Quick answer: identify the unknown before calculating
A heating investigation can hide a mathematical question inside a science story. One question asks for energy, another asks for the mass that could be warmed, another asks for the original value before a percentage reduction, and a fourth asks how many complete trials can fit within an energy or time limit. They are not the same problem. Under examination conditions, name the unknown and the condition that constrains it. Write one equation, solve it, attach the correct unit, and check whether the answer is physically or operationally sensible in the stated model.
The correct SEC Mathematics framework
SEAB’s 2027 SEC G3 directory, updated 26 October 2026, lists G3 Mathematics as K310, English as K300 and Physics as K323. The official K310 syllabus gives Paper 1 a duration of 2 hours 15 minutes, 90 marks and 50% weighting; Paper 2 is also 2 hours 15 minutes, 90 marks and 50%. Both require all questions to be answered; Paper 2 ends with a question applying mathematics to a real-world situation. An approved calculator may be used in both papers. These are official subject-paper details; our heating investigation is an original learning context, not a new cross-subject SEC exam paper.
How this chapter advances the sequence
Volume 0629 owns the Physics distinction between energy entering a heater and energy gained by the water. Volume 0626 already teaches unequal-duration energy logs; do that chapter first if your time-weighting is weak. Here we solve the inverse questions: what mass is possible, what change is required, what the percentage base really is, and how a plan changes at a constraint boundary. At primary level, PSLE Volume 0074 explains why the whole behind a percentage must be kept fixed. That idea becomes decisive here.
Use a fictional paper-only model
For every exercise unless another value is stated, use the supplied model c = 4200 J per kilogram per degree Celsius. In these problems this is a numerical constant, not a request to conduct a heating experiment. Imagine an authorised school display with paper records rather than apparatus that students are asked to build. A real laboratory may need additional controls and safety steps. Mathematics can solve a stated model exactly while the surrounding real-world conclusion remains conditional on the model’s assumptions. It is a sign of understanding, not weakness, to distinguish a computed value from a measured one.
Four lines of working that save many marks
Line one records the quantities with units. Line two gives the relationship connecting them. Line three substitutes numerical values without mixing grams and kilograms, minutes and seconds or percentages and decimal multipliers. Line four interprets the result with the required accuracy, rounding only when asked. This method is especially useful when a question switches from direct calculation to reverse calculation. If the task asks for a number of complete trials, the final line is a whole-number decision, not an unrounded decimal.
Consider Q = 4200mΔT. If Q and ΔT are known, divide Q by 4200ΔT to find m. If Q and m are known, divide Q by 4200m to find ΔT. The two rearrangements are mathematically different even though they start with the same physical equation. You do not get full value from memorising the equation if you insert the unknown into the wrong place. A quick estimate is a strong safeguard: heating more mass by the same number of degrees should require proportionally more energy.
Fractions, percentages and reverse percentages are three separate operations
Suppose one trial records 1600 J input and 1200 J in the specified water gain. The ratio is 1200/1600 = 0.75 and the percentage is 75%. If a later process reduces the input by 20%, then the new input is 0.80 times the original. If instead the reduced value is given and the original is required, divide by 0.80 rather than adding 20%. A common wrong answer confuses the final figure with the base figure. Write a sentence naming the whole each percentage uses; numbers without their denominators invite unjustified comparison.
Normalise comparisons only when it is fair
Two different water masses may show different temperature rises even when the energy gained per kilogram is the same. Normalising can make some comparisons useful. For example, compare energy per kilogram of water or joules per degree of rise, but explain exactly what is being held constant. An observed wattage at one time cannot be compared with a total number of joules until time is handled. Do not normalise away a relevant difference such as different starting temperatures when the question explicitly tests that difference.
Bounds and complete-count decisions
A maximum limit introduces an inequality. Suppose each fictional trial requires 1350 J and a supply budget allows at most 6000 J. The condition is 1350n ≤ 6000, so n ≤ 4.444… . Only four complete trials fit. The correct operational answer is four, with 600 J left unused under this model. Rounding 4.444 to four by ordinary decimal rounding happens to give the right count here, but the principle is not ordinary rounding: no fractional complete trial is allowed. If a different constraint is tighter, it wins.
Check graphs, scales and source statements
A calibration table can show temperature against time or cumulative energy. Its axes must identify which variable is independent, which is dependent and which unit is used. When readings are evenly spaced, a difference quotient gives an average rate for an interval. It need not be the instantaneous rate at any chosen moment. A smooth-looking graph may be an interpolation, not an observation. A claim that a curve ‘proves’ a universal relationship is stronger than the handful of measurements usually supports.
Worked questions 1–6: calculate input from power and time
Question 1: first-day classroom record
A paper model for the first-day classroom record gives a steady electrical power of 5 W for 120 seconds. Required: total input energy, not water temperature. The working is E = Pt = 5 × 120 = 600.0 J. A useful verification is to divide the answer by the stated time: the quotient recovers 5 W. That is a unit-and-meaning check, not merely a way to catch calculator mistakes. Because the model has specified constant power, summing smaller time intervals would agree with the single multiplication. If the power changed, this shortcut would need replacing by separate interval calculations. In the final sentence, put the time into seconds before calculating energy; only then could an examiner know that the chosen formula genuinely matches the data instead of an accidental numerical coincidence.
Question 2: short power test
A paper model for the short power test gives a steady electrical power of 8 W for 75 seconds. Required: total input energy, not water temperature. The working is E = Pt = 8 × 75 = 600.0 J. A useful verification is to divide the answer by the stated time: the quotient recovers 8 W. That is a unit-and-meaning check, not merely a way to catch calculator mistakes. Because the model has specified constant power, summing smaller time intervals would agree with the single multiplication. If the power changed, this shortcut would need replacing by separate interval calculations. In the final sentence, do not convert W into J without a duration; only then could an examiner know that the chosen formula genuinely matches the data instead of an accidental numerical coincidence.
Question 3: longer steady run
A paper model for the longer steady run gives a steady electrical power of 10 W for 180 seconds. Required: total input energy, not water temperature. The working is E = Pt = 10 × 180 = 1800.0 J. A useful verification is to divide the answer by the stated time: the quotient recovers 10 W. That is a unit-and-meaning check, not merely a way to catch calculator mistakes. Because the model has specified constant power, summing smaller time intervals would agree with the single multiplication. If the power changed, this shortcut would need replacing by separate interval calculations. In the final sentence, separate the two multiplied quantities; only then could an examiner know that the chosen formula genuinely matches the data instead of an accidental numerical coincidence.
Question 4: model heater check
A paper model for the model heater check gives a steady electrical power of 6.5 W for 200 seconds. Required: total input energy, not water temperature. The working is E = Pt = 6.5 × 200 = 1300.0 J. A useful verification is to divide the answer by the stated time: the quotient recovers 6.5 W. That is a unit-and-meaning check, not merely a way to catch calculator mistakes. Because the model has specified constant power, summing smaller time intervals would agree with the single multiplication. If the power changed, this shortcut would need replacing by separate interval calculations. In the final sentence, keep a decimal power value through the working; only then could an examiner know that the chosen formula genuinely matches the data instead of an accidental numerical coincidence.
Question 5: higher setting with shorter time
A paper model for the higher setting with shorter time gives a steady electrical power of 12 W for 90 seconds. Required: total input energy, not water temperature. The working is E = Pt = 12 × 90 = 1080.0 J. A useful verification is to divide the answer by the stated time: the quotient recovers 12 W. That is a unit-and-meaning check, not merely a way to catch calculator mistakes. Because the model has specified constant power, summing smaller time intervals would agree with the single multiplication. If the power changed, this shortcut would need replacing by separate interval calculations. In the final sentence, avoid judging total energy from watts alone; only then could an examiner know that the chosen formula genuinely matches the data instead of an accidental numerical coincidence.
Question 6: slow comparison run
A paper model for the slow comparison run gives a steady electrical power of 4 W for 240 seconds. Required: total input energy, not water temperature. The working is E = Pt = 4 × 240 = 960.0 J. A useful verification is to divide the answer by the stated time: the quotient recovers 4 W. That is a unit-and-meaning check, not merely a way to catch calculator mistakes. Because the model has specified constant power, summing smaller time intervals would agree with the single multiplication. If the power changed, this shortcut would need replacing by separate interval calculations. In the final sentence, verify whether the model assumes power stays constant; only then could an examiner know that the chosen formula genuinely matches the data instead of an accidental numerical coincidence.
Worked questions 7–12: recover missing mass from target gain
Question 7: a requested sample mass
The a requested sample mass is specified by a water-energy gain of 840 J and a temperature rise of 2 °C. In the given model Q = 4200mΔT. Solve for mass: m = 840/(4200 × 2) = 0.100 kg, or 100 g. The last conversion makes the scale of the sample easier to understand, but do not put grams into the formula while keeping a per-kilogram heat capacity. Check by multiplying the recovered mass by 4200 and 2: the result returns 840 J. If the question asks for a theoretical maximum mass from an electrical supply, it would need a stated assumption about how much of the input reaches the water. Here the unknown is determined only from the given thermal gain, and the algebra should not silently invent a conversion efficiency.
Question 8: a double-check of mass
The a double-check of mass is specified by a water-energy gain of 1260 J and a temperature rise of 3 °C. In the given model Q = 4200mΔT. Solve for mass: m = 1260/(4200 × 3) = 0.100 kg, or 100 g. The last conversion makes the scale of the sample easier to understand, but do not put grams into the formula while keeping a per-kilogram heat capacity. Check by multiplying the recovered mass by 4200 and 3: the result returns 1260 J. If the question asks for a theoretical maximum mass from an electrical supply, it would need a stated assumption about how much of the input reaches the water. Here the unknown is determined only from the given thermal gain, and the algebra should not silently invent a conversion efficiency.
Question 9: a sample-size proposal
The a sample-size proposal is specified by a water-energy gain of 2100 J and a temperature rise of 2.5 °C. In the given model Q = 4200mΔT. Solve for mass: m = 2100/(4200 × 2.5) = 0.200 kg, or 200 g. The last conversion makes the scale of the sample easier to understand, but do not put grams into the formula while keeping a per-kilogram heat capacity. Check by multiplying the recovered mass by 4200 and 2.5: the result returns 2100 J. If the question asks for a theoretical maximum mass from an electrical supply, it would need a stated assumption about how much of the input reaches the water. Here the unknown is determined only from the given thermal gain, and the algebra should not silently invent a conversion efficiency.
Question 10: a reverse planned trial
The a reverse planned trial is specified by a water-energy gain of 1890 J and a temperature rise of 1.5 °C. In the given model Q = 4200mΔT. Solve for mass: m = 1890/(4200 × 1.5) = 0.300 kg, or 300 g. The last conversion makes the scale of the sample easier to understand, but do not put grams into the formula while keeping a per-kilogram heat capacity. Check by multiplying the recovered mass by 4200 and 1.5: the result returns 1890 J. If the question asks for a theoretical maximum mass from an electrical supply, it would need a stated assumption about how much of the input reaches the water. Here the unknown is determined only from the given thermal gain, and the algebra should not silently invent a conversion efficiency.
Question 11: a display-test estimate
The a display-test estimate is specified by a water-energy gain of 2520 J and a temperature rise of 4 °C. In the given model Q = 4200mΔT. Solve for mass: m = 2520/(4200 × 4) = 0.150 kg, or 150 g. The last conversion makes the scale of the sample easier to understand, but do not put grams into the formula while keeping a per-kilogram heat capacity. Check by multiplying the recovered mass by 4200 and 4: the result returns 2520 J. If the question asks for a theoretical maximum mass from an electrical supply, it would need a stated assumption about how much of the input reaches the water. Here the unknown is determined only from the given thermal gain, and the algebra should not silently invent a conversion efficiency.
Question 12: a constrained laboratory schedule
The a constrained laboratory schedule is specified by a water-energy gain of 1575 J and a temperature rise of 2.5 °C. In the given model Q = 4200mΔT. Solve for mass: m = 1575/(4200 × 2.5) = 0.150 kg, or 150 g. The last conversion makes the scale of the sample easier to understand, but do not put grams into the formula while keeping a per-kilogram heat capacity. Check by multiplying the recovered mass by 4200 and 2.5: the result returns 1575 J. If the question asks for a theoretical maximum mass from an electrical supply, it would need a stated assumption about how much of the input reaches the water. Here the unknown is determined only from the given thermal gain, and the algebra should not silently invent a conversion efficiency.
Worked questions 13–18: recover temperature change instead of mass
Question 13: small water sample
A small water sample has mass 0.1 kg and a recorded gain of 840 J. Find the mean water-temperature change. Rearrange Q = mcΔT to ΔT = Q/(mc). Substitution gives 840/(0.1 × 4200) = 2.00 °C. The number is a change in temperature, not a predicted final temperature. To find the final temperature you would also need the initial reading; if it were 23.0 °C, the model would give 25.00 °C. Do not invent that initial temperature in an answer unless the task supplies it. Explain why a change in the mass would alter the result for the same Q: the same energy is shared among a different quantity of water. In a multi-stage question, preserve the units throughout the reverse division.
Question 14: medium sample
A medium sample has mass 0.2 kg and a recorded gain of 1680 J. Find the mean water-temperature change. Rearrange Q = mcΔT to ΔT = Q/(mc). Substitution gives 1680/(0.2 × 4200) = 2.00 °C. The number is a change in temperature, not a predicted final temperature. To find the final temperature you would also need the initial reading; if it were 23.0 °C, the model would give 25.00 °C. Do not invent that initial temperature in an answer unless the task supplies it. Explain why a change in the mass would alter the result for the same Q: the same energy is shared among a different quantity of water. In a multi-stage question, preserve the units throughout the reverse division.
Question 15: intermediate reading
A intermediate reading has mass 0.15 kg and a recorded gain of 1890 J. Find the mean water-temperature change. Rearrange Q = mcΔT to ΔT = Q/(mc). Substitution gives 1890/(0.15 × 4200) = 3.00 °C. The number is a change in temperature, not a predicted final temperature. To find the final temperature you would also need the initial reading; if it were 23.0 °C, the model would give 26.00 °C. Do not invent that initial temperature in an answer unless the task supplies it. Explain why a change in the mass would alter the result for the same Q: the same energy is shared among a different quantity of water. In a multi-stage question, preserve the units throughout the reverse division.
Question 16: large comparison sample
A large comparison sample has mass 0.25 kg and a recorded gain of 2100 J. Find the mean water-temperature change. Rearrange Q = mcΔT to ΔT = Q/(mc). Substitution gives 2100/(0.25 × 4200) = 2.00 °C. The number is a change in temperature, not a predicted final temperature. To find the final temperature you would also need the initial reading; if it were 23.0 °C, the model would give 25.00 °C. Do not invent that initial temperature in an answer unless the task supplies it. Explain why a change in the mass would alter the result for the same Q: the same energy is shared among a different quantity of water. In a multi-stage question, preserve the units throughout the reverse division.
Question 17: small precision trial
A small precision trial has mass 0.075 kg and a recorded gain of 1260 J. Find the mean water-temperature change. Rearrange Q = mcΔT to ΔT = Q/(mc). Substitution gives 1260/(0.075 × 4200) = 4.00 °C. The number is a change in temperature, not a predicted final temperature. To find the final temperature you would also need the initial reading; if it were 23.0 °C, the model would give 27.00 °C. Do not invent that initial temperature in an answer unless the task supplies it. Explain why a change in the mass would alter the result for the same Q: the same energy is shared among a different quantity of water. In a multi-stage question, preserve the units throughout the reverse division.
Question 18: three-tenths kilogram
A three-tenths kilogram has mass 0.3 kg and a recorded gain of 3780 J. Find the mean water-temperature change. Rearrange Q = mcΔT to ΔT = Q/(mc). Substitution gives 3780/(0.3 × 4200) = 3.00 °C. The number is a change in temperature, not a predicted final temperature. To find the final temperature you would also need the initial reading; if it were 23.0 °C, the model would give 26.00 °C. Do not invent that initial temperature in an answer unless the task supplies it. Explain why a change in the mass would alter the result for the same Q: the same energy is shared among a different quantity of water. In a multi-stage question, preserve the units throughout the reverse division.
Worked questions 19–24: denominator control and reverse percentages
Question 19: an 80% water-gain model
The an 80% water-gain model gives 2200 J as an electrical-input model and 1760 J as the estimated water gain. The fraction of input represented by the gain is 1760/2200 = 0.800; as a percentage it is 80.0%. The numerical difference is 440 J. Suppose the water-gain value is given along with that percentage, but input is hidden. Reverse the process by dividing 1760 by 0.800 to recover 2200 J. Do not add 20.0% to 1760: the missing proportion was defined relative to the original input, not the final amount. This algebra is useful for many non-science contexts too. Finish by saying that the percentage describes the stated model, not a proven route for every joule in an actual device.
Question 20: a three-quarter model
The a three-quarter model gives 1800 J as an electrical-input model and 1350 J as the estimated water gain. The fraction of input represented by the gain is 1350/1800 = 0.750; as a percentage it is 75.0%. The numerical difference is 450 J. Suppose the water-gain value is given along with that percentage, but input is hidden. Reverse the process by dividing 1350 by 0.750 to recover 1800 J. Do not add 25.0% to 1350: the missing proportion was defined relative to the original input, not the final amount. This algebra is useful for many non-science contexts too. Finish by saying that the percentage describes the stated model, not a proven route for every joule in an actual device.
Question 21: a larger trial
The a larger trial gives 3000 J as an electrical-input model and 2400 J as the estimated water gain. The fraction of input represented by the gain is 2400/3000 = 0.800; as a percentage it is 80.0%. The numerical difference is 600 J. Suppose the water-gain value is given along with that percentage, but input is hidden. Reverse the process by dividing 2400 by 0.800 to recover 3000 J. Do not add 20.0% to 2400: the missing proportion was defined relative to the original input, not the final amount. This algebra is useful for many non-science contexts too. Finish by saying that the percentage describes the stated model, not a proven route for every joule in an actual device.
Question 22: a lower-ratio run
The a lower-ratio run gives 2400 J as an electrical-input model and 1680 J as the estimated water gain. The fraction of input represented by the gain is 1680/2400 = 0.700; as a percentage it is 70.0%. The numerical difference is 720 J. Suppose the water-gain value is given along with that percentage, but input is hidden. Reverse the process by dividing 1680 by 0.700 to recover 2400 J. Do not add 30.0% to 1680: the missing proportion was defined relative to the original input, not the final amount. This algebra is useful for many non-science contexts too. Finish by saying that the percentage describes the stated model, not a proven route for every joule in an actual device.
Question 23: a high apparent fraction
The a high apparent fraction gives 4000 J as an electrical-input model and 3600 J as the estimated water gain. The fraction of input represented by the gain is 3600/4000 = 0.900; as a percentage it is 90.0%. The numerical difference is 400 J. Suppose the water-gain value is given along with that percentage, but input is hidden. Reverse the process by dividing 3600 by 0.900 to recover 4000 J. Do not add 10.0% to 3600: the missing proportion was defined relative to the original input, not the final amount. This algebra is useful for many non-science contexts too. Finish by saying that the percentage describes the stated model, not a proven route for every joule in an actual device.
Question 24: a first-day test
The a first-day test gives 1600 J as an electrical-input model and 1120 J as the estimated water gain. The fraction of input represented by the gain is 1120/1600 = 0.700; as a percentage it is 70.0%. The numerical difference is 480 J. Suppose the water-gain value is given along with that percentage, but input is hidden. Reverse the process by dividing 1120 by 0.700 to recover 1600 J. Do not add 30.0% to 1120: the missing proportion was defined relative to the original input, not the final amount. This algebra is useful for many non-science contexts too. Finish by saying that the percentage describes the stated model, not a proven route for every joule in an actual device.
Worked questions 25–30: find the feasible number of complete runs
Question 25: energy per trial and limited supply
A fictional school schedule uses 1350 J per complete modelled run, with at most 6000 J available for energy per trial and limited supply. The inequality is 1350n ≤ 6000. Division gives n ≤ 4.444, but n is an integer count of completed trials. Therefore at most 4 complete runs fit the energy budget, using 5400 J and leaving 600 J. If every run also takes 28 minutes, these 4 runs occupy 112 minutes excluding setup and breaks. The energy result alone does not prove the schedule fits a separate time limit; test the time inequality if one is provided. The examiner is looking for both the mathematical boundary and an interpretation of the number. The unused energy does not authorise a partial run when the question requires completion.
Question 26: fast models with a budget
A fictional school schedule uses 900 J per complete modelled run, with at most 5000 J available for fast models with a budget. The inequality is 900n ≤ 5000. Division gives n ≤ 5.556, but n is an integer count of completed trials. Therefore at most 5 complete runs fit the energy budget, using 4500 J and leaving 500 J. If every run also takes 26 minutes, these 5 runs occupy 130 minutes excluding setup and breaks. The energy result alone does not prove the schedule fits a separate time limit; test the time inequality if one is provided. The examiner is looking for both the mathematical boundary and an interpretation of the number. The unused energy does not authorise a partial run when the question requires completion.
Question 27: high-input demonstrations
A fictional school schedule uses 1600 J per complete modelled run, with at most 8500 J available for high-input demonstrations. The inequality is 1600n ≤ 8500. Division gives n ≤ 5.313, but n is an integer count of completed trials. Therefore at most 5 complete runs fit the energy budget, using 8000 J and leaving 500 J. If every run also takes 24 minutes, these 5 runs occupy 120 minutes excluding setup and breaks. The energy result alone does not prove the schedule fits a separate time limit; test the time inequality if one is provided. The examiner is looking for both the mathematical boundary and an interpretation of the number. The unused energy does not authorise a partial run when the question requires completion.
Question 28: short repeated lessons
A fictional school schedule uses 750 J per complete modelled run, with at most 3800 J available for short repeated lessons. The inequality is 750n ≤ 3800. Division gives n ≤ 5.067, but n is an integer count of completed trials. Therefore at most 5 complete runs fit the energy budget, using 3750 J and leaving 50 J. If every run also takes 19 minutes, these 5 runs occupy 95 minutes excluding setup and breaks. The energy result alone does not prove the schedule fits a separate time limit; test the time inequality if one is provided. The examiner is looking for both the mathematical boundary and an interpretation of the number. The unused energy does not authorise a partial run when the question requires completion.
Question 29: larger sample runs
A fictional school schedule uses 2100 J per complete modelled run, with at most 10000 J available for larger sample runs. The inequality is 2100n ≤ 10000. Division gives n ≤ 4.762, but n is an integer count of completed trials. Therefore at most 4 complete runs fit the energy budget, using 8400 J and leaving 1600 J. If every run also takes 35 minutes, these 4 runs occupy 140 minutes excluding setup and breaks. The energy result alone does not prove the schedule fits a separate time limit; test the time inequality if one is provided. The examiner is looking for both the mathematical boundary and an interpretation of the number. The unused energy does not authorise a partial run when the question requires completion.
Question 30: an uneven final block
A fictional school schedule uses 1200 J per complete modelled run, with at most 6700 J available for an uneven final block. The inequality is 1200n ≤ 6700. Division gives n ≤ 5.583, but n is an integer count of completed trials. Therefore at most 5 complete runs fit the energy budget, using 6000 J and leaving 700 J. If every run also takes 22 minutes, these 5 runs occupy 110 minutes excluding setup and breaks. The energy result alone does not prove the schedule fits a separate time limit; test the time inequality if one is provided. The examiner is looking for both the mathematical boundary and an interpretation of the number. The unused energy does not authorise a partial run when the question requires completion.
Worked questions 31–36: interpret temperature tables and average interval rates
Question 31: a steadily warming sample
A paper-based calibration table reads 0s: 22.0°C; 20s: 23.0°C; 40s: 24.1°C; 60s: 25.0°C. The overall change is 3.00 °C across 60 seconds, so the average rate of change is 3.00/60 = 0.0500 °C/s. It is not necessary that each shorter interval has this rate. Calculate adjacent differences if the question asks where warming was fastest, rather than assuming a constant gradient. A scientifically cautious answer notes that differences of 1.0, 1.1 and 0.9 °C. For a Mathematics response, label axes as time in seconds and temperature in degrees Celsius, choose an appropriate scale and clearly distinguish recorded points from joined segments. A straight line drawn through endpoints is a model, not evidence that every intermediate instant followed it.
Question 32: a slower second interval
A paper-based calibration table reads 0s: 23.0°C; 30s: 23.8°C; 60s: 24.2°C; 90s: 25.0°C. The overall change is 2.00 °C across 90 seconds, so the average rate of change is 2.00/90 = 0.0222 °C/s. It is not necessary that each shorter interval has this rate. Calculate adjacent differences if the question asks where warming was fastest, rather than assuming a constant gradient. A scientifically cautious answer notes that a rate need not be constant. For a Mathematics response, label axes as time in seconds and temperature in degrees Celsius, choose an appropriate scale and clearly distinguish recorded points from joined segments. A straight line drawn through endpoints is a model, not evidence that every intermediate instant followed it.
Question 33: a short observation
A paper-based calibration table reads 0s: 21.0°C; 15s: 21.6°C; 30s: 22.3°C; 45s: 23.1°C. The overall change is 2.10 °C across 45 seconds, so the average rate of change is 2.10/45 = 0.0467 °C/s. It is not necessary that each shorter interval has this rate. Calculate adjacent differences if the question asks where warming was fastest, rather than assuming a constant gradient. A scientifically cautious answer notes that the graph can have changing slopes. For a Mathematics response, label axes as time in seconds and temperature in degrees Celsius, choose an appropriate scale and clearly distinguish recorded points from joined segments. A straight line drawn through endpoints is a model, not evidence that every intermediate instant followed it.
Question 34: a near-flat middle passage
A paper-based calibration table reads 0s: 24.0°C; 25s: 24.5°C; 50s: 25.3°C; 75s: 25.9°C. The overall change is 1.90 °C across 75 seconds, so the average rate of change is 1.90/75 = 0.0253 °C/s. It is not necessary that each shorter interval has this rate. Calculate adjacent differences if the question asks where warming was fastest, rather than assuming a constant gradient. A scientifically cautious answer notes that avoid assuming a straight-line trend from endpoints. For a Mathematics response, label axes as time in seconds and temperature in degrees Celsius, choose an appropriate scale and clearly distinguish recorded points from joined segments. A straight line drawn through endpoints is a model, not evidence that every intermediate instant followed it.
Question 35: a possible recording irregularity
A paper-based calibration table reads 0s: 22.0°C; 20s: 22.9°C; 40s: 23.0°C; 60s: 24.2°C. The overall change is 2.20 °C across 60 seconds, so the average rate of change is 2.20/60 = 0.0367 °C/s. It is not necessary that each shorter interval has this rate. Calculate adjacent differences if the question asks where warming was fastest, rather than assuming a constant gradient. A scientifically cautious answer notes that repeat rather than quietly delete a point. For a Mathematics response, label axes as time in seconds and temperature in degrees Celsius, choose an appropriate scale and clearly distinguish recorded points from joined segments. A straight line drawn through endpoints is a model, not evidence that every intermediate instant followed it.
Question 36: a fine-time series
A paper-based calibration table reads 0s: 20.0°C; 10s: 20.7°C; 20s: 21.5°C; 30s: 22.2°C. The overall change is 2.20 °C across 30 seconds, so the average rate of change is 2.20/30 = 0.0733 °C/s. It is not necessary that each shorter interval has this rate. Calculate adjacent differences if the question asks where warming was fastest, rather than assuming a constant gradient. A scientifically cautious answer notes that round the interval rate appropriately. For a Mathematics response, label axes as time in seconds and temperature in degrees Celsius, choose an appropriate scale and clearly distinguish recorded points from joined segments. A straight line drawn through endpoints is a model, not evidence that every intermediate instant followed it.
Worked questions 37–42: diagnose six common reasoning traps
Question 37: a tempting but wrong shortcut
A learner says that 20% reduction means subtract one fifth of the new amount. The first repair is practical: Write a before-and-after table with original = 100% and new = 80%. The central mathematical principle is simple: Reverse the reduction by dividing the new amount by 0.80. Show how the method would work on a tiny invented example, then translate the answer into the original question’s units and constraints. In a worked examination answer, this diagnostic reasoning makes the correction visible. Saying ‘check your work’ is not enough, because it does not identify what went wrong. An effective self-check asks whether a number increased when it should have decreased, whether a unit changed halfway through a line, or whether a rounded decision exceeds the allowed limit. Repeated practice should build this reasoning habit, not just calculator speed.
Question 38: a tempting but wrong shortcut
A learner says 4.7 completed experiments can fit in the timetable. The first repair is practical: State that each trial must be finished and count whole intervals. The central mathematical principle is simple: Use the greatest integer satisfying every budget or time inequality. Show how the method would work on a tiny invented example, then translate the answer into the original question’s units and constraints. In a worked examination answer, this diagnostic reasoning makes the correction visible. Saying ‘check your work’ is not enough, because it does not identify what went wrong. An effective self-check asks whether a number increased when it should have decreased, whether a unit changed halfway through a line, or whether a rounded decision exceeds the allowed limit. Repeated practice should build this reasoning habit, not just calculator speed.
Question 39: a tempting but wrong shortcut
A learner compares a 200 g sample with a 100 g sample using only the temperature rise. The first repair is practical: List the masses beside their changes in temperature. The central mathematical principle is simple: Normalise only when the question’s comparison objective supports doing so. Show how the method would work on a tiny invented example, then translate the answer into the original question’s units and constraints. In a worked examination answer, this diagnostic reasoning makes the correction visible. Saying ‘check your work’ is not enough, because it does not identify what went wrong. An effective self-check asks whether a number increased when it should have decreased, whether a unit changed halfway through a line, or whether a rounded decision exceeds the allowed limit. Repeated practice should build this reasoning habit, not just calculator speed.
Question 40: a tempting but wrong shortcut
A learner adds watts to joules to calculate a total. The first repair is practical: Place units after every numerical term. The central mathematical principle is simple: Multiply power by time to obtain energy before adding compatible energy quantities. Show how the method would work on a tiny invented example, then translate the answer into the original question’s units and constraints. In a worked examination answer, this diagnostic reasoning makes the correction visible. Saying ‘check your work’ is not enough, because it does not identify what went wrong. An effective self-check asks whether a number increased when it should have decreased, whether a unit changed halfway through a line, or whether a rounded decision exceeds the allowed limit. Repeated practice should build this reasoning habit, not just calculator speed.
Question 41: a tempting but wrong shortcut
A learner rounds a required sample mass upward before checking a fixed energy cap. The first repair is practical: Substitute the rounded value into the original inequality. The central mathematical principle is simple: A rounded mass may violate the condition even when the unrounded calculation did not. Show how the method would work on a tiny invented example, then translate the answer into the original question’s units and constraints. In a worked examination answer, this diagnostic reasoning makes the correction visible. Saying ‘check your work’ is not enough, because it does not identify what went wrong. An effective self-check asks whether a number increased when it should have decreased, whether a unit changed halfway through a line, or whether a rounded decision exceeds the allowed limit. Repeated practice should build this reasoning habit, not just calculator speed.
Question 42: a tempting but wrong shortcut
A learner claims a correlation graph proves insulation caused all improvements. The first repair is practical: Distinguish the equation model from the causal experiment. The central mathematical principle is simple: A mathematical trend can support a prediction without isolating a physical cause. Show how the method would work on a tiny invented example, then translate the answer into the original question’s units and constraints. In a worked examination answer, this diagnostic reasoning makes the correction visible. Saying ‘check your work’ is not enough, because it does not identify what went wrong. An effective self-check asks whether a number increased when it should have decreased, whether a unit changed halfway through a line, or whether a rounded decision exceeds the allowed limit. Repeated practice should build this reasoning habit, not just calculator speed.
A full Paper 2-style decision task and solution
A proposed exhibition uses 1800 J of modelled electrical input per experiment. The experiment reports that 75% of this input appears as measured thermal gain by a specified water sample. The exhibition has an energy allowance of 8000 J and a total teaching-window limit of 70 minutes. Each experiment takes 18 minutes, and each one must be complete. Question (a): Find the gain per experiment. It is 0.75 × 1800 = 1350 J. Question (b): Find the maximum count by energy. 8000/1800 = 4.444…, so four. Question (c): Find the maximum count by time. 70/18 = 3.888…, so three. The combined feasible count is three, using 5400 J and 54 minutes. The tighter constraint is time.
Question (d): If the water mass for each run is 0.150 kg and c = 4200 J/(kg °C), find the modelled temperature rise. It is 1350/(0.150 × 4200) = 2.142857…, or approximately 2.14 °C. Question (e): A publicity note says the design ‘guarantees four demonstrations in a 70-minute visit’. Explain the error. Four fit the energy allowance, but 4 × 18 = 72 minutes exceeds the teaching-window limit, even before setup. The correct recommendation is three complete runs under these assumptions or a revised programme subject to a separately validated schedule.
For assessment, notice where the marks could be lost: an unexplained 75 instead of 0.75, a count of 3.9 experiments, omission of the 70-minute limit, changing the percentage base halfway, and offering a public claim that conflicts with the calculation. A good final sentence does not repeat all the working; it states the workable result and the condition it depends on. If new constraints arrive, solve the new problem rather than protecting the answer you already prefer.
Ten-minute retrieval practice
Retrieval 1. Explain why a percentage gain should be divided by the original denominator. After answering, check the relationship between the calculation and its context. If you do not know a variable, write an equation with that variable rather than inserting a guessed value. The strongest learner can explain why a method is valid, and also name a condition under which that method would no longer be valid.
Retrieval 2. Find the mass warmed by 1260 J through 3 °C with c = 4200 J/(kg °C). After answering, check the relationship between the calculation and its context. If you do not know a variable, write an equation with that variable rather than inserting a guessed value. The strongest learner can explain why a method is valid, and also name a condition under which that method would no longer be valid.
Retrieval 3. Recover an original input after a 15% decrease leaves 1700 J. After answering, check the relationship between the calculation and its context. If you do not know a variable, write an equation with that variable rather than inserting a guessed value. The strongest learner can explain why a method is valid, and also name a condition under which that method would no longer be valid.
Retrieval 4. State why an experiment count obtained as 5.9 must be interpreted as five if completion is mandatory. After answering, check the relationship between the calculation and its context. If you do not know a variable, write an equation with that variable rather than inserting a guessed value. The strongest learner can explain why a method is valid, and also name a condition under which that method would no longer be valid.
Retrieval 5. Describe one case where a larger wattage could supply less total energy than a smaller wattage. After answering, check the relationship between the calculation and its context. If you do not know a variable, write an equation with that variable rather than inserting a guessed value. The strongest learner can explain why a method is valid, and also name a condition under which that method would no longer be valid.
Retrieval 6. Explain why a maximum-energy condition does not prove a maximum-time condition is satisfied. After answering, check the relationship between the calculation and its context. If you do not know a variable, write an equation with that variable rather than inserting a guessed value. The strongest learner can explain why a method is valid, and also name a condition under which that method would no longer be valid.
Retrieval 7. Calculate average rate between two temperatures and distinguish it from an instantaneous rate. After answering, check the relationship between the calculation and its context. If you do not know a variable, write an equation with that variable rather than inserting a guessed value. The strongest learner can explain why a method is valid, and also name a condition under which that method would no longer be valid.
Retrieval 8. Identify the equation when the mass is known but the temperature rise is unknown. After answering, check the relationship between the calculation and its context. If you do not know a variable, write an equation with that variable rather than inserting a guessed value. The strongest learner can explain why a method is valid, and also name a condition under which that method would no longer be valid.
Retrieval 9. Say how you would test a second constraint after finding a first feasible value. After answering, check the relationship between the calculation and its context. If you do not know a variable, write an equation with that variable rather than inserting a guessed value. The strongest learner can explain why a method is valid, and also name a condition under which that method would no longer be valid.
Retrieval 10. Write one sentence that turns a precise model calculation into a qualified public conclusion. After answering, check the relationship between the calculation and its context. If you do not know a variable, write an equation with that variable rather than inserting a guessed value. The strongest learner can explain why a method is valid, and also name a condition under which that method would no longer be valid.
The route onward
Use this chapter for reverse calculations and constrained planning. Return to the physics chapter for the meaning of thermal data, and to Volume 0455 for an earlier example of graph intervals and thresholds. The speed–distance–time guide offers an accessible route back to rate language. The next volume will examine how a reader encounters claims in a visual text and decides whether its qualifying words match the numbers. For current examination rules, check the official K310 syllabus.
