2023 CCEA GCSE Science Single Award 1310 模拟试题及答案详解

(a) The fractions are separated by fractional distillation.
(b) As the number of carbon atoms in a fraction increases, its boiling point also increases: fraction W (C5–C10) boils over 40–100 °C, while fraction Z (over C25) boils above 350 °C, so boiling point rises as chain length rises.
(c) The hydrocarbons in fraction W belong to the alkanes, general formula \( C_nH_{2n+2} \).
(d) Balancing: 8 carbons need 8 CO2, 18 hydrogens need 9 H2O, giving 16 O atoms (from CO2) + 9 O atoms (from H2O) = 25 O atoms per octane molecule, so 12.5 O2; doubling everything to clear the fraction: \( 2C_8H_{18}(l) + 25O_2(g) \rightarrow 16CO_2(g) + 18H_2O(g) \). Check: C: 16=16; H: 36=36; O: 50=50. Balanced.
Final answer: fractional distillation; alkanes, \( C_nH_{2n+2} \); \( 2C_8H_{18}(l) + 25O_2(g) \rightarrow 16CO_2(g) + 18H_2O(g) \).

(a) [1] fractional distillation.
(b) [2] 1 mark: correct trend stated (more carbon atoms → higher boiling point); 1 mark: trend supported with data quoted from the table.
(c) [2] 1 mark: alkanes; 1 mark: \( C_nH_{2n+2} \) (accept CnH2n+2).
(d) [2] 1 mark: correct formulae and balancing (2C8H18 + 25O2 → 16CO2 + 18H2O, or any correctly balanced multiple); 1 mark: correct state symbols (l), (g), (g), (g). Accept H2O(l) with ECF only if consistent; reject unbalanced equations.

(a) The gas is hydrogen. Test: hold a lighted splint to the mouth of the test tube; a squeaky pop confirms hydrogen.
(b) Q reacts violently even with cold water, so it is most reactive; P reacts rapidly with acid but not water, so it is next; R reacts only slowly with acid, so it is less reactive than P; S does not react with either, so it is least reactive. Order: Q, P, R, S.
(c) S is unreactive with water and dilute acid, so it lies below hydrogen in the reactivity series (it is copper). Copper is still more reactive than silver, so it can displace silver from silver nitrate solution: this is a displacement reaction. Word equation: copper + silver nitrate → copper nitrate + silver.
(d) The reaction between metal P and hydrochloric acid is exothermic, so it releases heat energy to the surroundings (the test tube and its contents), raising the temperature.
(e) Q reacts vigorously with cold water, floating and igniting — this is characteristic behaviour of a Group 1 alkali metal (e.g. sodium or potassium).
Final answer: hydrogen (pop test); order Q > P > R > S; displacement reaction, copper + silver nitrate → copper nitrate + silver; exothermic; Group 1.

(a) [2] 1 mark: hydrogen; 1 mark: lighted splint gives a pop/squeaky pop.
(b) [1] correct order Q, P, R, S (all four correct for the mark).
(c) [2] 1 mark: displacement (reaction); 1 mark: correct word equation copper + silver nitrate → copper nitrate + silver (accept copper(II) nitrate).
(d) [1] exothermic reaction / heat energy released to surroundings.
(e) [1] Group 1 (alkali metals), with valid reasoning (vigorous reaction with cold water, catches fire); accept 'reactive metal group' only if Group 1 stated.

(a) The pH reaches exactly 7 (neutral) when 20 cm³ of acid has been added, so 20 cm³ is needed for exact neutralisation.
(b) At 0 cm³ added, pH = 13, which falls in the range pH 12–14, so the sodium hydroxide solution is a strong alkali.
(c) Sodium hydroxide reacts with hydrochloric acid to form the salt sodium chloride and water: \( NaOH(aq) + HCl(aq) \rightarrow NaCl(aq) + H_2O(l) \). Check: 1 Na, 1 O(from OH)+... atoms balance 1:1:1:1 on both sides, so the equation is already balanced.
(d) The acid supplies H+ ions and the alkali supplies OH- ions. These combine to form neutral water molecules: \( H^+(aq) + OH^-(aq) \rightarrow H_2O(l) \). Once the H+ and OH- ions have fully reacted, the solution is neither acidic nor alkaline, i.e. neutralised.
Final answer: 20 cm³; strong alkali; sodium chloride, \( NaOH(aq)+HCl(aq) \rightarrow NaCl(aq)+H_2O(l) \); H+ + OH- → H2O.

(a) [1] 20 cm³.
(b) [1] strong alkali (pH 12–14).
(c) [3] 1 mark: sodium chloride named; 1 mark: correct formulae NaOH + HCl → NaCl + H2O; 1 mark: correct state symbols (aq)(aq)→(aq)(l) and equation balanced.
(d) [2] 1 mark: H+ ions from acid combine with OH- ions from alkali to form water (ionic equation H+ + OH- → H2O accepted for the mark); 1 mark: explanation that this removes the excess ions responsible for acidity/alkalinity, giving a neutral solution.

(a) Comparing densities (2.7, 1.2, 1.05, 1.8 g/cm³), ABS plastic at 1.05 g/cm³ is the lowest.
(b) Only aluminium and graphene-coated polymer are good electrical conductors (needed for an antenna). Since cost is not the deciding factor here and a strong conductor is wanted, graphene-coated polymer is suitable, as graphene is known to be very strong as well as an excellent electrical conductor.
(c) A nanomaterial contains particles between 1 and 100 nm in size, where 1 nm = \( 1\times10^{-9} \) m.
(d) Nanoparticles are small enough to potentially enter cells in the body, so a possible risk is cell damage; there may also be unknown harmful effects on the environment.
(e) Unlike the materials in the table (whose properties are fixed), a smart material's properties (e.g. colour, for a thermochromic pigment) change reversibly in response to a change in its surroundings (e.g. temperature).
Final answer: ABS; graphene-coated polymer (strong + conducting); 1–100 nm particles; cell damage/environmental risk; properties change with surroundings.

(a) [1] ABS (standard plastic).
(b) [2] 1 mark: graphene-coated polymer identified; 1 mark: justified using conductivity data (and/or strength of graphene).
(c) [2] 1 mark: particle size 1–100 nm stated; 1 mark: correct unit/definition of nm (1×10⁻⁹ m).
(d) [1] any valid risk, e.g. potential cell damage in the body, or harmful/unknown effects on the environment.
(e) [1] properties change depending on a change in the surroundings (accept named example, e.g. colour changes with temperature).

Aluminium is extracted from the ore bauxite, which is purified to give aluminium oxide (alumina). Pure aluminium oxide has a very high melting point, so it is dissolved in molten cryolite to lower the operating temperature (and therefore the energy cost) while still allowing the mixture to conduct electricity, since electrolysis requires freely-moving ions.
At the negative electrode (cathode), positively charged aluminium ions are reduced (gain electrons): \( Al^{3+} + 3e^- \rightarrow Al \); the molten aluminium formed sinks and is tapped off.
At the positive electrode (anode), negatively charged oxide ions are oxidised (lose electrons): \( 2O^{2-} \rightarrow O_2 + 4e^- \).
The oxygen gas produced reacts with the hot carbon anodes, burning them away as carbon dioxide, so the anodes must be replaced periodically.
A continuous large direct current must be passed through the cell for the whole process, which is why aluminium extraction uses very large amounts of electrical energy.
Recycling aluminium only requires melting and reshaping the metal, not electrolysis, so it uses only a fraction of the energy needed to extract new aluminium from bauxite; it also saves the finite ore and reduces waste sent to landfill.
Final answer: bauxite is purified to aluminium oxide, dissolved in molten cryolite and electrolysed (Al³⁺+3e⁻→Al at the cathode, 2O²⁻→O2+4e⁻ at the anode); the carbon anodes burn away in the oxygen produced and must be replaced; recycling uses far less energy than extraction, which is why it is preferred.

Level of response marking (6 marks): Band A (5–6 marks): a full, logically ordered explanation covering at least 5–6 indicative points below with accurate use of specialist terms (electrolysis, electrode, ion, reduction/oxidation) and few errors. Band B (3–4 marks): a reasonably clear explanation covering 3–5 points with satisfactory use of specialist terms. Band C (1–2 marks): a basic, possibly disjointed answer covering 1–2 points with limited use of specialist terms. Band D (0 marks): no relevant content.
Indicative content: bauxite is the ore, purified to aluminium oxide (alumina); aluminium oxide is dissolved in molten cryolite to lower its melting point (rather than melting pure Al2O3, which would need far more energy); the mixture must be molten/dissolved so ions are free to move and carry charge; cathode (reduction): Al³⁺ + 3e⁻ → Al; anode (oxidation): 2O²⁻ → O2 + 4e⁻; oxygen produced reacts with the hot carbon anodes, burning them away, so anodes need regular replacement; the process needs a continuous large current, hence very high electrical energy use; recycling aluminium needs only melting, using a small fraction of the energy of extraction from ore, and saves ore/reduces waste.

(a) Magnesium's atomic number is 12, so a neutral atom has 12 protons and 12 electrons. Mass number = protons + neutrons, so neutrons = 24 − 12 = 12.
(b) With 12 electrons, the shells fill 2, 8, 2 (2+8+2=12, confirming the total). Magnesium has 2 electrons in its outer shell, so it is in Group 2; it has 3 occupied electron shells, so it is in Period 3 (elements in the same group have the same number of outer-shell electrons; elements in the same period have the same number of shells).
(c) A magnesium atom loses its 2 outer-shell electrons to achieve a full outer shell (like the noble gas neon), forming Mg²⁺. An oxygen atom (2,6) gains 2 electrons into its outer shell to become full (also like neon), forming O²⁻.
(d) The 2 outer-shell electrons of the magnesium atom (shown as dots) are transferred to the outer shell of the oxygen atom (shown as crosses), leaving Mg²⁺ with a full outer shell of 8 electrons (2,8) and O²⁻ with a full outer shell of 8 electrons (2,8); each ion would be drawn in its own square bracket with its charge shown outside. The ionic bond is the strong electrostatic force of attraction between the oppositely charged Mg²⁺ and O²⁻ ions.
Final answer: protons=12, neutrons=12, electrons=12; electronic structure 2,8,2, Group 2, Period 3; Mg loses 2 electrons to form Mg²⁺, O gains 2 electrons to form O²⁻; the ionic bond is the electrostatic attraction between Mg²⁺ and O²⁻.

(a) [3] 1 mark each: protons=12; neutrons=12; electrons=12.
(b) [3] 1 mark: electronic structure 2,8,2; 1 mark: Group 2 and Period 3 both correctly stated; 1 mark: correct reasoning linking outer electrons to group and number of shells to period.
(c) [2] 1 mark: Mg atom loses 2 electrons to form Mg²⁺ (full outer shell/noble gas structure); 1 mark: O atom gains 2 electrons to form O²⁻ (full outer shell/noble gas structure).
(d) [2] 1 mark: correct description of electron transfer from Mg to O (2 electrons moved, shown as dots/crosses, resulting ions each with full outer shells shown in brackets with charges); 1 mark: ionic bond correctly described as the electrostatic force of attraction between oppositely charged ions.

(a) calcium carbonate + hydrochloric acid → calcium chloride + water + carbon dioxide.
(b) The curve rises steeply at the start (fastest rate of gas production), then the gradient gradually decreases as the reaction proceeds (rate slows, as the concentration of acid falls), until the line becomes horizontal/flat from about 120 s onward (at 80 cm³), showing the reaction has finished (all the acid has reacted, since the marble is in excess).
(c) Mean rate = (volume of gas at 40 s − volume at 0 s) ÷ time = 42 cm³ ÷ 40 s = 1.05 cm³/s. Check by a second route: 1.05 cm³/s × 40 s = 42 cm³, matching the table value, so the calculation is confirmed correct.
(d) In the first 20 s, the 2.0 mol/dm³ acid produces 24 cm³ of gas compared with only 10 cm³ for the 1.0 mol/dm³ acid, so the more concentrated acid reacts faster from the start. This is because the 2.0 mol/dm³ acid contains more acid particles (H+ ions) in the same volume, so there are more frequent collisions between reacting particles per second, giving a higher frequency of successful collisions and therefore a faster rate of reaction.
(e) Doubling the concentration of the acid (while keeping its volume the same) doubles the number of moles (amount) of hydrochloric acid available to react. Since the marble chips were in excess in both experiments, all of the extra acid is able to react, so twice the amount, and therefore twice the volume, of carbon dioxide gas is produced before the reaction finishes.
(f) Repeat each experiment (at each concentration) and calculate a mean rate from the repeat readings, to improve reliability (or: ensure marble chips of the same size/surface area are used each time).
Final answer: calcium carbonate + hydrochloric acid → calcium chloride + water + carbon dioxide; steep-then-flattening curve reaching a plateau; mean rate = 1.05 cm³/s; higher concentration reacts faster due to more frequent particle collisions; doubling concentration doubles moles of acid so (with excess marble) doubles total gas produced; repeat and average to improve reliability.

(a) [1] correct word equation (all four species correct and in the right places).
(b) [2] 1 mark: steep/fast at the start, becoming less steep as the reaction proceeds; 1 mark: line becomes horizontal/flat (rate = zero) once the reaction is complete, with reference to the acid being used up.
(c) [2] 1 mark: correct method shown (change in volume ÷ change in time, using 42 cm³ and 40 s); 1 mark: correct final answer 1.05 cm³/s (accept equivalent, e.g. 1.05 cm³ s⁻¹; ECF from a mis-read table value with correct method).
(d) [3] 1 mark: correct comparative data quoted from the table (e.g. 24 cm³ vs 10 cm³ in the first 20 s, or any correctly matched pair showing 2.0 mol/dm³ faster); 1 mark: more particles (of acid/H+ ions) per unit volume at higher concentration; 1 mark: more frequent (successful) collisions between reacting particles, linked correctly to a faster rate.
(e) [2] 1 mark: doubling concentration (at constant volume) doubles the moles/amount of acid present; 1 mark: since marble is in excess, all the extra acid can react, so double the amount (volume) of CO2 gas is produced.
(f) [1] any valid method to improve reliability, e.g. repeat the experiment and calculate a mean, or standardise/control the size of the marble chips used.

(a) Rearranging \( v = f\lambda \) gives \( \lambda = v \div f = 340 \div 340 = 1.0 \) m. Check by a second route: \( f\lambda = 340 \times 1.0 = 340 \) m/s, matching the given speed, so the answer is confirmed.
(b) Sound is a longitudinal wave. In a longitudinal wave, the particles vibrate parallel to (along the same line as) the direction of energy transfer; in a transverse wave, the particles vibrate perpendicular (at right angles) to the direction of energy transfer.
(c) Since \( v = f\lambda \) and v is the same (constant) for both waves in a vacuum, f and \( \lambda \) must be inversely proportional: a lower frequency must go with a longer wavelength. So because a radio wave has a much lower frequency than a gamma ray, it must have a much longer wavelength.
(d) Gamma radiation is ionising and can damage or kill living cells, or cause mutations that lead to cancer.
Final answer: wavelength = 1.0 m; sound is longitudinal (particles vibrate parallel to energy transfer, unlike transverse where they vibrate perpendicular); radio waves have a much longer wavelength than gamma rays (since v is constant and f is much lower); gamma radiation can damage living cells/cause cancer.

(a) [2] 1 mark: correct rearrangement/substitution (\( \lambda = v/f \) or 340/340); 1 mark: correct answer 1.0 m with unit.
(b) [2] 1 mark: longitudinal correctly identified; 1 mark: correct comparison of particle vibration direction (parallel vs perpendicular to energy transfer).
(c) [2] 1 mark: recognises f and λ are inversely related when v is constant; 1 mark: correct conclusion that radio waves have a longer wavelength than gamma rays.
(d) [1] any valid danger, e.g. cell damage/death, causes cancer, causes mutation.

(a) In order of increasing wavelength (shortest to longest): X-rays, visible light, infrared, radio waves.
(b) Microwaves are absorbed by water molecules within the food. This absorbed energy increases the kinetic energy (vibration) of the water molecules, which raises the temperature of the food.
(c) The pulse travels down to the sea floor and back up in 0.80 s, so it covers a total distance of \( \text{distance} = \text{speed} \times \text{time} = 1500 \times 0.80 = 1200 \) m. Since this is a there-and-back journey, the depth is half this distance: \( 1200 \div 2 = 600 \) m. Check by a second route: one-way time = 0.80 ÷ 2 = 0.40 s, so depth = 1500 × 0.40 = 600 m — matches, confirming the answer.
Final answer: X-rays, visible light, infrared, radio waves; microwaves are absorbed by water molecules in the food, increasing their kinetic energy and heating the food; depth = 600 m.

(a) [1] correct full order (all four in the right sequence).
(b) [2] 1 mark: microwaves absorbed by water molecules in the food; 1 mark: this increases the molecules' kinetic energy/vibration, raising temperature.
(c) [3] 1 mark: use of distance = speed × time to find the total (there-and-back) distance; 1 mark: correct total distance of 1200 m; 1 mark: correct final depth of 600 m (halving the total distance), with unit. ECF applies if an arithmetic slip is carried through consistently.

(a) As speed increases, thinking distance increases in direct proportion: doubling the speed from 20 to 40 mph doubles the thinking distance from 6 m to 12 m, and tripling the speed to 60 mph triples it to 18 m.
(b) Braking distance increases much more steeply than in direct proportion to speed: doubling speed from 20 to 40 mph increases braking distance four-fold (6 m to 24 m), and increasing speed three-fold (20 to 60 mph) increases braking distance roughly nine-fold (6 m to 55 m). This is because braking distance depends on the car's kinetic energy, which increases with the square of speed, so much more energy (and distance) is needed to bring a faster car to rest.
(c) A wet, icy or loose road surface, or worn tyres/brakes, can increase braking distance (any one valid factor).
(d) Friction is the force between the tyres and the road surface that opposes the car's motion; it is this frictional force that decelerates the car when the brakes are applied. If the frictional force is reduced (e.g. on an icy road), there is less force available to slow the car down, so the braking distance increases.
Final answer: thinking distance is directly proportional to speed; braking distance rises much more steeply than speed because it depends on kinetic energy (∝ speed²); factors such as wet/icy roads or worn tyres/brakes increase it; friction provides the decelerating force, so reduced friction increases braking distance.

(a) [2] 1 mark: correct trend (thinking distance increases as speed increases); 1 mark: correct identification of direct proportionality, supported by data.
(b) [2] 1 mark: correct trend/data showing braking distance rises more steeply than speed; 1 mark: valid explanation linking to kinetic energy (∝ v²) or the greater energy/force needed to stop a faster car.
(c) [1] any one valid factor, e.g. wet/icy/loose road surface, worn tyres, worn brakes, vehicle mass/load.
(d) [2] 1 mark: friction identified as the (decelerating) force opposing motion between tyres and road; 1 mark: correct consequence that reduced friction increases braking distance.

(a) Acceleration = change in speed ÷ time taken = \( (24 - 0) \div 12 = 2 \) m/s². Check by a second route: \( 2 \text{ m/s}^2 \times 12\text{ s} = 24 \) m/s, matching the given final speed, confirming the answer.
(b) Resultant force = mass × acceleration = \( 1200 \times 2 = 2400 \) N. Check: \( 2400 \text{ N} \div 1200 \text{ kg} = 2 \) m/s², matching part (a), confirming the answer.
(c) The graph would be a horizontal (flat) straight line, because the speed is constant, so the gradient (which represents acceleration) is zero.
(d) The resultant force between 12 s and 20 s is 0 N. Because the car moves at a constant/steady speed in a straight line, the forces acting on it (the driving force forward and the resistive forces such as friction and air resistance backward) must be balanced (equal in size, opposite in direction), giving zero resultant force and therefore zero acceleration.
Final answer: acceleration = 2 m/s²; resultant force = 2400 N; the speed–time graph is a horizontal line between 12 s and 20 s; resultant force = 0 N because the forces on the car are balanced.

(a) [2] 1 mark: correct substitution into acceleration = change in speed ÷ time; 1 mark: correct answer 2 m/s² with unit.
(b) [2] 1 mark: correct substitution into resultant force = mass × acceleration; 1 mark: correct answer 2400 N with unit (ECF from part (a)).
(c) [1] horizontal/flat line described, with reference to zero gradient.
(d) [2] 1 mark: resultant force = 0 N; 1 mark: correct explanation that constant speed in a straight line means the forces are balanced.

(a) Half-life is the time taken for the activity (or the number of undecayed nuclei) of a radioactive source to fall to half of its original value.
(b) Activity falls from 800 to 100 counts per minute over 24 hours. Halving repeatedly: 800 → 400 (after 1 half-life) → 200 (after 2 half-lives) → 100 (after 3 half-lives). So 3 half-lives have passed in 24 hours, giving a half-life of \( 24 \div 3 = 8 \) hours. Check by a second route: using \( N = N_0 \times (0.5)^{t/T} \), \( 100 = 800 \times (0.5)^{24/T} \) gives \( (0.5)^{24/T} = 0.125 = (0.5)^3 \), so \( 24/T = 3 \), giving \( T = 8 \) hours — matching, confirming the answer.
(c) Alpha radiation is stopped/absorbed by a thin sheet of paper. Since radiation is still detected behind the paper, the source cannot be emitting alpha radiation alone; it must also (or instead) be emitting beta and/or gamma radiation, both of which can pass through paper.
Final answer: half-life = the time for activity to fall to half its original value; half-life of this isotope = 8 hours; the source must be emitting beta and/or gamma radiation (not alpha alone), since alpha cannot pass through paper.

(a) [1] correct definition referring to activity (or undecayed nuclei) falling to half its original value.
(b) [3] 1 mark: correct halving sequence shown (800→400→200→100); 1 mark: correct identification of 3 half-lives in 24 hours; 1 mark: correct final answer of 8 hours with working shown.
(c) [2] 1 mark: alpha is stopped by paper, so cannot be (solely) responsible for the continued detection; 1 mark: correct conclusion that beta and/or gamma radiation must also be present, since these penetrate paper.

(a) Mass is the amount of matter in an object, measured in kilograms (kg); it does not change with location. Weight is the force of gravity acting on an object, measured in newtons (N); it depends on the gravitational field strength, so it varies from place to place.
(b) Weight = mass × gravitational field strength = \( 80 \times 10 = 800 \) N. Check: \( 800 \div 10 = 80 \) kg, matching the given mass, confirming the answer.
(c) Weight on the Moon = \( 80 \times 1.6 = 128 \) N. Check by a second route: \( 1.6 \times 80 = 1.6 \times 8 \times 10 = 12.8 \times 10 = 128 \) N, matching, confirming the answer. The astronaut's mass is unchanged, remaining 80 kg, since mass does not depend on gravitational field strength.
(d) Light from distant galaxies shows red-shift (the further away a galaxy, the greater its red-shift), showing that galaxies are moving apart and the Universe is expanding from an earlier, more compact state — evidence for a Big Bang origin.
Final answer: mass (kg, constant) differs from weight (N, varies with gravity); weight on Earth = 800 N; weight on Moon = 128 N, mass unchanged at 80 kg; red-shift of distant galaxies is evidence for the Big Bang.

(a) [2] 1 mark: correct distinction between mass and weight; 1 mark: correct units (kg for mass, N for weight).
(b) [2] 1 mark: correct substitution W = mg = 80 × 10; 1 mark: correct answer 800 N with unit.
(c) [1] correct answer 128 N and mass stated as unchanged (80 kg); both parts needed for the mark.
(d) [1] any valid evidence, e.g. red-shift in light from distant galaxies, galaxies moving apart/further galaxies receding faster.

(a) Efficiency = useful output energy ÷ total input energy = \( 63000 \div 84000 = 0.75 = 75\% \). Check by a second route: \( 84000 \times 0.75 = 63000 \) J, matching the useful output given, confirming the answer.
(b) The Principle of Conservation of Energy states that energy cannot be created or destroyed, only changed from one form to another; the total amount of energy stays the same. The energy not usefully transferred (\( 84000 - 63000 = 21000 \) J) is not destroyed — it is transferred to the surroundings, mainly heating the kettle's body and the surrounding air, and becomes less useful (more spread out) rather than disappearing.
(c) Improving the insulation of the kettle body (or fitting/keeping a well-fitting lid) would reduce the heat energy lost to the surroundings, improving efficiency.
Final answer: efficiency = 75%; the 'wasted' 21 000 J is not destroyed but transferred to the surroundings (total energy conserved); better insulation would improve efficiency.

(a) [2] 1 mark: correct substitution 63000/84000; 1 mark: correct final answer 75% (accept 0.75).
(b) [2] 1 mark: correct statement of conservation of energy (not created/destroyed, only converted, total stays the same); 1 mark: correct explanation that the non-useful energy is transferred to the surroundings (not destroyed).
(c) [1] any valid suggestion, e.g. better insulation of the kettle body, using a lid to reduce heat loss.

Set up a circuit with the test wire connected in series with an ammeter and a variable resistor (rheostat), powered by a low-voltage power supply/battery, with a voltmeter connected in parallel across the wire only, to measure the voltage across it.
Using the variable resistor, adjust the circuit to give a series of different current settings. For each setting, record the current shown on the ammeter and the corresponding voltage shown on the voltmeter, building up a table of matching voltage and current readings across a suitable range.
Keep the current low, and only switch the circuit on briefly to take each reading, allowing the wire to cool between readings; this keeps the wire's temperature constant, since resistance changes with temperature and Ohm's law only applies at constant temperature. Repeat each reading and take a mean to improve reliability.
Plot a graph of voltage (y-axis) against current (x-axis). If the graph is a straight line passing through the origin, this shows that voltage and current are directly proportional for the wire at constant temperature, confirming Ohm's law; the gradient of the line is equal to the resistance of the wire, since voltage = current × resistance.
Final answer: connect the wire in series with an ammeter, a variable resistor and a power supply, with a voltmeter in parallel across the wire; vary and record matching current/voltage readings using brief, low currents to keep temperature constant; plot voltage against current — a straight line through the origin (gradient = resistance) confirms Ohm's law.

Level of response marking (6 marks): Band A (5–6 marks): a full, well-ordered account covering at least 5–6 indicative points with accurate specialist terms (ammeter, voltmeter, series, parallel, proportional, gradient/resistance) and few errors. Band B (3–4 marks): a reasonably clear account covering 3–5 points with satisfactory use of specialist terms. Band C (1–2 marks): a basic or disjointed account covering 1–2 points. Band D (0 marks): no relevant content.
Indicative content: circuit includes the test wire, an ammeter in series with the wire, and a voltmeter connected in parallel across the wire; power supply/battery and variable resistor (rheostat) used to vary the current; take a series of current/voltage reading pairs across a range of currents; repeat readings/take a mean to improve reliability; keep current low and switch on only briefly (allow wire to cool between readings) to keep temperature constant, since resistance depends on temperature; plot voltage (y-axis) against current (x-axis); straight line through the origin shows voltage and current are proportional — this is Ohm's law; the gradient of the V–I graph equals the resistance of the wire.

(a) The independent variable (the one changed/chosen by the student) is the material the rod is made from. The dependent variable (the one measured) is the time taken for the wax to melt and the pin to drop.
(b) To make this a fair test, variables that should be kept the same include: the length and diameter of each rod, the amount/type of wax used, the distance of the pin from the heated end, and the temperature of the water bath/heat source (any two).
(c) A shorter time to drop means heat is conducted faster, so is a better conductor. Times in increasing order are glass (>600 s), iron (130 s), aluminium (70 s), copper (45 s), so in order of increasing conductivity (worst to best): glass, iron, aluminium, copper.
(d) Copper is a metal, so it contains free (delocalised) electrons that can move through the metal's structure. When one end is heated, these free electrons gain kinetic energy and move quickly through the metal, transferring energy rapidly to other particles through frequent collisions — this is why metals conduct heat well. Glass is a non-metal with no free electrons; heat can only be transferred by the vibration of fixed particles passing energy to their neighbours through collisions, which is a much slower process than free-electron conduction. This is why copper conducts heat much better than glass.
(e) Loft insulation (or thick clothing) uses a material with low thermal conductivity, which traps air (a poor conductor); this reduces the rate of heat loss (by conduction) from a house (or the body), helping to keep it warm.
Final answer: independent variable = material, dependent variable = time for pin to drop; control e.g. rod dimensions and heat source temperature; increasing conductivity order: glass, iron, aluminium, copper; copper's free electrons carry heat energy quickly by collision, glass has none and relies on slow particle vibration; low-conductivity materials (e.g. loft insulation) trap air to reduce heat loss.

(a) [2] 1 mark: independent variable = material of the rod; 1 mark: dependent variable = time for the pin to drop.
(b) [2] 1 mark each for any two valid controlled variables (e.g. rod length, rod diameter, amount of wax, distance of pin from heat source, temperature of water bath).
(c) [1] correct order: glass, iron, aluminium, copper (all four correct for the mark).
(d) [3] 1 mark: copper (a metal) has free/delocalised electrons that can move through its structure; 1 mark: these free electrons carry kinetic/thermal energy quickly via frequent collisions; 1 mark: glass (non-metal, no free electrons) can only transfer heat by slower particle vibration, explaining the difference.
(e) [1] valid application named (e.g. loft insulation, thick clothing, insulated mugs) with correct reasoning that trapped air/low conductivity reduces heat loss.

(a) Average speed = distance ÷ time = \( 500 \div 25 = 20 \) m/s. Check by a second route: \( 20 \text{ m/s} \times 25\text{ s} = 500 \) m, matching the given distance, confirming the answer.
(b) To convert m/s to km/h, multiply by 3.6 (since 1000 m = 1 km and 3600 s = 1 h, so 1 m/s = \( 3600/1000 = 3.6 \) km/h): \( 20 \times 3.6 = 72 \) km/h. Check by a second route (dimensional analysis): \( 20 \tfrac{\text{m}}{\text{s}} \times \tfrac{1\,\text{km}}{1000\,\text{m}} \times \tfrac{3600\,\text{s}}{1\,\text{h}} = \tfrac{20 \times 3600}{1000} = 72 \) km/h, matching, confirming the answer.
(c) \( E_k = \frac{1}{2}mv^2 = 0.5 \times 900 \times 20^2 = 0.5 \times 900 \times 400 = 180000 \) J. Check by a second route: \( 900 \times 400 = 360000 \), and \( 360000 \div 2 = 180000 \) J, matching, confirming the answer.
Final answer: average speed = 20 m/s; 72 km/h; kinetic energy = 180 000 J.

(a) [2] 1 mark: correct substitution distance ÷ time = 500/25; 1 mark: correct answer 20 m/s with unit.
(b) [2] 1 mark: correct conversion method (× 3.6, or equivalent dimensional working); 1 mark: correct answer 72 km/h.
(c) [2] 1 mark: correct substitution into \( E_k = \tfrac{1}{2}mv^2 \) using v = 20 m/s (ECF from part (a)); 1 mark: correct final answer 180 000 J (accept 180 kJ) with unit.