\,\text{g}$ cart on a level track. The cart leaves at $4.0\,\text{m/s}$ . Use equations from the Physics Equations Sheet to determine the spring constant. Convert both measured quantities to SI units before substitution.
[5 marks]
Total for this question: 5
A spring with spring constant $80.0\,\text{N/m}$ stores $1.60\,\text{J}$ . Calculate its extension. The extension is then increased by $25.0\%$ without exceeding the limit of proportionality. Calculate the new elastic potential energy. Give both answers to $3$ significant figures and use equations from the Physics Equations Sheet.
[5 marks]
Total for this question: 5
A spring of spring constant $500\,\text{N/m}$ is compressed by $0.180\,\text{m}$ and launches a $0.400\,\text{kg}$ cart. The cart rises through a vertical height of $1.20\,\text{m}$ while $1.00\,\text{J}$ is transferred to thermal stores. Calculate the cart's speed at that height. Take $g=9.8\,\text{N/kg}$ and choose the relevant relationships from the Physics Equations Sheet.
[5 marks]
Total for this question: 5
A $1.20\,\text{kg}$ trolley initially moves at $3.00\,\text{m/s}$ . A motor then does $24.0\,\text{J}$ of work on the trolley as it rises through a vertical height of $1.00\,\text{m}$ . Its final speed is $4.00\,\text{m/s}$ . Calculate the energy transferred to thermal stores. Take $g=9.8\,\text{N/kg}$ and choose the relevant relationships from the Physics Equations Sheet.
[6 marks]
Total for this question: 6

4.1.1.3 · Energy changes in systems

Explanation

Specific heat capacity is the energy required to raise the temperature of $1\,\text{kg}$ of a substance by $1\,{}^\circ\text{C}$ .
Use $\Delta E=mc\Delta\theta$ , where energy is in joules, mass in kilograms, specific heat capacity in $\text{J/(kg }{}^\circ\text{C)}$ and temperature change in degrees Celsius.
In the required practical, measure mass, supply energy with an electrical heater using $E=Pt$ , and record the temperature rise.
Insulate the block, ensure good thermal contact and repeat readings because energy transferred to the surroundings otherwise makes the calculated value inaccurate.

Worked example

A $2.0\,\text{kg}$ block receives $18000\,\text{J}$ and warms by $15\,{}^\circ\text{C}$ . Calculate its specific heat capacity.

1.Rearrange $\Delta E=mc\Delta\theta$ to $c=\dfrac{\Delta E}{m\Delta\theta}$ .
2. $c=\dfrac{18000}{2.0\times15}$ .

Answer: $600\,\text{J/(kg }{}^\circ\text{C)}$

Common mistakes

Don't fall into the trap of substituting the final temperature instead of final minus initial temperature.
Don't fall into the trap of using mass in grams in the specific heat capacity equation.
Don't fall into the trap of assuming all heater energy reaches the block despite heating the surroundings.

Exam tip

In a practical evaluation, identify heat loss as making the calculated specific heat capacity too high.

Tier 1 · Easy

State what a specific heat capacity of $900\,\text{J/(kg}\,{}^\circ\text{C)}$ means.
[2 marks]
Total for this question: 2
During a specific heat capacity experiment, a block's temperatures at $0$ , $60$ , $120$ , $180$ and $240\,\text{s}$ are $20.1$ , $24.0$ , $27.9$ , $39.8$ and $35.8\,{}^\circ\text{C}$ . Identify the anomalous reading and state what the student should do before using the temperature trend.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A $1.5\,\text{kg}$ stone block has specific heat capacity $900\,\text{J/(kg}\,{}^\circ\text{C)}$ . Calculate the change in its thermal energy store when its temperature rises by $28\,{}^\circ\text{C}$ .
[2 marks]
Total for this question: 2
In the specific heat capacity required practical, a student leaves the metal block uninsulated and puts the heater and thermometer loosely into dry holes. State two changes to this setup and explain how each change improves the experiment.
[4 marks]
Total for this question: 4
A solid block of an unknown material has mass $0.400\,\text{kg}$ . Its cumulative energy input is $0$ , $1440$ , $2880$ and $4320\,\text{J}$ when its temperature is $18.0$ , $22.0$ , $26.0$ and $30.0\,{}^\circ\text{C}$ respectively. Use the first and last readings to determine the block's specific heat capacity. Explain why the temperature change, rather than the final temperature, is used.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A $75\,\text{W}$ heater warms a $0.30\,\text{kg}$ sample for $6.0\,\text{minutes}$ . The sample's temperature rises by $32\,{}^\circ\text{C}$ and $80\%$ of the electrical energy is transferred to its thermal energy store. Determine the sample's specific heat capacity.
[5 marks]
Total for this question: 5
A $540\,\text{g}$ metal block has specific heat capacity $390\,\text{J/(kg}\,{}^\circ\text{C)}$ . Its temperature must rise by $28\,{}^\circ\text{C}$ . A $70\,\text{W}$ heater transfers $78\%$ of its input energy to the block. Calculate the heating time in seconds and minutes.
[5 marks]
Total for this question: 5
A solid block of an unknown material has mass $0.600\,\text{kg}$ and warms by $15.0\,{}^\circ\text{C}$ after an electrical input of $10.8\,\text{kJ}$ . Measurements show that $1.80\,\text{kJ}$ was transferred to the heater, thermometer and surroundings instead of the block. Calculate the corrected specific heat capacity and the value obtained if this unwanted transfer were ignored. Explain the direction of the error in the uncorrected value.
[5 marks]
Total for this question: 5
A $0.300\,\text{kg}$ copper block at $95.0\,{}^\circ\text{C}$ is placed in $0.200\,\text{kg}$ of water at $18.0\,{}^\circ\text{C}$ in a well-insulated container. The final temperature is $T$ . Use $c_{\text{copper}}=385\,\text{J/(kg}\,{}^\circ\text{C)}$ and $c_{\text{water}}=4200\,\text{J/(kg}\,{}^\circ\text{C)}$ . By equating the energy released by the copper to the energy gained by the water, calculate $T$ and the energy transferred.
[6 marks]
Total for this question: 6
Three aluminium blocks have masses $0.500\,\text{kg}$ , $1.00\,\text{kg}$ and $1.50\,\text{kg}$ . Each block's temperature rises by $20.0\,{}^\circ\text{C}$ . A joulemeter records energy transfers of $9000\,\text{J}$ , $18\,000\,\text{J}$ and $27\,000\,\text{J}$ respectively. Calculate the specific heat capacity for each block and state what the results show about whether specific heat capacity depends on mass.
[6 marks]
Total for this question: 6

4.1.1.4 · Power

Explanation

Power is the rate of energy transfer or the rate of doing work. Use $P=E/t$ or $P=W/t$ , with energy or work in joules and time in seconds, to obtain power in watts.
One watt means one joule transferred each second.
Two devices can transfer the same amount of energy but have different powers: the device that completes the transfer in less time has greater power.
Conversely, two devices operating for the same time transfer different energies if their powers differ.
Power is a rate, not an energy store or a total amount of energy.

Worked example

A winch does $36000\,\text{J}$ of work in $45\,\text{s}$ . Calculate its power.

1.Use $P=\dfrac{W}{t}$ .
2. $P=\dfrac{36000}{45}$ .

Answer: $800\,\text{W}$

Common mistakes

Don't fall into the trap of giving power in joules rather than watts.
Don't fall into the trap of multiplying energy by time instead of dividing by time.
Don't fall into the trap of saying a more powerful device always transfers more energy without considering duration.

Exam tip

When comparing devices, state that the more powerful one transfers energy faster.

Tier 1 · Easy

A small motor transfers $600\,\text{J}$ of energy in $20\,\text{s}$ . Calculate its power.
[2 marks]
Total for this question: 2
A data table shows that a pulse welder transfers $18\,\text{kJ}$ in $6.0\,\text{s}$ at a power of $3000\,\text{W}$ , while a heater transfers $40\,\text{kJ}$ in $20\,\text{s}$ at a power of $2000\,\text{W}$ . State which device transfers energy at the greater rate and explain what one watt means.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Winch A and winch B each do $18\,\text{kJ}$ of work. A takes $30\,\text{s}$ and B takes $45\,\text{s}$ . Calculate both powers and compare them.
[4 marks]
Total for this question: 4
Appliance A operates at $650\,\text{W}$ for $30\,\text{s}$ . Appliance B operates at $480\,\text{W}$ for $45\,\text{s}$ . Calculate the energy transferred by each appliance in kilojoules. State which transfers more energy and by how much.
[3 marks]
Total for this question: 3
A joulemeter reads $12\,\text{kJ}$ at $0\,\text{s}$ , $36\,\text{kJ}$ at $40\,\text{s}$ , $75\,\text{kJ}$ at $100\,\text{s}$ and $105\,\text{kJ}$ at $160\,\text{s}$ . Calculate the mean power during each interval and identify the interval with the greatest mean power.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A hoist raises a $250\,\text{kg}$ load through $12\,\text{m}$ in $25\,\text{s}$ . Calculate the useful power of the hoist. Use $g=9.8\,\text{N/kg}$ . A second hoist performs the same lift in $18\,\text{s}$ ; calculate its useful power.
[5 marks]
Total for this question: 5
A machine operating at a constant power of $2.40\,\text{kW}$ transfers $1.62\,\text{MJ}$ . Determine the operating time in minutes. Explain whether this exceeds a $10.5\,\text{minutes}$ maintenance limit, and by how much.
[5 marks]
Total for this question: 5
An automated press completes $24$ cycles each minute while it is running and does $2.50\,\text{kJ}$ of work in every cycle. During one observation it runs for $6.0\,\text{minutes}$ and is then paused for $120\,\text{s}$ . Calculate the total work done and the mean power over the complete observation, including the pause.
[5 marks]
Total for this question: 5
A workshop circuit has a maximum power of $3.00\,\text{kW}$ . A cutter uses $1.20\,\text{kW}$ continuously and an extractor uses $0.85\,\text{kW}$ continuously. In every $60\,\text{s}$ , a compressor uses $2.00\,\text{kW}$ for $18\,\text{s}$ and is off for the remaining time. Calculate the mean total power over one minute and the greatest instantaneous total power. Evaluate the claim that the circuit is suitable because its mean power is below the limit.
[5 marks]
Total for this question: 5
A food mixer is labelled $720\,\text{W}$ . During a $60.0\,\text{s}$ run, a joulemeter reading rises from $12.4\,\text{kJ}$ to $55.6\,\text{kJ}$ . Calculate the mixer's mean power and evaluate the label. State one control variable and one safety precaution for the test.
[5 marks]
Total for this question: 5

4.1.2.1 · Energy transfers in a system

Explanation

Energy can be transferred usefully, stored or dissipated, but it cannot be created or destroyed. In a closed system, the total energy does not change.
Dissipated energy spreads into thermal stores of the surroundings and becomes less useful, rather than disappearing.
Reduce unwanted mechanical transfers with lubrication, which lowers friction.
Reduce transfer by heating with thermal insulation, including thicker walls and materials that trap air.
A material with a high thermal conductivity transfers energy through it rapidly; increasing wall thickness reduces the transfer rate for the same material and temperature difference.

Worked example

Explain two ways to reduce unwanted energy transfers from a heated water tank.

1.Add a thicker layer of low-thermal-conductivity insulation to reduce conduction.
2.Fit a lid to reduce energy transfer from the exposed water surface.

Answer: Use thick thermal insulation and a lid so energy is transferred to the surroundings more slowly.

Common mistakes

Don't fall into the trap of saying dissipated energy has been destroyed.
Don't fall into the trap of claiming a high thermal conductivity makes a material a good insulator.
Don't fall into the trap of suggesting lubrication reduces unwanted heating without linking it to reduced friction.

Exam tip

For an insulation explanation, name the transfer being reduced and the property or design feature that reduces its rate.

Tier 1 · Easy

Explain why adding oil to the axle of a turning wheel reduces unwanted energy transfers.
[2 marks]
Total for this question: 2
A table gives the thermal conductivity of two equal-thickness wall panels: panel A, $0.040\,\text{W/(m}\,{}^\circ\text{C)}$ ; panel B, $0.20\,\text{W/(m}\,{}^\circ\text{C)}$ . State which panel transfers energy more slowly by conduction and explain how the table supports your answer.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Wall X is twice as thick as wall Y and both are made from the same material. Explain which wall gives the lower rate of energy transfer by conduction. Then state how replacing the material with one of lower thermal conductivity affects the rate.
[3 marks]
Total for this question: 3
A student compares bubble wrap and felt as insulators. One layer of bubble wrap surrounds a $250\,\text{cm}^3$ glass beaker, while three layers of felt surround a $400\,\text{cm}^3$ metal beaker. The water starts at different temperatures, and each setup is tested once. State four changes needed for a valid and repeatable comparison.
[4 marks]
Total for this question: 4
A battery drives a fan inside a sealed, thermally insulated box. After the battery is depleted, the fan eventually stops. Explain why the total energy of the box and its contents has not changed, even though the stored energy has become less useful.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Plan an investigation to compare the effectiveness of three fabrics as thermal insulators around identical beakers of hot water. Include the measurements, control variables and how the results should be used.
[6 marks]
Total for this question: 6
Identical hot-water containers are tested with insulation. The table gives thickness, cost per container and three temperature drops after $12\,\text{minutes}$ : $5\,\text{mm}$ , £0.24, drops $18.2$ , $18.0$ , $18.1\,{}^\circ\text{C}$ ; $10\,\text{mm}$ , £0.39, drops $13.5$ , $13.7$ , $13.6\,{}^\circ\text{C}$ ; $15\,\text{mm}$ , £0.62, drops $11.8$ , $18.9$ , $12.0\,{}^\circ\text{C}$ . The largest value in the final row is anomalous. Choose the least expensive insulation that keeps the mean drop at or below $14.0\,{}^\circ\text{C}$ . Show how the data support your choice.
[5 marks]
Total for this question: 5
Three panels have equal area and thickness. Electric heaters keep their inner faces at the same constant temperature above the same room temperature. Over $20\,\text{minutes}$ , the energies needed to replace transfers through panels P, Q and R are $18.0$ , $11.0$ and $15.0\,\text{kJ}$ . Identify the best thermal insulator and explain your choice. When Q's thickness is doubled, the required energy falls to $6.60\,\text{kJ}$ . Calculate the percentage reduction and explain the effect of thickness.
[5 marks]
Total for this question: 5
A model flywheel has $18.0\,\text{J}$ in its kinetic energy store and then slows to rest. It takes $12.0\,\text{s}$ with a dry axle, $30.0\,\text{s}$ with oil and $24.0\,\text{s}$ with grease. Calculate the mean power transferred from the flywheel's kinetic store in each case and identify the best lubricant. Explain your choice in terms of friction.
[6 marks]
Total for this question: 6
A well-insulated container holds $0.500\,\text{kg}$ of water. Its temperature falls from $80.0$ to $72.0\,{}^\circ\text{C}$ during the first $5.00\,\text{minutes}$ and from $54.0$ to $51.0\,{}^\circ\text{C}$ during a later $5.00\,\text{minutes}$ . Use $c_{\text{water}}=4200\,\text{J/(kg}\,{}^\circ\text{C)}$ to calculate the mean power transferred from the water in each interval. Explain the difference and evaluate the claim that the insulation became more effective later.
[5 marks]
Total for this question: 5

4.1.2.2 · Efficiency

Explanation

Efficiency is the fraction of total input energy transferred usefully: $\text{efficiency}=\dfrac{\text{useful output energy}}{\text{total input energy}}$ .
The same ratio can use useful power output and total power input.
A decimal efficiency lies between $0$ and $1$ ; multiply by $100$ only when a percentage is required.
If efficiency is $0.72$ , then $72\%$ of the input is useful and $28\%$ is dissipated.
Improve efficiency by reducing unwanted transfers, such as using lubrication to reduce friction or thermal insulation to reduce heating of the surroundings.

A Sankey-style energy transfer showing useful output and dissipated energy.

Worked example

A device receives $500\,\text{J}$ and transfers $360\,\text{J}$ usefully. Calculate its efficiency.

1. $\text{efficiency}=\dfrac{360}{500}=0.72$ .
2.As a percentage, $0.72\times100=72\%$ .

Answer: $0.72$ or $72\%$

Common mistakes

Don't fall into the trap of dividing total input by useful output and obtaining an efficiency above $100\%$ .
Don't fall into the trap of multiplying by $100$ when the question requests a decimal efficiency.
Don't fall into the trap of mixing useful energy with total power in the same ratio.

Exam tip

Check that an ordinary device's efficiency is no greater than $1$ or $100\%$ .

Tier 1 · Easy

A device receives $90\,\text{J}$ and transfers $72\,\text{J}$ usefully. Calculate its efficiency as a decimal and as a percentage.
[2 marks]
Total for this question: 2
A Sankey diagram uses $30$ equal-width divisions for the input arrow and $21$ divisions for the useful-output arrow. A student says the device is $30\%$ efficient. Calculate the correct efficiency and the width of the dissipated-output arrow.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A pump has a total power input of $560\,\text{W}$ and a useful power output of $420\,\text{W}$ . Calculate its efficiency and its wasted power.
[3 marks]
Total for this question: 3
A winch provides $1.68\,\text{kW}$ of useful power and is $70\%$ efficient. Calculate its total input power and its dissipated power, both in watts.
[4 marks]
Total for this question: 4
A motor raises an $18.0\,\text{kg}$ load through $4.00\,\text{m}$ . Its total input energy is $960\,\text{J}$ . Calculate the useful energy transfer and the motor's efficiency as a percentage. Use $g=9.8\,\text{N/kg}$ and equations from the Physics Equations Sheet. Give both answers to $3$ significant figures.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Higher only: A motor transfers $80\%$ of its input energy to a rotating shaft. A generator then transfers $65\%$ of the shaft energy to electrical energy. Calculate the overall efficiency and the useful electrical energy produced from an initial input of $250\,\text{kJ}$ . Suggest one change that could increase the efficiency of this intended transfer and explain how it helps.
[6 marks]
Total for this question: 6
Higher only: Two bearing designs are tested. With dry bearings, input power is $1.50\,\text{kW}$ and useful output is $0.90\,\text{kW}$ . With lubricated bearings, input power is $1.40\,\text{kW}$ and useful output is $1.05\,\text{kW}$ . A student claims lubrication made the machine less efficient because the input fell. Calculate both efficiencies and evaluate the claim. Describe one further way to increase the efficiency of this machine and explain how it works.
[5 marks]
Total for this question: 5
A pump has a total input power of $1.50\,\text{kW}$ for $4.00\,\text{minutes}$ and an efficiency of $64.0\%$ . Calculate the total input energy, useful output energy and dissipated energy, all in kilojoules. Decide whether the useful output meets a target of $240\,\text{kJ}$ and state the shortfall or excess.
[5 marks]
Total for this question: 5
A motorised trolley of mass $40.0\,\text{kg}$ receives $1.60\,\text{kJ}$ of input energy. It rises through $2.50\,\text{m}$ and reaches a speed of $3.00\,\text{m/s}$ . Both increases in energy stores are useful. Calculate the total useful output, the efficiency and the energy dissipated. Take $g=9.8\,\text{N/kg}$ and choose the relevant relationships from the Physics Equations Sheet.
[5 marks]
Total for this question: 5
A boiler receives $18.0\,\text{MJ}$ from its fuel. It transfers $13.5\,\text{MJ}$ to the water as intended, $2.00\,\text{MJ}$ to the room's thermal store and the remainder to the exhaust and outside surroundings. A student counts the room heating as useful and reports an efficiency of $86.1\%$ . Evaluate this result by calculating the intended efficiency and accounting for all unwanted energy transfers.
[5 marks]
Total for this question: 5

4.1.3 · National and global energy resources

Explanation

Renewable resources are replenished as they are used: biofuel, wind, hydroelectricity, geothermal, tides, sunlight and water waves. Fossil fuels and nuclear fuel are non-renewable.
Compare resources for transport, heating and electricity generation using reliability, response to demand, environmental effects and cost.
Wind and solar output vary with conditions, while fossil-fuel stations can respond to demand but release greenhouse gases and pollutants.
Nuclear generation has low operational carbon emissions but produces radioactive waste.
Patterns of use reflect availability, technology, economics and political, social or ethical choices; science identifies impacts but does not make every decision.

Worked example

Compare wind power with natural gas for electricity generation.

1.Wind is renewable and produces no fuel emissions during operation, but its output is variable.
2.Natural gas is non-renewable and releases carbon dioxide, but generation is reliable and can respond quickly to demand.
3.A balanced conclusion depends on whether emissions, reliability or response time is prioritised.

Answer: Wind reduces fuel use and emissions but is intermittent; gas is controllable but non-renewable and carbon-emitting.

Common mistakes

Don't fall into the trap of calling nuclear fuel renewable because its operational carbon emissions are low.
Don't fall into the trap of claiming wind power is completely reliable because wind is free.
Don't fall into the trap of giving only one advantage without comparing the same criterion for both resources.

Exam tip

For ‘compare’, discuss both resources against the same criteria and give a justified conclusion.

Tier 1 · Easy

Classify wind, natural gas, geothermal and nuclear fuel as renewable or non-renewable energy resources.
[2 marks]
Total for this question: 2
A leaflet says, ‘Nuclear power is renewable because a station releases no carbon dioxide while generating electricity.’ Identify the error and give one correct environmental statement about nuclear generation.
[2 marks]
Total for this question: 2

Tier 2 · Standard

In a region, the share of transport energy supplied by biofuel rises from $4\%$ to $11\%$ , while the share of heating energy supplied by oil falls from $38\%$ to $25\%$ . Describe both trends. Explain one environmental reason for the change and one practical reason why oil may still be used.
[4 marks]
Total for this question: 4
Planning data give the minimum and maximum measured outputs of three stations: solar, $0$ and $62\,\text{MW}$ ; wind, $8$ and $70\,\text{MW}$ ; natural gas, $48$ and $52\,\text{MW}$ . Calculate the output range for each station, identify which was most reliable during the measurements and state one environmental disadvantage of that resource.
[4 marks]
Total for this question: 4
A council is considering wind turbines or a gas-fired power station. Scientists report that wind is renewable but its output varies, while gas generation is controllable but releases carbon dioxide. Residents disagree about cost, jobs and landscape changes. Give two conclusions that the scientific evidence supports and explain why science alone cannot make the final choice.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A country is choosing between expanding offshore wind and building a nuclear power station. Evaluate the two options for large-scale electricity supply. Your answer should consider reliability, environmental impacts and economic or social factors.
[6 marks]
Total for this question: 6
A region must choose one project with a cost no greater than £125 million, a firm output of at least $50\,\text{MW}$ and lifecycle emissions no greater than $20\,\text{g/kWh}$ . Planning estimates list cost, firm output and lifecycle emissions: solar, £42 million, $0\,\text{MW}$ , $48\,\text{g/kWh}$ ; natural gas, £58 million, $48\,\text{MW}$ , $420\,\text{g/kWh}$ ; tidal, £96 million, $34\,\text{MW}$ , $8\,\text{g/kWh}$ ; nuclear, £122 million, $58\,\text{MW}$ , $12\,\text{g/kWh}$ . Choose the only project meeting all three constraints, justify the choice from the table and explain one drawback not represented by the three columns.
[5 marks]
Total for this question: 5
A region has reliable access to geothermal energy, a windy coast and land that could grow biofuel crops. It needs a resource for transport, one for dependable heating and one for electricity generation. Propose a plan that uses geothermal energy, wind and biofuel once each. Justify every allocation, including a relevant limitation where appropriate.
[6 marks]
Total for this question: 6
A district needs a mean heating power of $46\,\text{kW}$ during a $30.0\,\text{minute}$ period of the night. A geothermal system supplies $32\,\text{kW}$ continuously and a biofuel boiler can supply up to $18\,\text{kW}$ . Calculate the energy supplied by each resource when the demand is met, show whether the boiler has sufficient power, and explain one environmental limitation of using biofuel for the remaining demand.
[5 marks]
Total for this question: 5
A council compares biofuel buses with battery buses charged from the regional electricity mix. For each $100\,\text{km}$ , the estimated lifecycle emissions are $17.0\,\text{kg}$ of carbon-dioxide equivalent for biofuel and $11.5\,\text{kg}$ for the battery option. Operating costs are £0.42 per kilometre for biofuel and £0.31 per kilometre for battery power. Each bus travels $28\,000\,\text{km}$ per year. Calculate the annual emissions and operating cost for each option. Evaluate which option the data support and give one relevant limitation not included in the figures.
[6 marks]
Total for this question: 6

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

4.1.1.1 · Energy stores and systems

Tier 1 · Easy

Mark scheme for 4.1.1.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	The toy's kinetic energy store decreases. Energy is transferred to thermal energy stores of the toy and its surroundings.	Choose the moving object as the system. Its speed falls, so its kinetic store decreases. Friction transfers energy mechanically to thermal stores in the wheels, floor and nearby air.	2
Total Question 1			2
02.1	The energy has not been lost or destroyed; total energy is conserved. It has been transferred from the flywheel's kinetic store to thermal stores of the flywheel and its surroundings.	Compare the decrease and increase: both are $260\,\text{J}$ . The flywheel's kinetic store has decreased, but an equal amount has been transferred mechanically, through friction, to thermal stores, so the energy account still totals $260\,\text{J}$ .	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.1.1.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	$8\,\text{kJ}$ It is mainly in thermal energy stores of the hotplate and surroundings.	Conservation of energy requires the output increases to total $96\,\text{kJ}$ . The accounted increase is $61+27=88\,\text{kJ}$ , so the remainder is $96-88=8\,\text{kJ}$ . This energy is dissipated to thermal stores of the apparatus and surroundings.	3
Total Question 1			3
02.1	Test C is inconsistent. Its output-store increases total $3.10\,\text{kJ}$ . The corrected input is $3.10\,\text{kJ}$ .	Audit each row by adding its store increases. Test A gives $1.10+0.90+0.40=2.40\,\text{kJ}$ and test B gives $1.30+1.00+0.40=2.70\,\text{kJ}$ . Test C gives $1.50+1.10+0.50=3.10\,\text{kJ}$ , so conservation requires a $3.10\,\text{kJ}$ input rather than $3.00\,\text{kJ}$ .	2
Total Question 2			2
03.1	As the trolley is released, the magnetic energy store decreases and the trolley's kinetic energy store increases. Energy is transferred mechanically by the magnetic force. As the trolley compresses the spring and stops, the trolley's kinetic store decreases and the spring's elastic potential store increases. This transfer is also mechanical because a force does work on the spring.	Follow the trolley through the two stages. Repulsion reduces the magnetic store of the magnet system and does mechanical work on the trolley, increasing its kinetic store. The moving trolley then does mechanical work compressing the spring, so the kinetic store decreases to zero while the elastic potential store increases.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.1.1.1 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	$36\,\text{J}$ is in thermal energy stores. The elastic potential store decreases by $240\,\text{J}$ ; the kinetic, gravitational potential and thermal stores increase by $150\,\text{J}$ , $54\,\text{J}$ and $36\,\text{J}$ respectively.	The total increase in stores must equal the original $240\,\text{J}$ . The known increases total $150+54=204\,\text{J}$ , leaving $240-204=36\,\text{J}$ in thermal stores. Checking on a common scale gives $150+54+36=240\,\text{J}$ , so energy is conserved.	4
Total Question 1			4
02.1	$32+23=55$ , not $50$ , so the drawn arrow widths do not conserve energy. For a $50\,\text{mm}$ input and a $32\,\text{mm}$ useful arrow, the dissipated arrow should be $18\,\text{mm}$ . The battery's chemical energy store decreases; there is no electrical energy store. Energy is transferred electrically to the motor, increasing the fan's kinetic store, with energy also dissipated to thermal stores and by sound.	First audit the widths: $32+23=55\,\text{mm}$ , so the outputs exceed the $50\,\text{mm}$ input. The missing consistent width is $50-32=18\,\text{mm}$ . Then repair the terminology: the battery has a chemical store, energy is transferred electrically, and the increases are in the fan's kinetic store and thermal stores of the equipment and surroundings; sound is a transfer by waves, not a store.	5
Total Question 2			5
03.1	For the water-only system, $600\,\text{J}$ crosses its boundary by heating and increases the water's thermal store. For the whole-container system, the block's thermal store decreases as energy is transferred by heating to the water, container and thermometer. The container and thermometer gain $720-600=120\,\text{J}$ in their thermal stores. The increases total $600+120=720\,\text{J}$ , equal to the decrease in the block's thermal store. All transfers are between parts of the whole-container system, so no net energy crosses its boundary.	Changing the system boundary changes whether a transfer is internal. With only the water inside the boundary, its $600\,\text{J}$ gain arrives by heating from the block outside that system. With the block, water, container and thermometer all inside the well-insulated boundary, the block's $720\,\text{J}$ thermal-store decrease is redistributed internally: $600\,\text{J}$ reaches the water and $120\,\text{J}$ reaches the container and thermometer. Since $600+120=720\,\text{J}$ , the whole-container energy total is unchanged.	5
Total Question 3			5
04.1	The increase in thermal stores during braking is $120\,\text{kJ}$ . During acceleration, energy is transferred electrically from the battery and mechanically to the van, increasing its kinetic store. During braking, energy is transferred mechanically and then electrically to the battery. Energy is dissipated to thermal stores because work is done against friction and resistive heating occurs. Over the complete journey, the battery's chemical store decreases by $640-310=330\,\text{kJ}$ and thermal stores increase by $210+120=330\,\text{kJ}$ , while the van's kinetic store has no net change.	Braking begins with a $430\,\text{kJ}$ decrease in the kinetic store. Of this, $310\,\text{kJ}$ increases the battery's chemical store, leaving $430-310=120\,\text{kJ}$ for thermal stores. Across both stages, the chemical-store change is $-640+310=-330\,\text{kJ}$ , the thermal-store change is $210+120=330\,\text{kJ}$ and the kinetic-store changes cancel. The transfers are electrical and mechanical during acceleration, then mechanical and electrical into the battery during braking. Energy is dissipated to thermal stores because work is done against friction and resistive heating occurs.	6
Total Question 4			6
05.1	$1.80\,\text{kJ}$ increases thermal stores while the battery is charged. Input arrow: $5.00\times20=100\,\text{mm}$ . Chemical-store arrow: $3.20\times20=64\,\text{mm}$ . Thermal-store arrow: $1.80\times20=36\,\text{mm}$ . The output widths account for the input because $64+36=100\,\text{mm}$ .	Conservation gives $5.00-3.20=1.80\,\text{kJ}$ transferred to thermal stores. At $20\,\text{mm}$ per kilojoule, the input arrow is $5.00\times20=100\,\text{mm}$ , the battery chemical-store arrow is $3.20\times20=64\,\text{mm}$ and the thermal-store arrow is $1.80\times20=36\,\text{mm}$ . The two output widths sum to the input width: $64+36=100\,\text{mm}$ .	6
Total Question 5			6

4.1.1.2 · Changes in energy

Tier 1 · Easy

Mark scheme for 4.1.1.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	$9.0\,\text{J}$	Use $E_k=\frac{1}{2}mv^2$ . Therefore $E_k=\frac{1}{2}\times2.0\times3.0^2=9.0\,\text{J}$ .	2
Total Question 1			2
02.1	The ratio of the slower trolley's kinetic energy to the faster trolley's is $1:4$ . The speed doubles and kinetic energy is proportional to speed squared for equal masses.	The masses are identical, so only the squared-speed factor changes. The speed ratio is $2.5:5.0=1:2$ , hence the kinetic-energy ratio is $1^2:2^2=1:4$ .	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.1.1.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	$1.4\times10^3\,\text{J}$	Use $E_p=mgh$ . Substitution gives $E_p=35\times9.8\times4.2=1440.6\,\text{J}$ , which to two significant figures is $1.4\times10^3\,\text{J}$ .	2
Total Question 1			2
02.1	At $1.5\,\text{m/s}$ : $0.90\,\text{J}$ . At $2.0\,\text{m/s}$ : $1.6\,\text{J}$ . At $2.5\,\text{m/s}$ : $2.5\,\text{J}$ . No: the increases are $0.70\,\text{J}$ and $0.90\,\text{J}$ , so they are not equal.	Choose $E_k=\frac{1}{2}mv^2$ . The three results are $0.5\times0.80\times1.5^2=0.90\,\text{J}$ , $0.5\times0.80\times2.0^2=1.6\,\text{J}$ and $0.5\times0.80\times2.5^2=2.5\,\text{J}$ . Their successive increases are $1.6-0.90=0.70\,\text{J}$ and $2.5-1.6=0.90\,\text{J}$ because speed is squared.	4
Total Question 2			4
03.1	$50.0\,\text{kg}$	Convert $7.35\,\text{kJ}=7350\,\text{J}$ . Rearrange $E_p=mgh$ to $m=E_p/(gh)$ . Therefore $m=7350/(9.8\times15.0)=50.0\,\text{kg}$ exactly.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.1.1.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	$4.0\,\text{m/s}$	The spring initially stores $E_e=\frac{1}{2}ke^2=\frac{1}{2}\times320\times0.15^2=3.6\,\text{J}$ . With no dissipation, $E_k=E_e$ , so $\frac{1}{2}\times0.45\times v^2=3.6$ . Hence $v^2=16$ and $v=4.0\,\text{m/s}$ .	4
Total Question 1			4
02.1	$800\,\text{N/m}$	Convert $180\,\text{g}=0.180\,\text{kg}$ and the compression $6.0\,\text{cm}=0.060\,\text{m}$ . The cart gains $E_k=\frac{1}{2}mv^2=0.5\times0.180\times4.0^2=1.44\,\text{J}$ . With a complete transfer, $\frac{1}{2}ke^2=1.44$ , so $k=2\times1.44/0.060^2=800\,\text{N/m}$ .	5
Total Question 2			5
03.1	Initial extension $=0.200\,\text{m}$ . New elastic potential energy $=2.50\,\text{J}$ .	Rearrange $E_e=\frac{1}{2}ke^2$ to $e=\sqrt{2E_e/k}$ . Thus $e=\sqrt{(2\times1.60)/80.0}=0.200\,\text{m}$ exactly. Increasing this by $25.0\%$ gives a new extension of $1.250\times0.200=0.250\,\text{m}$ . Since elastic energy is proportional to $e^2$ , the new energy is $1.60\times1.250^2=1.60\times1.5625=2.50\,\text{J}$ exactly.	5
Total Question 3			5
04.1	$3.46\,\text{m/s}$	The spring initially stores $E_e=\frac{1}{2}ke^2=0.5\times500\times0.180^2=8.10\,\text{J}$ . The gravitational store increases by $E_p=mgh=0.400\times9.8\times1.20=4.704\,\text{J}$ . After the $1.00\,\text{J}$ thermal transfer, the kinetic energy is $8.10-4.704-1.00=2.396\,\text{J}$ . Therefore $v=\sqrt{2E_k/m}=\sqrt{(2\times2.396)/0.400}=3.46\,\text{m/s}$ to three significant figures.	5
Total Question 4			5
05.1	$8.04\,\text{J}$	The initial kinetic energy is $0.5\times1.20\times3.00^2=5.40\,\text{J}$ . The motor adds $24.0\,\text{J}$ , giving $5.40+24.0=29.40\,\text{J}$ . The gravitational-store increase is $1.20\times9.8\times1.00=11.76\,\text{J}$ . The final kinetic energy is $0.5\times1.20\times4.00^2=9.60\,\text{J}$ . Therefore the energy transferred to thermal stores is $29.40-11.76-9.60=8.04\,\text{J}$ .	6
Total Question 5			6

4.1.1.3 · Energy changes in systems

Tier 1 · Easy

Mark scheme for 4.1.1.3 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	It takes $900\,\text{J}$ to raise the temperature of $1\,\text{kg}$ of the substance by $1\,{}^\circ\text{C}$ .	Interpret the unit one factor at a time: joules measure energy, 'per kilogram' fixes the mass at $1\,\text{kg}$ and 'per degree Celsius' fixes the temperature rise at $1\,{}^\circ\text{C}$ .	2
Total Question 1			2
02.1	$39.8\,{}^\circ\text{C}$ at $180\,\text{s}$ is anomalous. Repeat that measurement or the experiment; omit the value from the trend only if repeats show that it is anomalous.	The readings otherwise rise by about $4\,{}^\circ\text{C}$ each minute, so $39.8\,{}^\circ\text{C}$ breaks the pattern and even exceeds the later value. An unexpected point should be checked with a repeat rather than discarded automatically.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.1.1.3 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	$3.8\times10^4\,\text{J}$	Use $\Delta E=mc\Delta\theta$ . Therefore $\Delta E=1.5\times900\times28=37\,800\,\text{J}$ , which to two significant figures is $3.8\times10^4\,\text{J}$ .	2
Total Question 1			2
02.1	Wrap the block in thermal insulation, reducing energy transfer from the block to the surroundings. For the second change, either insert the heater fully so it is surrounded by the block, improving energy transfer to the block, or put a small amount of oil in the thermometer hole, improving thermal contact.	Repair the stated weaknesses directly. Insulation reduces the fraction of the heater's energy that warms the surroundings. The heater should be inserted fully so its heating element is surrounded by the block. A small amount of oil may be placed in the thermometer hole to fill air gaps, so the thermometer more closely follows the block's temperature.	4
Total Question 2			4
03.1	$900\,\text{J/(kg}\,{}^\circ\text{C)}$ The equation uses the rise in temperature, here $30.0-18.0=12.0\,{}^\circ\text{C}$ , not the final temperature.	Between the first and last readings, $\Delta E=4320\,\text{J}$ and $\Delta\theta=30.0-18.0=12.0\,{}^\circ\text{C}$ . Rearrange $\Delta E=mc\Delta\theta$ to $c=\Delta E/(m\Delta\theta)$ . Hence $c=4320/(0.400\times12.0)=900\,\text{J/(kg}\,{}^\circ\text{C)}$ .	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.1.1.3 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	$2.3\times10^3\,\text{J/(kg}\,{}^\circ\text{C)}$	Convert the time: $6.0\,\text{min}=360\,\text{s}$ . The heater transfers $E=Pt=75\times360=27\,000\,\text{J}$ , so the sample receives $0.80\times27\,000=21\,600\,\text{J}$ . Rearranging $\Delta E=mc\Delta\theta$ gives $c=21\,600/(0.30\times32)=2250\,\text{J/(kg}\,{}^\circ\text{C)}$ , or $2.3\times10^3\,\text{J/(kg}\,{}^\circ\text{C)}$ to two significant figures.	5
Total Question 1			5
02.1	$108\,\text{s}$ , or $1.80\,\text{minutes}$	Convert the mass: $540\,\text{g}=0.540\,\text{kg}$ . The block needs $\Delta E=mc\Delta\theta=0.540\times390\times28=5896.8\,\text{J}$ . This is $78\%$ of the heater input, so the unrounded input is $5896.8/0.78=7560\,\text{J}$ . Hence $t=E/P=7560/70=108\,\text{s}$ , and $108/60=1.80\,\text{minutes}$ .	5
Total Question 2			5
03.1	Corrected specific heat capacity $=1000\,\text{J/(kg}\,{}^\circ\text{C)}$ . Ignoring the unwanted transfer gives $1200\,\text{J/(kg}\,{}^\circ\text{C)}$ . The uncorrected value is too high because it treats energy transferred elsewhere as if it all increased the block's thermal store.	Convert the energies to joules. The block receives $10\,800-1800=9000\,\text{J}$ , so $c=9000/(0.600\times15.0)=1000\,\text{J/(kg}\,{}^\circ\text{C)}$ . Without the correction, $c=10\,800/(0.600\times15.0)=1200\,\text{J/(kg}\,{}^\circ\text{C)}$ . Including energy that did not heat the block makes the numerator too large and therefore overestimates $c$ .	5
Total Question 3			5
04.1	Final temperature $=27.3\,{}^\circ\text{C}$ . Energy transferred $=7.82\,\text{kJ}$ .	The copper cools by $95.0-T$ and the water warms by $T-18.0$ . Conservation in the insulated system gives $0.300\times385(95.0-T)=0.200\times4200(T-18.0)$ . Therefore $115.5(95.0-T)=840(T-18.0)$ , so $26\,092.5=955.5T$ and $T=27.3077\,{}^\circ\text{C}$ . The water gains $0.200\times4200\times(27.3077-18.0)=7818\,\text{J}=7.82\,\text{kJ}$ to three significant figures.	6
Total Question 4			6
05.1	$0.500\,\text{kg}$ block: $c=9000/(0.500\times20.0)=900\,\text{J/(kg }{}^\circ\text{C)}$ . $1.00\,\text{kg}$ block: $c=18\,000/(1.00\times20.0)=900\,\text{J/(kg }{}^\circ\text{C)}$ . $1.50\,\text{kg}$ block: $c=27\,000/(1.50\times20.0)=900\,\text{J/(kg }{}^\circ\text{C)}$ . The three equal values show that specific heat capacity is a property of aluminium and does not depend on the block's mass. Repeat the measurements and compare the values; differences could indicate random variation or unequal energy transfer to the surroundings.	Use $c=\Delta E/(m\Delta\theta)$ for each block. The calculations give $9000/(0.500\times20.0)=900$ , $18\,000/(1.00\times20.0)=900$ and $27\,000/(1.50\times20.0)=900\,\text{J/(kg }{}^\circ\text{C)}$ . Since all three values are equal, the results support that specific heat capacity is a property of the material, not its mass. Repeats would test consistency and help reveal variation caused by energy transferred to the surroundings.	6
Total Question 5			6

4.1.1.4 · Power

Tier 1 · Easy

Mark scheme for 4.1.1.4 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	$30\,\text{W}$	Use $P=E/t$ . Thus $P=600/20=30\,\text{W}$ .	2
Total Question 1			2
02.1	The pulse welder transfers energy at the greater rate: $3000\,\text{W}$ . One watt means one joule of energy transferred per second.	Read the power column rather than comparing the total energies: $3000\,\text{W}>2000\,\text{W}$ , so the welder transfers energy faster. The unit watt is equivalent to joules per second, so it describes a transfer rate.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.1.1.4 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Winch A: $600\,\text{W}$ Winch B: $400\,\text{W}$ A is more powerful because it does the same work in less time.	Convert $18\,\text{kJ}$ to $18\,000\,\text{J}$ . Then $P_A=18\,000/30=600\,\text{W}$ and $P_B=18\,000/45=400\,\text{W}$ . Since both do the same work, A's shorter time gives it the greater power.	4
Total Question 1			4
02.1	Appliance A transfers $19.5\,\text{kJ}$ . Appliance B transfers $21.6\,\text{kJ}$ . Appliance B transfers $2.1\,\text{kJ}$ more energy.	Rearrange $P=E/t$ to $E=Pt$ . Appliance A transfers $650\times30=19\,500\,\text{J}=19.5\,\text{kJ}$ and appliance B transfers $480\times45=21\,600\,\text{J}=21.6\,\text{kJ}$ . The difference is $21.6-19.5=2.1\,\text{kJ}$ , so B transfers more.	3
Total Question 2			3
03.1	$0$ to $40\,\text{s}$ : $600\,\text{W}$ . $40$ to $100\,\text{s}$ : $650\,\text{W}$ . $100$ to $160\,\text{s}$ : $500\,\text{W}$ . The greatest mean power occurs from $40$ to $100\,\text{s}$ .	Use the change in the cumulative reading for each interval. From $0$ to $40\,\text{s}$ , $36-12=24\,\text{kJ}=24\,000\,\text{J}$ , so $P=24\,000/40=600\,\text{W}$ . From $40$ to $100\,\text{s}$ , $75-36=39\,\text{kJ}=39\,000\,\text{J}$ and $100-40=60\,\text{s}$ , so $P=39\,000/60=650\,\text{W}$ . From $100$ to $160\,\text{s}$ , $105-75=30\,\text{kJ}=30\,000\,\text{J}$ and $160-100=60\,\text{s}$ , so $P=30\,000/60=500\,\text{W}$ . The middle interval has the greatest value.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.1.1.4 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	First hoist: $1.2\times10^3\,\text{W}$ Second hoist: $1.6\times10^3\,\text{W}$	The useful work is the gain in gravitational potential energy: $W=mgh=250\times9.8\times12=29\,400\,\text{J}$ . The first power is $29\,400/25=1176\,\text{W}=1.2\times10^3\,\text{W}$ to two significant figures. The second power is $29\,400/18=1633\,\text{W}=1.6\times10^3\,\text{W}$ to two significant figures.	5
Total Question 1			5
02.1	$11.25\,\text{minutes}$ . It exceeds the limit by $0.75\,\text{minutes}$ .	Convert $2.40\,\text{kW}=2400\,\text{W}$ and $1.62\,\text{MJ}=1\,620\,000\,\text{J}$ . Rearrange $P=E/t$ to $t=E/P$ : $t=1\,620\,000/2400=675\,\text{s}$ . Then $675/60=11.25\,\text{minutes}$ . The excess is $11.25-10.5=0.75\,\text{minutes}$ .	5
Total Question 2			5
03.1	Total work done $=360\,\text{kJ}$ . Mean power over the complete observation $=750\,\text{W}$ .	The press completes $24\times6.0=144$ cycles, so it does $144\times2.50=360\,\text{kJ}=360\,000\,\text{J}$ of work. The full elapsed time is $6.0\times60+120=480\,\text{s}$ . Therefore the mean power is $P=W/t=360\,000/480=750\,\text{W}$ .	5
Total Question 3			5
04.1	Mean total power $=2.65\,\text{kW}$ . Greatest instantaneous total power $=4.05\,\text{kW}$ . The circuit is not suitable: while the compressor runs, the limit is exceeded by $1.05\,\text{kW}$ even though the one-minute mean is below the limit.	In one minute, the cutter transfers $1.20\times60=72.0\,\text{kJ}$ , the extractor transfers $0.85\times60=51.0\,\text{kJ}$ and the compressor transfers $2.00\times18=36.0\,\text{kJ}$ . The total is $159\,\text{kJ}$ , so the mean power is $159\,\text{kJ}/60\,\text{s}=2.65\,\text{kW}$ . When all three operate, their powers add to $1.20+0.85+2.00=4.05\,\text{kW}$ , which is $4.05-3.00=1.05\,\text{kW}$ above the limit.	5
Total Question 4			5
05.1	The energy transferred is $55.6-12.4=43.2\,\text{kJ}=43\,200\,\text{J}$ . Mean power $=43\,200/60.0=720\,\text{W}$ . The measured mean power agrees exactly with the $720\,\text{W}$ label for this run. Keep the dough mass constant. Keep hands and loose clothing away from moving parts and switch off before changing the dough or attachments.	The joulemeter change is $55.6-12.4=43.2\,\text{kJ}=43\,200\,\text{J}$ . Therefore $P=\Delta E/t=43\,200/60.0=720\,\text{W}$ , exactly matching the label for this run. Keeping the dough mass fixed makes the test comparable, and the mixer must be switched off before any adjustment because its moving parts are hazardous.	5
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4.1.2.1 · Energy transfers in a system

Tier 1 · Easy

Mark scheme for 4.1.2.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	The oil reduces friction at the axle. Less energy is transferred mechanically to thermal energy stores of the axle and surroundings.	Link the change to the transfer pathway: lubrication lowers the frictional force, so less work is done against friction and less energy is dissipated to thermal stores.	2
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02.1	Panel A transfers energy more slowly by conduction. Its thermal conductivity is lower: $0.040<0.20\,\text{W/(m}\,{}^\circ\text{C)}$ .	The panels have equal thickness, so compare their thermal conductivities directly. Panel A has the smaller value, so energy is transferred through it at the lower rate.	2
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Tier 2 · Standard

Mark scheme for 4.1.2.1 Tier 2 · Standard
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01.1	Wall X gives the lower rate because it is thicker. A lower thermal conductivity reduces the rate of energy transfer further.	Hold the material and temperature difference constant first: increasing thickness increases the distance through which conduction occurs, reducing the transfer rate. Holding thickness constant next, a lower thermal conductivity means the material conducts energy more slowly.	3
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02.1	Use identical beakers made from the same material. Use equal volumes or masses of water at the same starting temperature. Use the same insulation thickness, with equal coverage. Repeat each test under the same room conditions and calculate mean temperature drops over the same time.	Change only the insulating material. Standardise the container, water mass, initial temperature, insulation thickness and measurement time. Repeats and means reduce the effect of random variation and make the comparison repeatable.	4
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03.1	The box and its contents form a closed system, so no net energy is transferred across the boundary. The battery's chemical store and the fan's kinetic store decrease as energy is transferred to thermal stores inside the box. The energy is conserved but dissipated, so it is spread out in a less useful thermal store.	Choose the insulated box and all its contents as the system. Energy is redistributed from the battery's chemical store, through the fan's kinetic store, into thermal stores by electrical and mechanical transfers. Because none leaves the closed system, its total is unchanged; the final thermal energy is simply less useful for driving the fan.	3
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Tier 3 · Hard

Mark scheme for 4.1.2.1 Tier 3 · Hard
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01.1	Use identical beakers containing equal masses of water at the same initial temperature. Wrap each beaker with the same thickness or number of layers of one fabric, with one unwrapped beaker if a control is wanted. Measure temperature at fixed time intervals for the same total time. Control beaker shape, water mass, starting temperature, fabric area and room conditions. Repeat and calculate mean temperature drops or cooling rates. The smallest mean temperature drop or lowest cooling rate identifies the best insulator.	Set up identical beakers with equal water masses and equal starting temperatures. Change only the fabric type, keeping fabric thickness and coverage fixed. Record each temperature at regular intervals with the same type of thermometer, repeat the trials and calculate a mean temperature drop over a fixed time. Compare the mean drops: the fabric giving the smallest drop has reduced the unwanted energy transfer most effectively.	6
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02.1	The $5\,\text{mm}$ mean drop is $18.1\,{}^\circ\text{C}$ . The $10\,\text{mm}$ mean drop is $13.6\,{}^\circ\text{C}$ . Omitting the stated anomaly, the $15\,\text{mm}$ mean drop is $11.9\,{}^\circ\text{C}$ . Choose $10\,\text{mm}$ insulation: it meets the $14.0\,{}^\circ\text{C}$ limit and costs less than the $15\,\text{mm}$ option.	Calculate the row means: $(18.2+18.0+18.1)/3=18.1\,{}^\circ\text{C}$ and $(13.5+13.7+13.6)/3=13.6\,{}^\circ\text{C}$ . The prompt identifies $18.9\,{}^\circ\text{C}$ as anomalous, so the final mean is $(11.8+12.0)/2=11.9\,{}^\circ\text{C}$ . Both $10\,\text{mm}$ and $15\,\text{mm}$ meet the limit, but £0.39 is less than £0.62; the $10\,\text{mm}$ option is therefore the least expensive valid choice.	5
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03.1	Panel Q is the best thermal insulator. At equal area, thickness, time and temperature difference, it needs the least replacement energy, so it has the lowest rate of energy transfer. Percentage reduction $=(11.0-6.60)/11.0\times100=40.0\%$ . Increasing the thickness reduces the rate of energy transfer by conduction.	Because the inner temperatures are held constant, the heater energy replaces the energy transferred through each panel. Q needs only $11.0\,\text{kJ}$ in the same time, so its transfer rate is the lowest. Doubling Q's thickness reduces the energy by $11.0-6.60=4.40\,\text{kJ}$ , giving $4.40/11.0\times100=40.0\%$ . A thicker panel increases the distance over which conduction occurs, reducing the transfer rate.	5
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04.1	Dry axle: $P=18.0/12.0=1.50\,\text{W}$ . Oil: $P=18.0/30.0=0.600\,\text{W}$ . Grease: $P=18.0/24.0=0.750\,\text{W}$ . Oil is the best lubricant because it gives the lowest mean power transfer. With oil, the least work is done against friction each second, so energy is transferred to thermal stores at the lowest rate.	The same $18.0\,\text{J}$ is transferred in each case, so use $P=E/t$ . The mean powers are $18.0/12.0=1.50\,\text{W}$ for the dry axle, $18.0/30.0=0.600\,\text{W}$ for oil and $18.0/24.0=0.750\,\text{W}$ for grease. Oil gives the lowest rate of energy transfer, so the least work is done against friction each second.	6
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05.1	First interval: $56.0\,\text{W}$ . Later interval: $21.0\,\text{W}$ . The later transfer rate is lower because the temperature difference between the water and its surroundings is smaller. The data do not show that the insulation changed; the falling transfer rate is expected with the same insulation as the water cools.	For the first interval, $\Delta E=mc\Delta\theta=0.500\times4200\times8.0=16\,800\,\text{J}$ and $t=300\,\text{s}$ , so $P=16\,800/300=56.0\,\text{W}$ . Later, $\Delta E=0.500\times4200\times3.0=6300\,\text{J}$ , giving $P=6300/300=21.0\,\text{W}$ . The insulation is unchanged; the smaller water-to-surroundings temperature difference gives a lower transfer rate.	5
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4.1.2.2 · Efficiency

Tier 1 · Easy

Mark scheme for 4.1.2.2 Tier 1 · Easy
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01.1	$0.80$ $80\%$	Efficiency is useful output divided by total input: $72/90=0.80$ . Multiplying by $100$ gives $80\%$ .	2
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02.1	Efficiency $=70\%$ Dissipated width $=9$ divisions	The useful fraction is $21/30=0.70=70\%$ . Conservation fixes the remaining width at $30-21=9$ divisions. The student's $30\%$ is the dissipated fraction, not the efficiency.	2
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Tier 2 · Standard

Mark scheme for 4.1.2.2 Tier 2 · Standard
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01.1	Efficiency $=75\%$ Wasted power $=140\,\text{W}$	Efficiency $=420/560=0.75=75\%$ . Conservation of energy per second gives wasted power $=560-420=140\,\text{W}$ .	3
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02.1	Total input power $=2400\,\text{W}$ Dissipated power $=720\,\text{W}$	Convert the useful power: $1.68\,\text{kW}=1680\,\text{W}$ . Since $0.70=1680/P_{\text{input}}$ , $P_{\text{input}}=1680/0.70=2400\,\text{W}$ . The dissipated power is $2400-1680=720\,\text{W}$ .	4
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03.1	Useful energy transfer $=706\,\text{J}$ to $3$ significant figures. Efficiency $=73.5\%$ to $3$ significant figures.	The useful output is the load's increase in gravitational potential energy: $E_p=mgh=18.0\times9.8\times4.00=705.6\,\text{J}$ , which is $706\,\text{J}$ to three significant figures. Using the unrounded value, efficiency $=705.6/960=0.735$ , which is $73.5\%$ to three significant figures.	4
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Tier 3 · Hard

Mark scheme for 4.1.2.2 Tier 3 · Hard
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01.1	Overall efficiency $=52\%$ $130\,\text{kJ}$ For example, lubricating the bearings reduces energy transferred by friction to thermal stores, so a greater fraction reaches the useful electrical output.	Successive efficiencies multiply: $0.80\times0.65=0.52$ , so the overall efficiency is $52\%$ . The useful output is $0.52\times250=130\,\text{kJ}$ . To improve the intended transfer, reduce a named unwanted pathway: lubricating bearings reduces work done against friction and therefore reduces dissipation to thermal stores, increasing the useful fraction.	6
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02.1	Dry-bearing efficiency $=60\%$ . Lubricated-bearing efficiency $=75\%$ . The claim is incorrect: the efficiency increased by $15$ percentage points. For example, fit low-friction ball bearings. This reduces friction, so less power is dissipated to thermal stores and a greater fraction is transferred usefully.	Use useful output divided by total input for each design: $0.90/1.50=0.60=60\%$ and $1.05/1.40=0.75=75\%$ . The input falling does not by itself show lower efficiency; the useful fraction has risen by $75-60=15$ percentage points. A further valid change must reduce an unwanted pathway: low-friction ball bearings reduce work done against friction and therefore reduce dissipation to thermal stores.	5
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03.1	Total input energy $=360\,\text{kJ}$ . Useful output energy $=230.4\,\text{kJ}$ . Dissipated energy $=129.6\,\text{kJ}$ . The target is not met; the shortfall is $9.6\,\text{kJ}$ .	Convert $1.50\,\text{kW}=1500\,\text{W}$ and $4.00\,\text{min}=240\,\text{s}$ . The input energy is $Pt=1500\times240=360\,000\,\text{J}=360\,\text{kJ}$ . The useful output is $0.640\times360=230.4\,\text{kJ}$ and the dissipated energy is $360-230.4=129.6\,\text{kJ}$ . The useful output misses the target by $240-230.4=9.6\,\text{kJ}$ .	5
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04.1	Total useful output $=1.16\,\text{kJ}$ . Efficiency $=72.5\%$ . Energy dissipated $=440\,\text{J}$ .	The gravitational-store increase is $mgh=40.0\times9.8\times2.50=980\,\text{J}$ . The kinetic-store increase is $0.5mv^2=0.5\times40.0\times3.00^2=180\,\text{J}$ . Both are useful, so the useful total is $980+180=1160\,\text{J}=1.16\,\text{kJ}$ . Efficiency is $1160/1600=0.725=72.5\%$ , and the dissipated energy is $1600-1160=440\,\text{J}$ .	5
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05.1	The intended efficiency is $13.5/18.0=0.750=75.0\%$ . The exhaust and outside surroundings receive $18.0-13.5-2.00=2.50\,\text{MJ}$ . The total unwanted transfer is $2.00+2.50=4.50\,\text{MJ}$ . The reported value counts an unintended room transfer as useful; useful output is defined by the intended transfer to the water, so $86.1\%$ is not the boiler's intended efficiency.	Only the $13.5\,\text{MJ}$ delivered to the water is the intended useful output, giving $13.5/18.0=0.750$ . The remaining input is unwanted: $2.00\,\text{MJ}$ reaches the room and $18.0-13.5-2.00=2.50\,\text{MJ}$ reaches the exhaust and outside surroundings. These sum to $4.50\,\text{MJ}$ and close the energy account because $13.5+4.50=18.0\,\text{MJ}$ .	5
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4.1.3 · National and global energy resources

Tier 1 · Easy

Mark scheme for 4.1.3 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Renewable: wind and geothermal. Non-renewable: natural gas and nuclear fuel.	Apply the definition: wind and geothermal resources are replenished as they are used, whereas natural gas and nuclear fuel come from finite reserves.	2
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02.1	Nuclear power is non-renewable because nuclear fuel comes from finite reserves. A station has low carbon dioxide emissions during operation, but it produces radioactive waste.	Separate resource classification from operational emissions. Low carbon dioxide release while generating does not make a fuel renewable; the fuel is finite. A valid environmental comparison can credit low operational carbon emissions while recognising radioactive waste.	2
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Tier 2 · Standard

Mark scheme for 4.1.3 Tier 2 · Standard
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01.1	Biofuel's share rises by $7$ percentage points. Oil's share falls by $13$ percentage points. Biofuel is renewable and regrowth can absorb carbon dioxide, whereas burning oil adds carbon dioxide from a finite fuel. Oil may remain in use because it is reliable, easy to store and transport, or supported by existing heating infrastructure.	Read each change in percentage points: $11-4=7$ for biofuel and $25-38=-13$ for oil. Link the direction of change to an environmental driver, such as replacing a non-renewable fossil fuel with a replenishable resource whose crop regrowth absorbs carbon dioxide. Then give a distinct practical constraint: oil can be stored and supplied on demand and existing boilers and distribution systems already use it.	4
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02.1	Solar range $=62\,\text{MW}$ ; wind range $=62\,\text{MW}$ ; natural-gas range $=4\,\text{MW}$ . Natural gas was the most reliable during these measurements because its output varied least and stayed high. Natural gas is non-renewable and releases carbon dioxide when burned.	Subtract minimum from maximum: solar $62-0=62\,\text{MW}$ , wind $70-8=62\,\text{MW}$ and natural gas $52-48=4\,\text{MW}$ . The gas station's small range and high minimum support the limited conclusion that it was most reliable in this dataset. Burning gas uses a finite fuel and emits carbon dioxide.	4
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03.1	Wind can be replenished as it is used, but its variable output makes it less reliable without storage or backup. Gas generation can respond to demand, but it uses a non-renewable fuel and releases carbon dioxide. Science can identify effects such as reliability and emissions. The final choice also depends on economic and social priorities, such as cost, employment and acceptance of landscape changes, which science cannot decide by itself.	Separate evidence from value judgements. Scientific evidence supports conclusions about renewability, variation in output, controllability and carbon dioxide emissions. Choosing how to balance those effects against cost, jobs and local preferences requires political, social and economic decisions as well as science.	4
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Tier 3 · Hard

Mark scheme for 4.1.3 Tier 3 · Hard
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01.1	Offshore wind is renewable and produces no carbon dioxide while operating, but its output is intermittent and needs backup, storage or a wide network. Wind farms can be built in stages, but they affect seascapes, habitats and navigation and require many turbines for a large output. Nuclear fuel is non-renewable, but a station can provide a large, reliable output with low operational carbon dioxide emissions. Nuclear power creates radioactive waste, has accident and decommissioning concerns, and has high construction cost and long build time. A justified decision depends on the value placed on dependable output, cost, local effects and long-term waste management.	Build a balanced comparison around the stated criteria. Credit wind for renewability and low operational emissions but recognise variable output and local impacts. Credit nuclear for dependable high output and low operational emissions but recognise finite fuel, radioactive waste, cost and public concern. Finish with a conditional judgement: nuclear better meets continuous output, while wind avoids fuel use and long-lived waste if variability can be managed.	6
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02.1	Choose nuclear power. It passes the constraints by margins of £3 million on cost, $8\,\text{MW}$ on firm output and $8\,\text{g/kWh}$ on lifecycle emissions. Solar fails firm output by $50\,\text{MW}$ and emissions by $28\,\text{g/kWh}$ ; natural gas fails output by $2\,\text{MW}$ and emissions by $400\,\text{g/kWh}$ ; tidal fails output by $16\,\text{MW}$ . A drawback not shown is radioactive waste, finite fuel, high decommissioning cost, accident concern or a long construction time.	Apply every constraint to each row. Nuclear is the unique satisfier: $125-122=3$ million pounds of cost headroom, $58-50=8\,\text{MW}$ of firm-output headroom and $20-12=8\,\text{g/kWh}$ of emissions headroom. Solar misses output by $50-0=50\,\text{MW}$ and emissions by $48-20=28\,\text{g/kWh}$ ; gas misses them by $50-48=2\,\text{MW}$ and $420-20=400\,\text{g/kWh}$ ; tidal misses output by $50-34=16\,\text{MW}$ . A balanced decision must also recognise an omitted disadvantage such as radioactive-waste management or long build and decommissioning times.	5
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03.1	Use biofuel for transport because a fuel can be stored and carried in vehicles. Biofuel is renewable if crops are replaced, although growing it uses land and burning it releases pollutants and carbon dioxide. Use geothermal energy for heating because the local source can provide a dependable supply without relying on weather. Geothermal energy is renewable, but suitable sites are geographically limited. Use wind for electricity generation because it is renewable and releases no fuel emissions while operating. Wind output varies, so storage, backup generation or a wider network would be needed for reliable supply.	Match each resource to the stated use and then qualify the choice. A portable biofuel suits transport, dependable local geothermal energy suits heating, and coastal wind can generate electricity without consuming fuel. A complete plan also recognises constraints: land and combustion impacts for biofuel, site availability for geothermal energy, and variable output for wind.	6
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04.1	Total heating demand $=82.8\,\text{MJ}$ . Geothermal energy supplied $=57.6\,\text{MJ}$ . Biofuel energy required $=82.8-57.6=25.2\,\text{MJ}$ , corresponding to a mean power of $14.0\,\text{kW}$ . The $18\,\text{kW}$ boiler is sufficient, with $4.0\,\text{kW}$ spare power. Biofuel can be replenished if crops are replaced, but growing it uses land and combustion releases carbon dioxide and other pollutants.	Convert $30.0\,\text{minutes}$ to $1800\,\text{s}$ . The heating demand is $46\,000\times1800=82.8\,\text{MJ}$ . Geothermal energy contributes $32\,000\times1800=57.6\,\text{MJ}$ , leaving $82.8-57.6=25.2\,\text{MJ}$ . This requires $25.2\times10^6/1800=14.0\,\text{kW}$ from biofuel. Since $18-14=4\,\text{kW}$ , the boiler can cover the shortfall. The calculation does not remove land-use and combustion impacts from the resource choice.	5
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05.1	Biofuel annual emissions $=4760\,\text{kg}$ ; battery annual emissions $=3220\,\text{kg}$ of carbon-dioxide equivalent. Biofuel annual operating cost $=£11\,760$ ; battery annual operating cost $=£8680$ . The battery option saves $1540\,\text{kg}$ of carbon-dioxide equivalent and £3080 per bus each year, so the supplied data support that option. A relevant omitted limitation is the cost and environmental impact of batteries and charging infrastructure, or that emissions change if the electricity-generation mix changes. Biofuel also uses agricultural land and produces exhaust pollutants when burned.	Each bus completes $28\,000/100=280$ blocks of $100\,\text{km}$ . Emissions are therefore $280\times17.0=4760\,\text{kg}$ for biofuel and $280\times11.5=3220\,\text{kg}$ for battery power, a difference of $1540\,\text{kg}$ . Costs are $28\,000\times0.42=£11\,760$ and $28\,000\times0.31=£8680$ , a difference of £3080. Both numerical criteria favour battery buses, but a complete evaluation must also consider an omitted environmental, infrastructure or resource constraint.	6
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