Cambridge IGCSE · Thinka-original Practice Paper

2023 Cambridge IGCSE Sciences - Co-ordinated (Double) (0654) Practice Paper with Answers

Thinka Nov 2023 (V2) Cambridge International A Level-Style Mock — Sciences - Co-ordinated (Double) (0654)

220 marks255 mins2023

An original Thinka practice paper modelled on the structure and difficulty of the Nov 2023 (V2) Cambridge International A Level Sciences - Co-ordinated (Double) (0654) paper. Not affiliated with or reproduced from Cambridge.

Section Theory Core / Extended

Answer all questions. Show your working and write your answers in the spaces provided.

12 Question · 120 marks

Question 1 · Structured

10 marks

A student investigates water transport in plants.

(a) Define the term transpiration. [2]

(b) Describe and explain the effect of increased humidity on the rate of transpiration. [3]

(c) Xylem vessels are adapted to their functions of transport and support. Describe how xylem vessels are adapted for:
(i) transport of water [2]
(ii) support of the plant [1]

(d) State two uses of water within a plant, other than transport or support. [2]

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Worked solution

(a) Transpiration is the loss of water vapour from plant leaves by evaporation of water at the surfaces of the mesophyll cells followed by diffusion of water vapour through the stomata. [2 marks]

(b) Increased humidity decreases the rate of transpiration [1]. This is because increased humidity increases the concentration of water vapour in the air outside the leaf, reducing the water vapour concentration gradient between the air spaces inside the leaf and the external atmosphere [1]. Consequently, the rate of diffusion of water vapour out through the stomata is reduced [1].

(c) (i) Transport adaptations (any two): long continuous tubes / no end walls [1]; dead cells / no cytoplasm to obstruct water flow [1]; thick/lignified walls prevent collapse under tension [1]. (ii) Support adaptations: walls contain lignin which is strong/rigid to support the stem [1].

(d) Uses of water (any two): reactant in photosynthesis [1]; solvent for chemical reactions / transport of mineral ions or sucrose [1]; maintaining turgidity of cells [1].

Marking scheme

(a)
- loss of water vapour from plant leaves by evaporation at the surfaces of the mesophyll cells [1]
- followed by diffusion of water vapour through the stomata [1]

(b)
- rate of transpiration decreases [1]
- reduces the water vapour concentration gradient (between inside and outside of leaf) [1]
- slower rate of diffusion of water vapour out of stomata [1]

(c)
(i) Any two from:
- no end walls / forms continuous tube [1]
- hollow / dead cells / no cytoplasm (allowing unimpeded flow) [1]
- lignified walls (preventing collapse) [1]
(ii)
- walls strengthened with lignin (providing support/rigidity) [1]

(d) Any two from:
- photosynthesis [1]
- keeping cells turgid [1]
- solvent for metabolic reactions [1]

Question 2 · Structured

10 marks

A student investigates the rate of reaction between dilute hydrochloric acid and calcium carbonate (marble chips):

$$\text{CaCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CaCl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})$$

(a) Describe a method the student can use to measure the rate of this reaction, naming the key apparatus used to measure the rate. [3]

(b) The student repeats the experiment using the same mass of calcium carbonate, but as a fine powder instead of large marble chips.
(i) State how the rate of reaction changes. [1]
(ii) Explain this change in rate in terms of particles and collisions. [3]

(c) State and explain the effect of increasing the temperature of the acid on the rate of reaction. [3]

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Worked solution

(a) Method 1: Gas collection. Place the acid and calcium carbonate in a conical flask connected to a gas syringe using a delivery tube [1]. Start a stopwatch and measure the volume of carbon dioxide gas collected at regular time intervals [1]. The rate is determined by volume of gas per unit time [1].
Method 2: Mass loss. Place the reaction flask on a digital balance [1]. Add cotton wool to the neck of the flask to prevent liquid splash but allow gas to escape [1]. Measure the mass at regular time intervals [1].

(b) (i) The rate of reaction increases [1].
(ii) Using a fine powder increases the surface area [1]. This means more particles of calcium carbonate are exposed to the acid [1], resulting in a higher frequency of collisions (more collisions per unit time) [1].

(c) Increasing the temperature increases the rate of reaction [1]. This is because particles gain kinetic energy and move faster, leading to more frequent collisions [1]. Also, a greater proportion of colliding particles have energy equal to or greater than the activation energy, resulting in a higher percentage of successful collisions [1].

Marking scheme

(a)
- measure volume of gas collected / measure mass of flask and contents [1]
- use of gas syringe / balance [1]
- take readings at regular time intervals using a stopwatch/timer [1]

(b)
(i) rate increases [1]
(ii)
- powder has greater surface area [1]
- more reactant particles exposed to collision [1]
- increased frequency of collisions / more collisions per unit time [1] (do not accept "more collisions" alone)

(c)
- rate increases [1]
- particles have more kinetic energy / move faster, leading to more frequent collisions [1]
- more particles have energy \ge activation energy, so higher proportion of collisions are successful [1]

Question 3 · Structured

10 marks

A toy car of mass $ 2.5 \text{ kg} $ starts from rest and accelerates uniformly to a speed of $ 6.0 \text{ m/s} $ in $ 4.0 \text{ s} $. It then travels at a constant speed of $ 6.0 \text{ m/s} $ for a further $ 8.0 \text{ s} $.

(a) Calculate the acceleration of the toy car during the first $ 4.0 \text{ s} $. State the unit. [3]

(b) Calculate the total distance travelled by the toy car during the entire $ 12.0 \text{ s} $ journey. [3]

(c) Calculate the kinetic energy of the car when it is travelling at its constant speed. [2]

(d) Calculate the useful power required to accelerate the car from rest to $ 6.0 \text{ m/s} $ in $ 4.0 \text{ s} $. Assume there are no energy losses to the surroundings. [2]

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Worked solution

(a) Acceleration is given by the formula:
$$a = \frac{v - u}{t}$$
$$a = \frac{6.0 \text{ m/s} - 0 \text{ m/s}}{4.0 \text{ s}} = 1.5 \text{ m/s}^2$$
Award 1 mark for formula/working, 1 mark for correct value $ 1.5 $, and 1 mark for the correct unit $ \text{m/s}^2 $.

(b) Total distance is the area under the speed-time graph:
- For the first $ 4.0 \text{ s} $ (constant acceleration from rest):
$$\text{Distance}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4.0 \text{ s} \times 6.0 \text{ m/s} = 12.0 \text{ m}$$ [1 mark]
- For the next $ 8.0 \text{ s} $ (constant speed of $ 6.0 \text{ m/s} $):
$$\text{Distance}_2 = \text{speed} \times \text{time} = 6.0 \text{ m/s} \times 8.0 \text{ s} = 48.0 \text{ m}$$ [1 mark]
- Total distance:
$$\text{Total Distance} = 12.0 \text{ m} + 48.0 \text{ m} = 60.0 \text{ m}$$ [1 mark]

(c) Kinetic energy formula:
$$\text{KE} = \frac{1}{2} m v^2$$
$$\text{KE} = \frac{1}{2} \times 2.5 \text{ kg} \times (6.0 \text{ m/s})^2 = 1.25 \times 36 = 45 \text{ J}$$
Award 1 mark for working/formula, 1 mark for the correct value $ 45 \text{ J} $.

(d) Useful Power is work done (change in kinetic energy) divided by time:
$$\text{Power} = \frac{\Delta \text{KE}}{t}$$
$$\text{Power} = \frac{45 \text{ J}}{4.0 \text{ s}} = 11.25 \text{ W}$$
Award 1 mark for formula/working, 1 mark for correct value $ 11.25 \text{ W} $ (accept $ 11 \text{ W} $ or $ 11.3 \text{ W} $).

Marking scheme

(a)
- formula: a = \frac{v-u}{t} or \frac{6.0}{4.0} [1]
- 1.5 [1]
- \text{m/s}^2 [1]

(b)
- distance in first part = 12 \text{ m} [1]
- distance in second part = 48 \text{ m} [1]
- total distance = 60 \text{ m} [1]

(c)
- formula: \text{KE} = \frac{1}{2}mv^2 or working: 0.5 \times 2.5 \times 6^2 [1]
- 45 (J) [1]

(d)
- formula: P = \frac{W}{t} or \frac{\Delta \text{KE}}{t} or working: \frac{45}{4.0} [1]
- 11.25 (W) / 11 / 11.3 [1]

Question 4 · Structured

10 marks

This question is about atoms, isotopes, and bonding. (a) Define the term isotopes. [2] (b) An atom of chlorine-37 has a proton number of 17 and a nucleon number of 37. (i) State the number of neutrons in this atom. [1] (ii) Describe the formation of a chloride ion, Cl-, from a chlorine atom in terms of electrons and charge. [2] (c) Carbon reacts with chlorine to form tetrachloromethane, CCl4, which is a covalent compound. Describe the sharing of electrons in a molecule of CCl4, including the number of covalent bonds and the behavior of outer shell electrons. [3] (d) Explain why tetrachloromethane has a low boiling point. [2]

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Worked solution

(a) Isotopes are defined as atoms of the exact same element containing the exact same number of protons (atomic number) but a different number of neutrons (mass number). (b)(i) Number of neutrons = Nucleon number - Proton number = 37 - 17 = 20. (ii) A chlorine atom has 7 electrons in its outer shell. To become stable, it gains 1 electron from another atom to achieve a full outer shell of 8 electrons. Since it gains a negatively charged electron, it becomes a negatively charged chloride ion (Cl-). (c) Carbon has 4 outer electrons and needs 4 more to complete its outer shell. Each chlorine atom has 7 outer electrons and needs 1 more. Carbon shares one electron with each of the four chlorine atoms, and each chlorine shares one electron back. This forms 4 single covalent bonds, leaving 8 electrons in the outer shell of all bonded atoms. (d) CCl4 is a simple covalent substance. Although the covalent bonds within the molecule are strong, the forces between molecules (intermolecular forces) are very weak and require very little thermal energy to overcome, resulting in a low boiling point.

Marking scheme

(a) 1 mark for stating: atoms with the same proton number / same element. 1 mark for stating: different number of neutrons / different nucleon number. (b)(i) 1 mark for: 20. (b)(ii) 1 mark for: gains one electron. 1 mark for: to achieve a full outer shell / resulting in a net negative charge. (c) 1 mark for: 4 single covalent bonds formed. 1 mark for: carbon shares 4 electrons (one with each chlorine). 1 mark for: chlorine atoms each share 1 electron (all atoms obtain a full outer shell). (d) 1 mark for: weak intermolecular forces (forces between molecules). 1 mark for: requires little energy to overcome.

Question 5 · Structured

10 marks

This question is about electric circuits. (a) A student sets up a circuit with a 12 V d.c. power supply, a 5.0 ohm resistor, and an unknown resistor R connected in series. The current in the circuit is 1.5 A. (i) Calculate the total resistance of the circuit. [2] (ii) Determine the resistance of resistor R. [1] (iii) Calculate the potential difference across the 5.0 ohm resistor. [2] (b) A second identical 5.0 ohm resistor is now connected in parallel with the first 5.0 ohm resistor. (i) State the effect of this change on the combined resistance of these two parallel resistors. [1] (ii) Explain how this change affects the total current drawn from the 12 V power supply, assuming resistor R remains in series with this parallel combination. [2] (c) Calculate the electrical energy transferred by the power supply in 4.0 minutes when the current is 1.5 A and the potential difference is 12 V. [2]

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Worked solution

(a)(i) Total resistance R_total = V / I = 12 V / 1.5 A = 8.0 ohms. (ii) For series circuits, R_total = R1 + R2. Thus, 8.0 = 5.0 + R, meaning R = 8.0 - 5.0 = 3.0 ohms. (iii) Potential difference V = I * R = 1.5 A * 5.0 ohms = 7.5 V. (b)(i) When an identical resistor is added in parallel, the combined resistance of the parallel branch is halved (becomes 2.5 ohms), which is less than the original 5.0 ohms. (ii) Since the parallel section's resistance decreases, the total resistance of the entire series-parallel circuit decreases. Since Current = Voltage / Resistance, a lower total resistance results in a larger total current flowing from the power supply. (c) Electrical energy E = V * I * t. Time t = 4.0 minutes = 4.0 * 60 = 240 seconds. E = 12 V * 1.5 A * 240 s = 4320 J (or 4.3 kJ).

Marking scheme

(a)(i) 1 mark for formula R = V / I or correct working. 1 mark for correct value with unit (8.0 ohms). (a)(ii) 1 mark for: 3.0 ohms (allow follow-through from a(i)). (a)(iii) 1 mark for formula V = I * R or correct working. 1 mark for: 7.5 V. (b)(i) 1 mark for: resistance decreases (or halves / becomes 2.5 ohms). (b)(ii) 1 mark for: total circuit resistance decreases. 1 mark for: total current increases (since I = V / R). (c) 1 mark for converting time to seconds (240 s). 1 mark for: 4320 J (or 4.32 kJ).

Question 6 · Structured

10 marks

This question is about reproduction and germination in plants. (a) State the difference between self-pollination and cross-pollination. [2] (b) Explain how the structure of a pollen grain from an insect-pollinated flower differs from that of a wind-pollinated flower, and how each is adapted to its method of transfer. [4] (c) Germination of seeds requires specific environmental conditions. Explain why the following conditions are necessary for the process of germination: (i) warmth [2] (ii) oxygen [2]

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Worked solution

(a) Self-pollination involves the transfer of pollen grains from the anther of a flower to the stigma of the same flower or another flower on the same plant. Cross-pollination involves the transfer of pollen grains from the anther of a flower to the stigma of a flower on a completely different plant of the same species. (b) Pollen grains of insect-pollinated flowers have a sticky or spiky outer wall (exine) which allows them to easily stick to the hairs and bodies of visiting insects. In contrast, pollen grains of wind-pollinated flowers are small, light, dry, and smooth-surfaced, which allows them to be suspended in the air and carried over long distances by light wind currents. (c)(i) Warmth (suitable temperature) is required because seed germination relies on chemical reactions catalysed by enzymes. Warmth provides the optimum temperature for these enzymes to work efficiently to break down food stores in the cotyledons. (ii) Oxygen is required for aerobic respiration. The germinating embryo needs a continuous supply of energy to divide cells, grow, and break through the seed coat, and respiration converts stored sugars into usable energy.

Marking scheme

(a) 1 mark for: self-pollination is within the same flower/plant. 1 mark for: cross-pollination is to a different plant (of the same species). (b) 1 mark for: insect-pollinated pollen is sticky/spiky. 1 mark for: adaptation explanation (clings to insect bodies). 1 mark for: wind-pollinated pollen is light/smooth/small. 1 mark for: adaptation explanation (easily carried by wind/air currents). (c)(i) 1 mark for: provides optimum temperature for enzyme activity. 1 mark for: to speed up growth/metabolic chemical reactions. (c)(ii) 1 mark for: needed for (aerobic) respiration. 1 mark for: to release energy for cell division / growth of radicle/plumule.

Question 7 · Structured

10 marks

a) Describe how water is absorbed by root hair cells and state two structural features of root hair cells that adapt them for this function. [3]

b) An experiment is set up to measure the rate of water uptake of a leafy shoot using a potometer.

i) Explain why the leafy shoot must be cut underwater before being fitted to the potometer. [2]

ii) State and explain the effect of increasing light intensity on the rate of transpiration. [3]

c) Distinguish between the transport of substances in xylem vessels and phloem tubes, in terms of what is transported and the direction of transport. [2]

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Worked solution

a) Water is absorbed into root hair cells by osmosis, moving from an area of higher water potential in the soil to a lower water potential in the cell, across a partially permeable membrane. Two adaptations: large surface area (to increase rate of absorption) and thin cell walls (for a short diffusion distance).

b) i) The shoot must be cut underwater to prevent air from entering the xylem vessels. An air bubble would break the continuous column of water, stopping water transport up the stem.

ii) Increasing light intensity increases the rate of transpiration. In brighter light, stomata open wider to allow more carbon dioxide to enter for photosynthesis, which also allows more water vapour to diffuse out of the leaf.

c) Xylem vessels transport water and dissolved mineral ions upwards from the roots to the leaves. Phloem tubes transport dissolved sugars (sucrose) and amino acids both upwards and downwards (from sources to sinks).

Marking scheme

a) Osmosis mentioned [1]; movement down water potential gradient / from high to low water potential [1]; any two adaptations (large surface area, thin cell wall) [1]. (Max 3 marks)

b) i) Prevents air bubbles entering xylem [1]; prevents blockage / maintains continuous water column [1].

ii) Transpiration rate increases [1]; stomata open wider [1]; to allow CO2 entry / leading to greater diffusion of water vapour [1].

c) Xylem transports water/minerals and phloem transports sucrose/amino acids [1]; xylem transport is upwards only, phloem is both directions [1].

Question 8 · Structured

10 marks

a) Potassium is a Group I alkali metal.

i) Describe two physical properties of potassium that are different from typical transition metals like iron. [2]

ii) Describe what is observed when a small piece of potassium is added to a trough of cold water containing universal indicator. [3]

b) Chlorine and bromine are Group VII halogens.

i) State and explain the trend in reactivity as you descend Group VII. [2]

ii) An aqueous solution of chlorine is added to a solution of potassium bromide. Write a word equation for this reaction and explain why a reaction occurs. [3]

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Worked solution

a) i) Potassium is much softer (can be cut easily with a knife) and has a lower density (it floats on water) compared to transition metals like iron which are hard and dense.

ii) The potassium floats on the water, moves rapidly across the surface, melts into a shiny ball, and catches fire with a lilac flame. The universal indicator turns purple/blue as an alkaline solution (potassium hydroxide) is formed.

b) i) Reactivity decreases down the group. This is because as you go down, the outer shell is further from the nucleus, meaning there is weaker attraction to gain the incoming electron.

ii) Word equation: chlorine + potassium bromide -> potassium chloride + bromine.

Explanation: Chlorine is more reactive than bromine, so it successfully displaces the bromide ions from the potassium bromide solution.

Marking scheme

a) i) Soft / can be cut with a knife [1]; low density / floats [1]; low melting point [1]. (Any two, max 2 marks)

ii) Floats/darts on water [1]; lilac flame [1]; indicator turns purple/blue [1]. (Max 3 marks)

b) i) Reactivity decreases down the group [1]; due to weaker attraction on the incoming electron because of increased distance from the nucleus [1].

ii) Correct word equation [1]; chlorine is more reactive than bromine [1]; so it displaces bromine/bromide [1].

Question 9 · Structured

10 marks

a) A ray of light in air enters a semi-circular glass block at an angle of incidence of $35^{\circ}$. The angle of refraction in the glass is $22^{\circ}$.

i) Calculate the refractive index of the glass. Show your working. [2]

ii) Explain what is meant by the term 'critical angle'. [2]

b) Electromagnetic waves have many applications.

i) State one use of infrared waves and one use of ultraviolet waves. [2]

ii) State one danger to human health of exposure to ultraviolet waves. [1]

c) A sound wave traveling through water has a frequency of $250\text{ Hz}$ and a wavelength of $6.0\text{ m}$.

i) Calculate the speed of sound in this water. State the formula used and show your working. [2]

ii) State how the speed of sound in air compares to the speed of sound in water. [1]

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Worked solution

a) i) Using Snell's Law:
$n = \frac{\sin i}{\sin r}$
$n = \frac{\sin 35^{\circ}}{\sin 22^{\circ}} = \frac{0.5736}{0.3746} = 1.53$
Refractive index of the glass is 1.53.

ii) The critical angle is the angle of incidence in the optically denser medium for which the angle of refraction is $90^{\circ}$ in the less dense medium.

b) i) Infrared: television remote controls / optical fiber communication. Ultraviolet: sunbeds / sterilization of water.

ii) Ultraviolet rays can cause skin cancer, premature aging of the skin, or damage to eyes (cataracts).

c) i) Formula: $v = f \lambda$

Calculation: $v = 250 \times 6.0 = 1500\text{ m/s}$.

ii) The speed of sound in air is much slower than the speed of sound in water.

Marking scheme

a) i) Formula and substitution: $n = \frac{\sin 35}{\sin 22}$ [1]; Correct calculation: $1.53$ (allow 1.5) [1].

ii) Angle of incidence in denser medium [1]; that results in an angle of refraction of $90^{\circ}$ / boundary limit before total internal reflection [1].

b) i) Correct use of infrared (e.g. remote controls, thermal imaging) [1]; correct use of ultraviolet (e.g. sterilising, tanning) [1].

ii) Correct danger of UV (e.g. skin cancer, sunburn, eye damage) [1].

c) i) Formula: $v = f \lambda$ and calculation: $250 \times 6.0 = 1500\text{ m/s}$ [2] (1 mark for formula/substitution, 1 mark for correct value with unit).

ii) Speed in air is lower/slower than in water [1].

Question 10 · Structured

10 marks

(a) Define the term transpiration. [2]

(b) Explain the mechanism by which water moves from the root cortex cells into the xylem, and then up to the leaves. [3]

(c) A student uses a potometer to investigate the rate of transpiration in a leafy shoot.

(i) State two precautions the student must take when setting up the potometer to ensure accurate measurements. [2]

(ii) A fan is switched on near the shoot. Predict and explain the effect of this on the rate of movement of the air bubble in the potometer. [3]

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Worked solution

(a) Transpiration is the loss of water vapour from plant leaves by evaporation of water at the surfaces of the mesophyll cells followed by diffusion of water vapour through the stomata.

(b) Water moves across the root cortex cells by osmosis, down a water potential gradient. It enters the xylem vessels and is pulled up the stem to the leaves by a transpiration pull, which is a tension force created by the evaporation of water from the leaves.

(c)(i) Cut the shoot under water to prevent air locks in the xylem. Seal all joints with petroleum jelly (Vaseline) to ensure the apparatus is airtight.

(c)(ii) The bubble moves faster / rate of movement increases. This is because the wind from the fan moves water vapour away from the leaf surface, which increases the water potential gradient between the inside and outside of the leaf, speeding up diffusion.

Marking scheme

(a)
1 mark: evaporation of water at mesophyll cell surfaces.
1 mark: diffusion of water vapour through stomata.

(b)
1 mark: water moves across cortex by osmosis / down a water potential gradient.
1 mark: enters xylem and is pulled up by transpiration pull / tension.
1 mark: mention of cohesive forces of water / continuous water column.

(c)(i)
1 mark: cut shoot under water (to prevent air bubble blockages).
1 mark: seal joints with petroleum jelly / grease / wax (to make airtight).

(c)(ii)
1 mark: rate / speed of bubble increases.
1 mark: air flow removes water vapour from the leaf surface.
1 mark: increases the water potential / concentration gradient.

Question 11 · Structured

10 marks

A student investigates the rate of reaction between dilute hydrochloric acid and excess calcium carbonate chips:

$\text{CaCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CaCl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})$

(a) State how the rate of this reaction can be measured in terms of product formation. [2]

(b) The experiment is repeated using the same volume and concentration of acid, and the same mass of calcium carbonate, but using powdered calcium carbonate instead of chips.

(i) Describe and explain, using collision theory, the effect of using powdered calcium carbonate on the rate of reaction. [3]

(ii) Explain how the slope of the curve (of volume of gas against time) and the final volume of gas produced change when powdered calcium carbonate is used instead of chips. [3]

(c) Explain, in terms of activation energy, how adding a catalyst increases the rate of a chemical reaction. [2]

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Worked solution

(a) Measure the volume of carbon dioxide gas produced per unit time using a gas syringe, or measure the loss in mass of the flask over time using a balance.

(b)(i) The rate of reaction increases because powdered calcium carbonate has a much larger surface area than chips. This results in a higher frequency of successful collisions between reactant particles per unit time.

(b)(ii) The slope of the curve becomes steeper because the initial rate of reaction is greater. The final volume of gas remains the same because the limiting reactant (dilute hydrochloric acid) and the amount of calcium carbonate are unchanged, producing the same amount of product.

(c) A catalyst provides an alternative reaction pathway with a lower activation energy. This means a greater proportion of colliding particles have energy equal to or greater than the activation energy, increasing the rate of successful collisions.

Marking scheme

(a)
1 mark: measure volume of gas / loss of mass.
1 mark: per unit time / at regular time intervals.

(b)(i)
1 mark: rate of reaction increases.
1 mark: larger surface area (for powder).
1 mark: higher collision frequency / more collisions per unit time.

(b)(ii)
1 mark: slope / gradient is steeper.
1 mark: final volume of gas remains the same.
1 mark: same amount of limiting reactant / reactants are used up.

(c)
1 mark: provides an alternative pathway / mechanism.
1 mark: with a lower activation energy.

Question 12 · Structured

10 marks

A ship uses marine sonar to determine the depth of the seabed.

(a) State whether sound waves are longitudinal or transverse waves, and describe how they transfer energy through a medium. [3]

(b) The sonar system emits a pulse of sound of frequency $25\text{ kHz}$.

(i) Explain why humans cannot hear this sound. [1]

(ii) The speed of sound in seawater is $1500\text{ m/s}$. Calculate the wavelength of this sound wave. Show your working. [2]

(c) The sound pulse is reflected from the seabed and detected by a receiver on the ship $0.80\text{ s}$ after transmission. Calculate the depth of the seabed. Show your working. [3]

(d) State the effect, if any, on the speed of sound in water if the temperature of the water increases. [1]

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Worked solution

(a) Sound waves are longitudinal waves. They transfer energy by causing the particles of the medium to vibrate back and forth (parallel to the direction of wave travel), creating a series of compressions (high pressure regions) and rarefactions (low pressure regions).

(b)(i) The frequency of $25\text{ kHz}$ (which is $25000\text{ Hz}$) is above the upper limit of human hearing, which is $20000\text{ Hz}$ (or $20\text{ kHz}$).

(b)(ii) Use $v = f \lambda$, so $\lambda = \frac{v}{f}$.
$\lambda = \frac{1500}{25000} = 0.06\text{ m}$.

(c) Total distance travelled by the pulse = $v \times t = 1500 \times 0.80 = 1200\text{ m}$.
Since the sound travels to the seabed and back, the depth is half the total distance:
Depth = $\frac{1200}{2} = 600\text{ m}$.

(d) The speed of sound increases.

Marking scheme

(a)
1 mark: longitudinal wave.
1 mark: particles vibrate parallel to the direction of energy transfer / wave propagation.
1 mark: mention of compressions and rarefactions.

(b)(i)
1 mark: limit of human hearing is $20\text{ kHz}$ / $20000\text{ Hz}$ (sound is ultrasound).

(b)(ii)
1 mark: $v = f \lambda$ or correct rearrangement.
1 mark: correct calculation $0.06\text{ m}$ (accept $6\text{ cm}$).

(c)
1 mark: distance = $v \times t$ (or $1500 \times 0.80$ or $1200\text{ m}$).
1 mark: depth = $\text{distance} / 2$ (or $600$).
1 mark: unit of meters (m).

(d)
1 mark: speed of sound increases.

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