4.1 Energy — printable coverage pack

4.1.1.1 · Energy stores and systems

A system is one object or a group of objects; when it changes, energy is transferred between stores within the system or between the system and its surroundings.
Describe a change by naming the store that decreases, the store that increases and the transfer pathway: mechanically, electrically, by heating or by radiation.
For a complete redistribution, the increases in all stores equal the decrease in the original store because total energy is conserved.
A common error is to say that an object 'contains energy' or that energy is used up; name a store and track where the energy is transferred instead.

Tier 1 · Easy

1. 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]
Answer
- The toy's kinetic energy store decreases.
- Energy is transferred to thermal energy stores of the toy and its surroundings.
Method: 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.

Tier 2 · Standard

1. 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]
Answer
- $8\,\text{kJ}$
- It is mainly in thermal energy stores of the hotplate and surroundings.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- $36\,\text{J}$ 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.
Method: 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.

4.1.1.2 · Changes in energy

Use $E_k=\frac{1}{2}mv^2$ for kinetic energy, $E_e=\frac{1}{2}ke^2$ for elastic potential energy below the limit of proportionality, and $E_p=mgh$ for gravitational potential energy.
Convert mass to kilograms, speed to metres per second, extension to metres and height to metres before substituting; use the given value of $g$ .
If energy is transferred between stores without dissipation, equate the decrease in one store to the increase in the other and then solve for the unknown.
A common error is to forget that speed and extension are squared, or to use the spring's total length instead of its extension.

Tier 1 · Easy

1. Calculate the kinetic energy of a $2.0\,\text{kg}$ cart moving at $3.0\,\text{m/s}$ .[2 marks]
Answer
- $9.0\,\text{J}$
Method: Use $E_k=\frac{1}{2}mv^2$ . Therefore $E_k=\frac{1}{2}\times2.0\times3.0^2=9.0\,\text{J}$ .

Tier 2 · Standard

1. 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]
Answer
- $1.4\times10^3\,\text{J}$
Method: Use $E_p=mgh$ . Substitution gives $E_p=35\times9.8\times4.2=1440.6\,\text{J}$ , which to two significant figures is $1.4\times10^3\,\text{J}$ .

Tier 3 · Hard

1. 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]
Answer
- $4.0\,\text{m/s}$
Method: 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 $v=4.0\,\text{m/s}$ .

4.1.1.3 · Energy changes in systems

A temperature change transfers energy to or from a system's thermal energy store according to $\Delta E=mc\Delta\theta$ .
For a specific heat capacity investigation, measure the mass, supply a known energy using $E=Pt$ , record the temperature change and calculate $c=E/(m\Delta\theta)$ .
Specific heat capacity is the energy needed to raise the temperature of $1\,\text{kg}$ of a substance by $1\,{}^\circ\text{C}$ .
A common error is to substitute the final temperature for $\Delta\theta$ ; subtract the initial temperature and account for energy transferred to the surroundings.

Tier 1 · Easy

1. State what a specific heat capacity of $900\,\text{J/(kg}\,{}^\circ\text{C)}$ means.[2 marks]
Answer
- It takes $900\,\text{J}$ to raise the temperature of $1\,\text{kg}$ of the substance by $1\,{}^\circ\text{C}$ .
Method: 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 $1\,{}^\circ\text{C}$ .

Tier 2 · Standard

1. 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]
Answer
- $3.8\times10^4\,\text{J}$
Method: Use $\Delta E=mc\Delta\theta$ . Therefore $\Delta E=1.5\times900\times28=37\,800\,\text{J}$ , which to two significant figures is $3.8\times10^4\,\text{J}$ .

Tier 3 · Hard

1. 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]
Answer
- $2.3\times10^3\,\text{J/(kg}\,{}^\circ\text{C)}$
Method: 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.

4.1.1.4 · Power

Power is the rate of energy transfer or the rate of doing work: $P=E/t$ and $P=W/t$ .
Use joules and seconds to obtain watts, and rearrange before substituting when energy, work or time is unknown.
Two devices can transfer the same energy but have different powers; the device taking less time has the greater power.
A common error is to treat power as an amount of energy; $1\,\text{W}$ means an energy transfer of $1\,\text{J}$ every second.

Tier 1 · Easy

1. A small motor transfers $600\,\text{J}$ of energy in $20\,\text{s}$ . Calculate its power.[2 marks]
Answer
- $30\,\text{W}$
Method: Use $P=E/t$ . Thus $P=600/20=30\,\text{W}$ .

Tier 2 · Standard

1. 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]
Answer
- Winch A: $600\,\text{W}$
- Winch B: $400\,\text{W}$
- A is more powerful because it does the same work in less time.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- First hoist: $1.2\times10^3\,\text{W}$
- Second hoist: $1.6\times10^3\,\text{W}$
Method: 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.

4.1.2.1 · Energy transfers in a system

Energy can be transferred usefully, stored or dissipated, but it cannot be created or destroyed; a closed system has no net change in total energy.
Reduce unwanted mechanical transfers with lubrication, and reduce transfers by heating with insulation such as thicker walls or trapped air layers.
A material with greater thermal conductivity transfers energy by conduction at a greater rate; thicker walls reduce the rate of transfer through a given material.
A common error is to call dissipated energy 'lost'; it remains in thermal stores but is spread through the surroundings and is less useful.

Tier 1 · Easy

1. Explain why adding oil to the axle of a turning wheel reduces unwanted energy transfers.[2 marks]
Answer
- The oil reduces friction at the axle.
- Less energy is transferred mechanically to thermal energy stores of the axle and surroundings.
Method: 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.

Tier 2 · Standard

1. 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]
Answer
- Wall X gives the lower rate because it is thicker.
- A lower thermal conductivity reduces the rate of energy transfer further.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- 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.
Method: 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.

4.1.2.2 · Efficiency

Efficiency is $\text{useful output energy transfer}/\text{total input energy transfer}$ or $\text{useful power output}/\text{total power input}$ .
Calculate the ratio using matching quantities, then multiply by $100$ only when a percentage efficiency is required.
An efficiency of $0.72$ means $72\%$ of the input is transferred usefully and $28\%$ is dissipated; at Higher Tier, improve efficiency by reducing unwanted transfers, for example with lubrication or thermal insulation.
A common error is to divide total input by useful output, producing a value greater than $1$ or $100\%$ for an ordinary device.

Tier 1 · Easy

1. A device receives $90\,\text{J}$ and transfers $72\,\text{J}$ usefully. Calculate its efficiency as a decimal and as a percentage.[2 marks]
Answer
- $0.80$
- $80\%$
Method: Efficiency is useful output divided by total input: $72/90=0.80$ . Multiplying by $100$ gives $80\%$ .

Tier 2 · Standard

1. 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]
Answer
- Efficiency $=75\%$
- Wasted power $=140\,\text{W}$
Method: Efficiency $=420/560=0.75=75\%$ . Conservation of energy per second gives wasted power $=560-420=140\,\text{W}$ .

Tier 3 · Hard

1. 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]
Answer
- 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.
Method: 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.

4.1.3 · National and global energy resources

Renewable resources include biofuel, wind, hydroelectricity, geothermal, tides, sunlight and water waves; fossil fuels and nuclear fuel are non-renewable.
Compare resources for their uses in transport, heating and electricity generation, considering reliability, response to demand and environmental effects.
Trends in resource use can reflect cost, technology, availability, environmental targets and political, social or ethical choices, not science alone.
A common error is to call a resource renewable because it produces little carbon dioxide while operating; renewability depends on replenishment as it is used.

Tier 1 · Easy

1. Classify wind, natural gas, geothermal and nuclear fuel as renewable or non-renewable energy resources.[2 marks]
Answer
- Renewable: wind and geothermal.
- Non-renewable: natural gas and nuclear fuel.
Method: Apply the definition: wind and geothermal resources are replenished as they are used, whereas natural gas and nuclear fuel come from finite reserves.

Tier 2 · Standard

1. 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]
Answer
- 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.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- 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.
Method: 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.