AQA GCSE Physics Paper 1, Higher TierPaper 1 — every question, answered
We read the actual downloaded question papers and mark schemes for every AQA Physics Paper 1 Higher Tier sitting we have available, including the 2020 paper, which prints Wednesday 20 May 2020 as its exam date on the cover even though its mark scheme is titled June 2020, a genuine artefact of the pandemic disruption to that year's exams. Below is what each recurring question type has actually asked, what the real circuits, graphs and data showed, and a complete worked answer for the higher tariff questions. This is the closest you can get to seeing exactly what a full mark answer looks like without a real exam paper in front of you.
Questions © AQA, quoted for analysis. Diagrams, circuits and graphs described in our own words, not reproduced. Mark scheme content translated into plain English, not copied. PrepWise is independent and not endorsed by AQA.
The paper is a template
Same 5 question types, every sitting“Describe a method to determine [density / the I-V characteristics of a component]”
Three of the four sittings we analysed ask for a full written method worth 6 marks, marked on a three level scheme (5 to 6, 3 to 4, 1 to 2). The apparatus and the exact required practical change each year, but the marking always rewards the same thing: a method that would actually work, with the key steps identified and put in a logical order, not just a list of correct-sounding equipment.
1×asked“Describe a method the student could use to obtain these results.”
“Describe a method the student could use to obtain these results.”
What it’s really asking
The question follows a graph of current against potential difference for a filament lamp and asks for the full circuit and method a student could use to generate that data, including how to get the negative values shown on the graph.
What the sources actually showed — June 2023
A graph showing current in amps plotted against potential difference in volts for a filament lamp, with a curved line passing through the origin and flattening out at higher voltages in both the positive and negative direction, referenced as Q06.1 in the real paper.
The full worked answer — June 2023
Set up a circuit with the filament lamp connected in series with an ammeter, a variable resistor and a power supply, and connect a voltmeter in parallel with the filament lamp. Close the switch and use the variable resistor to set the potential difference across the lamp to a chosen value, then record the current shown on the ammeter and the potential difference shown on the voltmeter.
Repeat this measurement using the variable resistor to change the potential difference in steps of about 1 volt, from 0 volts up to 6 volts, recording the current at each value. To obtain the negative values needed for the graph, reverse the connections to the power supply so that the current flows through the lamp in the opposite direction, then repeat the same range of readings with the reversed connections.
Take repeat readings of the current at each value of potential difference and calculate a mean, discarding any anomalous results, to improve the reliability of the data before plotting the final graph of current against potential difference.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise IV characteristics and density questionsWhat the sources actually showed — June 2022
A photograph of a small rock found on a beach that the student wants to identify by calculating its density, referenced as Q02.1 in the real paper.
The full worked answer — June 2022
Measure the mass of the rock using a balance, making sure the balance reads zero before the rock is placed on it. To find the volume of the rock, partly fill a measuring cylinder with a known volume of water and record this initial volume.
Lower the rock into the measuring cylinder using a thread so that it is fully submerged, and record the new, higher volume reading. The volume of the rock is equal to the final volume reading minus the initial volume reading, since the rock displaces its own volume of water.
Use the mass and the calculated volume in the equation density equals mass divided by volume to calculate the density of the rock, and compare this value with a table of known densities for different rock types to identify it.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise IV characteristics and density questionsWhat the sources actually showed — June 2021
A circuit diagram showing a variable resistor, resistor R, an ammeter in series and a voltmeter connected in parallel with resistor R, alongside a battery and switch, referenced as Q03.1 in the real paper.
The full worked answer — June 2021
Close the switch and use the variable resistor to set a chosen current in the circuit. Record the current shown on the ammeter, which is connected in series, and the potential difference across resistor R, which is shown on the voltmeter connected in parallel with R.
Repeat this for a range of different current values by adjusting the variable resistor, keeping the current low enough throughout that the temperature of resistor R does not increase significantly, since a change in temperature would affect the resistance being measured. To obtain negative current and potential difference values, reverse the connections to the power supply and repeat the same range of readings.
Switch the circuit off between readings to avoid unnecessary heating of resistor R, and take repeat readings at each value of current, discarding any anomalies, before plotting a graph of current against potential difference.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise IV characteristics and density questionsWhat the mark scheme rewarded
- Naming the ammeter, voltmeter, balance or measuring cylinder in the correct position or role
- Describing how to get a full range of readings by varying the independent variable
- Stating a specific method to obtain negative values, where a graph shows negative data, such as reversing the power supply connections
- Naming a reliability step such as repeat readings, calculating a mean, discarding anomalies, or switching off between readings to avoid unwanted heating
- Ammeters are always connected in series with the component being measured, voltmeters always in parallel
- Reversing the connections to a power supply reverses the direction of current, giving negative values on a current-potential difference graph
- Repeating readings and calculating a mean reduces the effect of random error
- Displacement of water in a measuring cylinder gives the volume of an irregular solid, since the solid pushes aside its own volume of water
- Density is calculated using density equals mass divided by volume, with mass measured on a balance
Traps examiners saw
- Naming the right apparatus but never explaining how it is actually used to take the measurement
- Forgetting the specific instruction in the question, such as how to obtain negative values, and describing only a generic method
- Writing the steps in a jumbled order rather than the sequence a person would really carry them out in
- Missing the reliability step entirely, since a method with no repeats or anomaly check cannot reach the top level
Full-mark self-check 0 of 4
The method for every 6-mark practical method — same every sittingMark bands, steps, timing
What this question type rewards
The topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
- Naming the correct apparatus for the measurement being taken
- Describing HOW each measurement is taken, not just naming the instrument
- Sequencing the steps so a stranger could follow and repeat the method
- Including a way to get a repeatable, reliable result, such as taking repeat readings or reversing connections to get a full range of values
The steps
- Identify exactly what has to be measured and what apparatus measures it
- Write the steps in the order you would actually do them in the lab, from setting up to taking the final reading
- State how you get a RANGE of results, not just one reading, since the mark scheme always credits varying a control input
- State how you make the results more reliable, such as taking repeat readings or reversing a connection to get negative values
- If a circuit is involved, describe it in words precisely enough that an examiner reading only your writing could still draw it correctly
What does an I-V characteristic graph show for a component?
This question always wants a method that would genuinely work if you followed it, in the right order, with a way to get repeatable results. Practise writing full methods for the required practicals, not just naming the equipment.
Practise IV characteristics and density questions“Write down the equation for efficiency, then calculate the useful power output or total power input”
Every single sitting we analysed asks students to state the efficiency equation and then apply it to a real energy transfer, whether that is a National Grid, an LED, solar cells, or a wind turbine. The equation itself is worth one mark, and the calculation that follows is worth up to three more, so this pairing is worth mastering as one combined skill.
1×asked“Write down the equation which links efficiency, total energy input and useful energy output. Then: the total energy input to the National Grid from one power station is 34.2 GJ, and the National Grid has an efficiency of 0.992. Calculate the useful energy output from this power station to consumers in GJ.”
“Write down the equation which links efficiency, total energy input and useful energy output. Then: the total energy input to the National Grid from one power station is 34.2 GJ, and the National Grid has an efficiency of 0.992. Calculate the useful energy output from this power station to consumers in GJ.”
What it’s really asking
State the efficiency equation, then substitute the given efficiency and total energy input to find the useful energy output, referenced as Q01.4 and Q01.5 in the real paper.
What the sources actually showed — June 2023
A diagram of the National Grid showing a power station, two transformers and consumers connected by transmission cables, with the total energy input from one power station given as 34.2 GJ and the National Grid's efficiency given as 0.992.
The full worked answer — June 2023
Efficiency equals useful energy output divided by total energy input.
Substituting the given values, 0.992 equals useful energy output divided by 34.2. Rearranging, useful energy output equals 0.992 multiplied by 34.2, which gives 33.9 GJ.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise efficiency calculationsWhat the mark scheme rewarded
- The efficiency equation stated correctly, in words or symbols, as its own separate mark
- Correct substitution of the given efficiency and total energy value
- Correct rearrangement to isolate useful energy output
- An answer given to more than 3 significant figures, which the real mark scheme explicitly allows
- Efficiency has no unit, since it is a ratio of two energy or two power values
- Efficiency can never be greater than 1, since useful output can never exceed total input
- The same equation can use power values instead of energy values, since power and energy differ only by time
- Wasted energy equals total energy input minus useful energy output
Traps examiners saw
- Writing the equation upside down, with total input on top and useful output on the bottom
- Substituting the efficiency value as a percentage instead of a decimal without converting it first
- Rounding the final answer to fewer significant figures than the mark scheme's stated allowance
Full-mark self-check 0 of 4
1×asked“Write down the equation which links efficiency, total power input and useful power output. Then: the tiles are used to power LED lights in the pavement. An LED light has a total power input of 4.0 W and the efficiency of the LED light is 0.85. Calculate the useful power output of the LED light.”
“Write down the equation which links efficiency, total power input and useful power output. Then: the tiles are used to power LED lights in the pavement. An LED light has a total power input of 4.0 W and the efficiency of the LED light is 0.85. Calculate the useful power output of the LED light.”
What it’s really asking
State the efficiency equation using power rather than energy, then substitute the given power input and efficiency to find the useful power output, referenced as Q03.3 and Q03.4 in the real paper.
What the sources actually showed — June 2022
A photograph of pavement tiles that generate electricity when walked on, used to power LED lights, with the total power input to an LED light given as 4.0 W and its efficiency given as 0.85.
The full worked answer — June 2022
Efficiency equals useful power output divided by total power input.
Substituting the given values, 0.85 equals useful power output divided by 4.0. Rearranging, useful power output equals 0.85 multiplied by 4.0, which gives 3.4 W.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise efficiency calculationsWhat the mark scheme rewarded
- The efficiency equation stated correctly using power rather than energy, since this question gives power values
- Correct substitution of 0.85 and 4.0 into the equation
- Correct rearrangement and calculation of the final answer in watts
- Power is the rate of energy transfer, measured in watts, so the efficiency equation works equally with power or energy values
- An efficiency of 0.85 means 85 percent of the input power becomes useful output, with the rest wasted, usually as heat
- Total power input minus useful power output gives the wasted power, here 4.0 minus 3.4 equals 0.6 W wasted
Traps examiners saw
- Confusing the power version and the energy version of the equation and mixing up which values are being substituted
- Forgetting that 0.85 is already a decimal efficiency and multiplying by 100 unnecessarily
Full-mark self-check 0 of 3
1×asked“At one time, the total power input to the solar cells was 7.8 kW. The efficiency of the solar cells was 0.15. Calculate the useful power output of the solar cells.”
“At one time, the total power input to the solar cells was 7.8 kW. The efficiency of the solar cells was 0.15. Calculate the useful power output of the solar cells.”
What it’s really asking
Substitute the given total power input and efficiency into the efficiency equation to find the useful power output, after first converting kilowatts to watts, referenced as Q05.3 in the real paper.
What the sources actually showed — June 2021
A diagram of a house with solar cells on its roof feeding a large battery via a cable, with the total power input to the solar cells given as 7.8 kW and their efficiency given as 0.15, referenced as Figure 5 in the real paper.
The full worked answer — June 2021
First convert 7.8 kW into watts, giving 7800 W, since the efficiency equation needs consistent units throughout the calculation.
Substituting into the efficiency equation, 0.15 equals useful power output divided by 7800. Rearranging, useful power output equals 0.15 multiplied by 7800, which gives 1170 W.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise efficiency calculationsWhat the mark scheme rewarded
- Converting 7.8 kW to 7800 W before substituting
- Correct substitution of the efficiency and converted power input
- Correct rearrangement and calculation giving 1170 W as the useful power output
- Kilo means one thousand, so kilowatts must be multiplied by 1000 to convert to watts
- An efficiency of 0.15 for solar cells reflects the real limitation that only a fraction of incoming light energy is converted to electrical energy, the rest is reflected or converted to heat
- This same low efficiency is why a very large surface area of solar cells is needed to generate a meaningful amount of power
Traps examiners saw
- Forgetting to convert kilowatts to watts before substituting, which gives an answer 1000 times too small
- Substituting efficiency as 15 instead of 0.15
Full-mark self-check 0 of 3
1×asked“Write down the equation which links efficiency, total power input and useful power output. Then: the total power input to the LED was 0.24 W. The efficiency of the LED was 0.75. Calculate the useful power output of the LED.”
“Write down the equation which links efficiency, total power input and useful power output. Then: the total power input to the LED was 0.24 W. The efficiency of the LED was 0.75. Calculate the useful power output of the LED.”
What it’s really asking
State the efficiency equation, then substitute the given total power input and efficiency to find the useful power output, referenced as Q02.5 and Q02.6 in the real paper.
What the sources actually showed — May 2020
A photograph of an LED torch, with the total power input to its LED given as 0.24 W and its efficiency given as 0.75.
The full worked answer — May 2020
Efficiency equals useful power output divided by total power input.
Substituting the given values, 0.75 equals useful power output divided by 0.24. Rearranging, useful power output equals 0.75 multiplied by 0.24, which gives 0.18 W.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise efficiency calculationsWhat the mark scheme rewarded
- The efficiency equation stated correctly, using power values
- Correct substitution of 0.75 and 0.24 into the equation
- Correct rearrangement and calculation, giving 0.18 W as the final answer
- An LED is far more efficient than an old-style filament bulb, since filament bulbs waste most of their input power as heat rather than light
- The 0.75 efficiency here means 25 percent of the power input to the LED is still wasted, typically as heat in the diode itself
Traps examiners saw
- Multiplying 0.24 by 0.75 correctly but then rounding incorrectly to more or fewer significant figures than the calculation actually supports
- Mixing up which value is total input and which is efficiency when both are given as decimals under 1
Full-mark self-check 0 of 3
The method for every Efficiency formula and calculation — same every sittingMark bands, steps, timing
What this question type rewards
The topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
- Stating the efficiency equation exactly, in either words or symbols
- Substituting the correct given values into the equation in the right place
- Rearranging correctly to find whichever quantity is missing
- Giving the final answer to the correct number of significant figures where the question asks for it
The steps
- Write down efficiency equals useful power output divided by total power input, or the energy version of the same equation, as your first line
- Identify which of the three quantities, efficiency, useful output or total input, is missing
- Substitute the two known values into the equation before you rearrange, this is often worth its own mark even if the final answer is wrong
- Rearrange to make the missing quantity the subject, then calculate the final value
- Check the units and significant figures match what the question specifically asks for
Which equation correctly defines efficiency?
This pairing, state the equation then calculate, appears on every single sitting in some form. Learn the equation cold and practise substituting real numbers from different energy transfer contexts.
Practise efficiency calculations“Describe the difference between irradiation and contamination, or explain how the risk from internal contamination differs from external irradiation”
Two of the four sittings we have full papers for test this exact distinction, once as a shorter 2-mark definition question and once as a fuller 5-mark explanation question. Both real mark schemes reward the same underlying idea: irradiation is exposure to radiation from outside, contamination is radioactive material actually getting onto or inside a person, and the type of radiation involved changes which situation is more dangerous.
1×asked“Describe the difference between irradiation and contamination.”
“Describe the difference between irradiation and contamination.”
What it’s really asking
Give a clear, correctly worded definition of each term, referenced as Q05.5 in the real paper.
What the sources actually showed — June 2023
A statement that people who work in the nuclear power industry need to be aware of irradiation and contamination, with no further data given.
The full worked answer — June 2023
Irradiation is the exposure of an object or person to radiation from a source that stays outside the body, whereas contamination is the unwanted presence of radioactive material or atoms on an object or, if the material gets inside a person, within the body itself.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise radiation hazard questionsWhat the mark scheme rewarded
- Irradiation defined as exposure of a person or object to radiation from an external source
- Contamination defined as the unwanted presence of radioactive material on an object or, allowably, inside a person
- A worker can be irradiated by standing near a radioactive source without any radioactive material actually touching or entering their body
- Contamination means actual radioactive atoms are present, which continue emitting radiation for as long as they remain there or until they decay
Traps examiners saw
- Defining only one of the two terms and leaving the other undefined
- Describing what causes irradiation or contamination rather than defining what the words actually mean
Full-mark self-check 0 of 3
1×asked“Internal contamination of the human body means radioactive material is inside the human body. Explain how the risk from internal contamination is different to the risk from external irradiation by a source of alpha radiation.”
“Internal contamination of the human body means radioactive material is inside the human body. Explain how the risk from internal contamination is different to the risk from external irradiation by a source of alpha radiation.”
What it’s really asking
Explain, using the properties of alpha radiation specifically, why the same type of radiation is low risk when the source stays outside the body but high risk once the radioactive material is inside the body, referenced as Q07.3 in the real paper.
What the sources actually showed — June 2022
A statement that alpha particles, beta particles and gamma rays are types of nuclear radiation, followed by a specific question about internal contamination versus external irradiation from a source of alpha radiation.
The full worked answer — June 2022
Alpha radiation has a low penetrating ability, so when the alpha source stays outside the body, the radiation is stopped by the outer layer of skin before it can reach living cells underneath. This means the external irradiation risk from an alpha source is low.
Once inside the body, however, alpha radiation is absorbed directly by living tissue and organs, since there is no layer of dead skin cells to stop it internally. Alpha radiation is also highly ionising, meaning it transfers a large amount of energy to a small area of tissue over a short range.
This means internal contamination by alpha radiation causes a much greater risk of harm to cells, tissues and DNA than external irradiation from the same type of source, since the highly ionising radiation is now acting directly on living cells at close range rather than being absorbed by dead skin first.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise radiation hazard questionsWhat the mark scheme rewarded
- Alpha radiation has a low penetrating ability, so external irradiation risk is low, since it is stopped by skin
- Internally, alpha radiation is absorbed by living tissue and organs, since there is no skin barrier to stop it
- Alpha radiation is highly ionising, which increases the harm caused once it is absorbed by living cells
- Internal contamination causes a greater risk of harm to cells, tissues, organs or DNA than external irradiation from the same source
- Alpha particles are the largest and least penetrating type of nuclear radiation, stopped by a few centimetres of air or a sheet of paper
- Ionising radiation can damage DNA within cells, which is linked to an increased risk of cancer developing later
- Workers in the nuclear industry are monitored specifically to detect any internal contamination, since this is far harder to spot than external irradiation
Traps examiners saw
- Explaining only the internal risk and forgetting to also explain why the external risk is comparatively low
- Describing alpha radiation generically without linking its specific properties, low penetration and high ionising power, to the two different risk situations
- Confusing irradiation and contamination and mixing up which situation the question is actually asking about
Full-mark self-check 0 of 3
The method for every Irradiation and contamination — same every sittingMark bands, steps, timing
What this question type rewards
The topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
- Defining irradiation as exposure to radiation from a source outside the body
- Defining contamination as radioactive material being present on or inside a person or object
- Explaining which type of radiation, usually alpha, is more dangerous once inside the body due to its high ionising power
- Explaining why the same type of radiation is comparatively low risk from outside the body, since it cannot penetrate skin
The steps
- State clearly which situation is irradiation, exposure from an outside source, and which is contamination, radioactive material on or in the body
- Identify the type of radiation involved in the specific scenario given
- Explain why that type of radiation is low risk from outside the body, usually because it cannot penetrate far, such as alpha being stopped by skin
- Explain why the same radiation becomes high risk once inside the body, since it can then irradiate living cells directly at short range with its high ionising power
- Link this back to the specific health risk named in the question, such as cancer or cell damage
Which type of radiation is used in smoke detectors?
Irradiation and contamination sound similar but the physics of why they carry different risks is exactly what the mark scheme wants explained. Practise linking radiation properties to real risk scenarios.
Practise radiation hazard questions“Calculate the specific heat capacity, the mass of a substance, or the specific latent heat, using data from a graph or a described experiment”
All four sittings we analysed include a calculation built around the specific heat capacity equation, energy transferred equals mass multiplied by specific heat capacity multiplied by change in temperature, or its specific latent heat equivalent. The numbers, the missing quantity and the practical context all change, but the same equation and the same rearranging skill are tested every time.
1×asked“The energy transferred to the iron block between 5 and 10 minutes was 26 000 J. The mass of the iron block was 2.0 kg. Calculate the specific heat capacity of iron. Use information from the given graph and the Physics Equations Sheet.”
“The energy transferred to the iron block between 5 and 10 minutes was 26 000 J. The mass of the iron block was 2.0 kg. Calculate the specific heat capacity of iron. Use information from the given graph and the Physics Equations Sheet.”
What it’s really asking
Read the temperature change between 5 and 10 minutes off the given graph, then rearrange the specific heat capacity equation to find the specific heat capacity of iron, referenced as Q02.2 in the real paper.
What the sources actually showed — June 2023
A graph showing the temperature of an iron block in degrees Celsius plotted against time in minutes, rising from about 20 degrees Celsius at 0 minutes to about 80 degrees Celsius at 15 minutes, with the temperatures at 5 minutes and 10 minutes readable from the curve.
The full worked answer — June 2023
Reading from the graph, the temperature at 5 minutes is 28 degrees Celsius and at 10 minutes is 54 degrees Celsius, giving a change in temperature of 54 minus 28, which equals 26 degrees Celsius.
Substituting into the equation, 26 000 equals 2.0 multiplied by specific heat capacity multiplied by 26. Rearranging, specific heat capacity equals 26 000 divided by the result of 2.0 multiplied by 26, which gives 500 J/kg degrees Celsius.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise specific heat capacity questionsWhat the mark scheme rewarded
- Correctly reading the temperature change between 5 and 10 minutes from the graph as 26 degrees Celsius
- Correct substitution of 26 000 J, 2.0 kg and the temperature change into the specific heat capacity equation
- Correct rearrangement to isolate specific heat capacity
- The final calculated value, 500 J/kg degrees Celsius, or an answer consistent with the student's own graph reading
- Specific heat capacity is the energy needed to raise the temperature of 1 kg of a substance by 1 degree Celsius
- A flatter section on a temperature-time graph during heating usually indicates a change of state rather than a temperature change
- Reading two points from a graph and subtracting is a very common way exam questions test whether a student can extract data accurately
Traps examiners saw
- Misreading the graph's scale and taking the wrong two temperature values
- Substituting the temperature values themselves into the equation instead of the calculated temperature change
- Forgetting to convert units if the graph or question gives data in a different form
Full-mark self-check 0 of 4
1×asked“For one type of insulating material, the temperature of the water decreased from 85.0 degrees Celsius to 65.0 degrees Celsius. The energy transferred from the water was 10.5 kJ. Specific heat capacity of water = 4200 J/kg degrees Celsius. Calculate the mass of water in the can.”
“For one type of insulating material, the temperature of the water decreased from 85.0 degrees Celsius to 65.0 degrees Celsius. The energy transferred from the water was 10.5 kJ. Specific heat capacity of water = 4200 J/kg degrees Celsius. Calculate the mass of water in the can.”
What it’s really asking
Convert the given energy from kilojoules to joules, calculate the temperature change from the two given temperatures, then rearrange the specific heat capacity equation to find the mass of water, referenced as Q04.4 in the real paper.
What the sources actually showed — June 2022
A labelled diagram of the apparatus used, a thermometer through a lid on a metal can of hot water with insulating material wrapped around the can, referenced as Figure 4 in the real paper, alongside a description of a student investigating the insulating properties of different materials by wrapping them around the can and timing how quickly the temperature decreased, with one material's water cooling from 85.0 degrees Celsius to 65.0 degrees Celsius while transferring 10.5 kJ of energy.
The full worked answer — June 2022
First convert 10.5 kJ to joules, giving 10 500 J, since the specific heat capacity given is in joules per kilogram per degree Celsius. The change in temperature is 85.0 minus 65.0, which equals 20 degrees Celsius.
Substituting into the equation, 10 500 equals mass multiplied by 4200 multiplied by 20. Rearranging, mass equals 10 500 divided by the result of 4200 multiplied by 20, which gives 0.125 kg.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise specific heat capacity questionsWhat the mark scheme rewarded
- Converting 10.5 kJ to 10 500 J before substituting
- Correctly calculating the temperature change as 20 degrees Celsius
- Correct substitution and rearrangement of the specific heat capacity equation to find mass
- The final calculated value, 0.125 kg, or an answer consistent with an incorrectly converted energy value carried through correctly
- Kilo means one thousand, so kilojoules must be multiplied by 1000 to convert to joules before use in the standard equation
- A larger mass of water would need more energy for the same temperature drop, since specific heat capacity relates energy to mass directly
- This kind of investigation compares different insulating materials by measuring how quickly heat is lost from an otherwise identical setup
Traps examiners saw
- Forgetting to convert kilojoules to joules, giving a mass answer 1000 times too large
- Subtracting the temperatures in the wrong order and getting a negative value that then confuses the rest of the calculation
- Rearranging the equation incorrectly, for example dividing by the temperature change instead of the full product of specific heat capacity and temperature change
Full-mark self-check 0 of 4
1×asked“Electricity is used to increase the temperature of 4.0 g of deuterium by 50 000 000 degrees Celsius. Specific heat capacity of deuterium = 5200 J/kg degrees Celsius. Calculate the energy needed to increase the temperature of the deuterium by 50 000 000 degrees Celsius.”
“Electricity is used to increase the temperature of 4.0 g of deuterium by 50 000 000 degrees Celsius. Specific heat capacity of deuterium = 5200 J/kg degrees Celsius. Calculate the energy needed to increase the temperature of the deuterium by 50 000 000 degrees Celsius.”
What it’s really asking
Convert the given mass from grams to kilograms, then substitute directly into the specific heat capacity equation to calculate the energy needed, referenced as Q02.2 in the real paper.
What the sources actually showed — June 2021
A statement that nuclear fusion of deuterium is difficult to achieve on Earth because of the very high temperature needed, with a mass of 4.0 g of deuterium and a specific heat capacity of 5200 J/kg degrees Celsius given, alongside the huge required temperature increase of 50 000 000 degrees Celsius.
The full worked answer — June 2021
First convert 4.0 g to kilograms, giving 0.004 kg, since the specific heat capacity given is in joules per kilogram per degree Celsius.
Substituting into the equation, energy equals 0.004 multiplied by 5200 multiplied by 50 000 000, which gives 1.04 times 10 to the power 9 J.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise specific heat capacity questionsWhat the mark scheme rewarded
- Converting 4.0 g to 0.004 kg before substituting
- Correct substitution of mass, specific heat capacity and the given temperature increase into the equation
- The final calculated value, 1.04 times 10 to the power 9 J, or an equivalent correctly rounded answer
- Grams must be divided by 1000 to convert to kilograms, the standard SI unit used in the specific heat capacity equation
- Nuclear fusion requires extremely high temperatures because the nuclei involved are positively charged and repel each other strongly, so huge kinetic energy is needed to force them close enough to fuse
- This is why fusion happens naturally in stars, where gravity provides the enormous pressure and temperature needed, but is very difficult to sustain on Earth
Traps examiners saw
- Forgetting to convert grams to kilograms, which gives an answer 1000 times too large
- Getting confused by the very large numbers involved and making an arithmetic slip when multiplying three large or small values together
Full-mark self-check 0 of 3
1×asked“The energy transferred to the water in 100 seconds was 155 000 J. Specific heat capacity of water = 4200 J/kg degrees Celsius. Determine the mass of water in the kettle. Use the given graph. Give your answer to 2 significant figures.”
“The energy transferred to the water in 100 seconds was 155 000 J. Specific heat capacity of water = 4200 J/kg degrees Celsius. Determine the mass of water in the kettle. Use the given graph. Give your answer to 2 significant figures.”
What it’s really asking
Read the temperature change over 100 seconds off the given graph, then rearrange the specific heat capacity equation to find the mass of water, giving the final answer to 2 significant figures, referenced as Q07.2 in the real paper.
What the sources actually showed — May 2020
A graph showing the temperature of water inside an electric kettle in degrees Celsius plotted against time in seconds after the kettle was switched on, starting at about 22 degrees Celsius and rising to 100 degrees Celsius by 100 seconds, with an initial flatter section before the temperature begins to rise steadily.
The full worked answer — May 2020
Reading from the graph, the temperature increased from about 22 degrees Celsius at 0 seconds to 100 degrees Celsius at 100 seconds, giving a change in temperature of 100 minus 22, which equals 78 degrees Celsius.
Substituting into the equation, 155 000 equals mass multiplied by 4200 multiplied by 78. Rearranging, mass equals 155 000 divided by the result of 4200 multiplied by 78, which gives 0.4731 kg, or 0.47 kg to 2 significant figures.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise specific heat capacity questionsWhat the mark scheme rewarded
- Correctly reading the temperature change over 100 seconds from the graph as 78 degrees Celsius
- Correct substitution of the given energy, the specific heat capacity of water and the temperature change
- Correct rearrangement to isolate mass
- The final answer rounded to exactly 2 significant figures, as specifically instructed in the question
- The flatter section at the very start of a kettle's heating graph reflects the time taken for the heating element itself to heat up before the water temperature starts rising steadily
- A kettle transfers electrical energy to thermal energy in the water via a heating element, following the same specific heat capacity equation used for any substance
- Rounding to the exact number of significant figures asked for is worth checking carefully, since an unrounded or wrongly rounded correct answer can still lose the final mark
Traps examiners saw
- Misreading the graph's starting temperature as 0 degrees Celsius instead of the actual starting value of about 22 degrees Celsius
- Calculating the correct unrounded mass but then giving it to the wrong number of significant figures
- Forgetting that the specific heat capacity equation cannot be used across a graph section where the substance is changing state rather than changing temperature
Full-mark self-check 0 of 4
The method for every Specific heat capacity and specific latent heat calculations — same every sittingMark bands, steps, timing
What this question type rewards
The topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
- Identifying the correct change in temperature, often by reading two values off a graph and subtracting them
- Substituting the given energy, mass, specific heat capacity or change in temperature correctly into the equation
- Rearranging the equation correctly for whichever quantity is missing
- Giving the final answer with the correct unit
The steps
- Read off any values you need from a graph carefully, especially a change in temperature between two labelled points
- Write down the equation, energy transferred equals mass multiplied by specific heat capacity multiplied by change in temperature, or energy transferred equals mass multiplied by specific latent heat for a change of state
- Substitute every given value into the equation, including any value you calculated in a previous step
- Rearrange to isolate whichever quantity is missing
- Calculate the final value and check the unit matches what is being asked for
What does the specific heat capacity of a substance measure?
Specific heat capacity calculations appear on every sitting we have, usually needing a value read carefully from a graph first. Practise the full chain: read the data, substitute, rearrange, calculate.
Practise specific heat capacity questions“Determine the resistance of the filament lamp / component at a stated potential difference or current”
Three of the four sittings we have full papers for ask students to find resistance using potential difference equals current multiplied by resistance, with the twist that the current or potential difference usually has to be read off a graph first rather than being given directly in the question text.
1×asked“Determine the resistance of the filament lamp when the potential difference across it is +3.0 V. Use the given graph and the Physics Equations Sheet.”
“Determine the resistance of the filament lamp when the potential difference across it is +3.0 V. Use the given graph and the Physics Equations Sheet.”
What it’s really asking
Read the current at +3.0 V off the given current-potential difference graph, then use V = IR rearranged to find resistance, referenced as Q06.2 in the real paper.
What the sources actually showed — June 2023
A graph showing current in amps plotted against potential difference in volts for a filament lamp, curving through the origin and flattening out at higher voltages, with a current of about 0.16 A readable at +3.0 V.
The full worked answer — June 2023
Reading from the graph, the current at a potential difference of +3.0 V is 0.16 A.
Substituting into V = IR, 3.0 equals 0.16 multiplied by resistance. Rearranging, resistance equals 3.0 divided by 0.16, which gives 18.75 ohms.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise resistance and Ohm's law questionsWhat the mark scheme rewarded
- Reading a current value in the range 0.15 A to 0.17 A at +3.0 V from the graph
- Correct substitution of the read-off current and the given potential difference into V = IR
- Correct rearrangement to isolate resistance
- The final answer, 18.75 ohms, or the allowed rounded alternatives of 19 ohms or 18.8 ohms
- A filament lamp's resistance increases as its temperature increases, which is why its current-potential difference graph curves rather than forming a straight line through the origin
- V = IR can be rearranged to R = V divided by I whenever resistance is the unknown quantity
- Reading a value from a curve rather than a straight line always carries a small range of acceptable answers, since the exact point is harder to pin down precisely
Traps examiners saw
- Reading the current at the wrong potential difference value on the graph
- Rearranging V = IR incorrectly, for example calculating current multiplied by potential difference instead of dividing
Full-mark self-check 0 of 4
1×asked“The potential difference across the battery is 480 V. There is a current of 15 A in the circuit connecting the battery to the motor of the electric car. Calculate the resistance of the motor.”
“The potential difference across the battery is 480 V. There is a current of 15 A in the circuit connecting the battery to the motor of the electric car. Calculate the resistance of the motor.”
What it’s really asking
Substitute the given potential difference and current directly into a rearranged version of V = IR to find the resistance, referenced as Q01.5 in the real paper.
What the sources actually showed — June 2021
A description of an electric car being charged, with the potential difference across its battery given as 480 V and the current in the circuit connecting the battery to the motor given directly as 15 A.
The full worked answer — June 2021
Substituting into V = IR, 480 equals 15 multiplied by resistance. Rearranging, resistance equals 480 divided by 15, which gives 32 ohms.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise resistance and Ohm's law questionsWhat the mark scheme rewarded
- Correct substitution of 480 and 15 into V = IR
- Correct rearrangement to isolate resistance
- The final calculated value, 32 ohms
- The equation V = IR links potential difference, current and resistance for any component in a circuit, not only for filament lamps
- A motor with a fixed resistance experiences a proportional relationship between potential difference and current, unlike a filament lamp whose resistance changes with temperature
Traps examiners saw
- Substituting the values into the wrong positions in the equation, for example dividing 15 by 480 instead of 480 by 15
- Making an arithmetic slip when dividing 480 by 15
Full-mark self-check 0 of 3
1×asked“Determine the resistance of the filament lamp when the potential difference across it is 1.0 V. Use data from the given graph.”
“Determine the resistance of the filament lamp when the potential difference across it is 1.0 V. Use data from the given graph.”
What it’s really asking
Read the current at 1.0 V off the given current-potential difference graph, then use V = IR rearranged to find resistance, referenced as Q01.4 in the real paper.
What the sources actually showed — May 2020
A graph showing current in amps plotted against potential difference in volts for a filament lamp, curving through the origin from negative 8 V to positive 8 V, with a current of about 0.08 A readable at 1.0 V.
The full worked answer — May 2020
Reading from the graph, the current at a potential difference of 1.0 V is 0.08 A.
Substituting into V = IR, 1.0 equals 0.08 multiplied by resistance. Rearranging, resistance equals 1.0 divided by 0.08, which gives 12.5 ohms.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise resistance and Ohm's law questionsWhat the mark scheme rewarded
- Reading a current value of 0.08 A at 1.0 V from the graph
- Correct substitution of the read-off current and the given potential difference into V = IR
- Correct rearrangement to isolate resistance
- The final calculated value, 12.5 ohms, or an answer consistent with the student's own graph reading
- At low potential differences, a filament lamp's resistance is lower because its temperature is closer to room temperature, which is why the curve is steeper near the origin
- This is the same underlying equation and skill tested with the resistor circuit in other sittings, only here the value has to be read from a curved rather than a straight-line graph
Traps examiners saw
- Reading the current at a nearby point on the graph rather than exactly at 1.0 V
- Forgetting that the graph is not a straight line, so resistance calculated at 1.0 V will be different from resistance calculated at a higher potential difference
Full-mark self-check 0 of 4
The method for every Resistance from a graph or from V = IR — same every sittingMark bands, steps, timing
What this question type rewards
The topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
- Reading the correct current or potential difference value off the given graph at the stated point
- Writing down V = IR or a correctly rearranged version of it
- Substituting the read-off value and any other given value correctly
- Calculating the final resistance value correctly, in ohms
The steps
- Find the exact point on the graph the question asks about, usually a stated potential difference or current value
- Read off the corresponding current or potential difference as precisely as the graph allows
- Write V = IR, or state that resistance equals potential difference divided by current
- Substitute the read-off value and any other given value, then rearrange to make resistance the subject if needed
- Calculate the final value in ohms
Which of the following best describes electrical resistance?
Resistance calculations often hide the real difficulty in the graph-reading step before V = IR is even used. Practise reading current or potential difference precisely off a curved graph.
Practise resistance and Ohm's law questionsThe question types that keep coming up
Across the 4 sittings we have full papers for, these are the recurring question archetypes and topics with the most marks at stake.
Not the primary focus in the 4 sittings we have full papers for
Detailed calculations involving background radiation dose comparisons as the main focus of a whole question, rather than as one part of a larger question · Static electricity questions asking for a full explanation of charge transfer by rubbing as the sole focus of a 6-mark question
These areas have not been the main focus of a Paper 1 question in the papers we analysed, but they remain fully in scope for the specification, so do not skip them entirely.
Before you revise
Are these real mark-scheme answers?
The sources are described in our own words, not reproduced, and the worked answers are written entirely by us, aimed at what the real AQA mark scheme rewarded for each sitting. They are not copied from AQA's own exemplar materials, since that would breach copyright, but they are built to hit exactly what the real mark scheme credited that year. PrepWise is independent of AQA and not endorsed by them.
Will the exact same questions come up again this year?
Sometimes the numbers and context change but the underlying calculation or method stays the same, which is exactly what this page tracks. Efficiency calculations, specific heat capacity calculations and describe-a-method practicals have appeared in some form in almost every sitting we have full papers for. Use this page to learn the SKILL behind each recurring question type, since the exact numbers will always be different on the day.
Why does the 2020 paper look different from the others?
AQA's usual summer exams were cancelled across the UK in 2020 due to the pandemic. The paper we analysed for that year prints Wednesday 20 May 2020 as its exam date on the question paper cover, but its accompanying mark scheme is titled June 2020, a real mismatch from that disrupted exam series, not an error on our part. We have flagged this honestly rather than guessing which date is correct.
Is PrepWise free to use for this?
Yes, PrepWise is free during alpha. You can practise every topic on this page without paying anything right now.
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