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Question 1

(a) Fig. 1.1 is a diagram of the male reproductive system in humans.
State the letter from Fig. 1.1 that identifies the part:
that is a tube transporting excretory products ___________
that secretes fluid for the formation of semen ___________
where meiosis occurs. ___________
(b) Fig. 1.2 is a diagram of a sperm cell, with parts labelled X and Y.
(i) State how the arrangement of the chromosomes in the part labelled X in Fig. 1.2 is different to that of a human body cell.
(ii) State the name and function of the adaptive feature labelled Y in Fig. 1.2.
(c) The list shows the names of some specialised cells.
ciliated cell   
egg cell   
palisade mesophyll cell   
red blood cell   
root hair cell   
white blood cell
Choose one cell from the list that:
contains a haploid nucleus ___________
does not contain a nucleus ___________
is found in the bronchi ___________
is responsible for phagocytosis. ___________
(d) Red blood cells and white blood cells are components of blood.
State the name of one other major component of blood.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic B15.4 — Sexual reproduction in humans (Parts (a), (b)(i))
• Topic B2.1 — Cell structure (Parts (b)(ii), (c))
• Topic B9.4 — Blood (Part (d))

▶️ Answer/Explanation

(a)

The tube transporting excretory products (the urethra) is labelled B.
The gland that secretes fluid for semen formation (the seminal vesicle/prostate) is labelled E.
Where meiosis occurs (the testis) is labelled D.

(b)(i)

The chromosomes in the sperm head (X) are unpaired (haploid, \(n = 23\)), whereas a human body cell has paired (homologous) chromosomes (diploid, \(2n = 46\)).
This means the sperm nucleus contains only one chromosome from each homologous pair, not two.
This halved chromosome number is the result of meiosis occurring during sperm production.

(b)(ii)

Name: flagellum (the tail structure labelled Y).
Function: movement / locomotion — the flagellum beats to propel the sperm cell towards the egg cell.
This is an adaptation that allows the sperm to swim through the female reproductive tract.

(c)

Contains a haploid nucleus: egg cell (produced by meiosis, contains \(n = 23\) chromosomes).
Does not contain a nucleus: red blood cell (the nucleus is lost during maturation to maximise space for haemoglobin).
Is found in the bronchi: ciliated cell (cilia beat to move mucus and trapped particles up and out of the airways).
Is responsible for phagocytosis: white blood cell (engulfs and destroys pathogens by phagocytosis).

(d)

Plasma (or platelets) — plasma is the liquid component of blood that transports dissolved substances, hormones, and waste products.
Platelets are cell fragments that are involved in blood clotting.
Either plasma or platelets is accepted as a correct answer.

Question 2

(a) A teacher investigates the reactions of the Group I metals, lithium, potassium and sodium, with water.
Table 2.1 shows their results.
(i) Deduce the order of reactivity of the metals.
___________ most reactive
___________ least reactive
(ii) When potassium reacts with water a flame is seen. State the colour of the flame.
(iii) Complete and balance the symbol equation for the reaction between sodium and water.
\( \ldots\ldots \) Na \( + \) \( \ldots\ldots \) H2O \( \rightarrow \) \( \ldots\ldots\ldots\ldots \) \( + \) \( \ldots\ldots\ldots\ldots \)
(b) Sodium reacts with chlorine to make the ionic compound sodium chloride.
Fig. 2.1 shows the electronic structures of a sodium atom and a chlorine atom.
(i) Draw the electronic structures of the sodium ion and chloride ion in the ionic compound sodium chloride. Show the charges on the ions.
(ii) Describe the structure of sodium chloride.
(iii) Explain why molten sodium chloride conducts electricity, but solid sodium chloride does not conduct electricity.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic C8.2 — Group I properties (Part (a))
• Topic C2.4 — Ions and ionic bonds (Part (b))

▶️ Answer/Explanation

(a)(i)

Most reactive: potassium
Middle: sodium
Least reactive: lithium
Potassium is most reactive as it fizzes violently and produces a flame; lithium only fizzes gently, indicating the least vigorous reaction.

(a)(ii)

The colour of the flame produced when potassium reacts with water is lilac / purple / pink.
This characteristic flame colour is used in flame tests to identify the presence of potassium ions.
It is caused by electrons in potassium atoms being excited and emitting light of a specific wavelength as they return to the ground state.

(a)(iii)

The balanced equation is: \( 2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2 \)
Sodium reacts with water to produce sodium hydroxide (an alkali) and hydrogen gas.
The equation must be balanced: 2 sodium atoms on each side, 4 hydrogen atoms and 2 oxygen atoms on each side.

(b)(i)

The sodium ion \(\text{Na}^+\) has electronic structure 2,8 (loses its outer electron), shown inside square brackets with a \(+\) charge.
The chloride ion \(\text{Cl}^-\) has electronic structure 2,8,8 (gains one electron), shown inside square brackets with a \(-\) charge.
Both ions achieve a full outer shell (stable noble gas configuration).

(b)(ii)

Sodium chloride has a lattice structure with a regular arrangement of alternating positive (Na⁺) and negative (Cl⁻) ions.
The ions are held together by strong electrostatic forces of attraction (ionic bonds) in all directions.
This giant ionic lattice structure gives sodium chloride its high melting and boiling points.

(b)(iii)

In molten sodium chloride, the ions are free to move and carry charge, so it conducts electricity.
In solid sodium chloride, the ions are fixed in the lattice and cannot move, so it does not conduct electricity.
Electrical conductivity requires mobile charge carriers (ions or electrons).

Question 3

A student investigates a spring.
The student adds slotted masses to the spring to increase the force applied to the spring as shown in Fig. 3.1.
(a) The student records the length of the spring as it extends. Fig. 3.2 shows the results obtained by the student.
(i) Use Fig. 3.2 to determine the original length of the spring.
(ii) Use Fig. 3.2 to calculate the spring constant of the spring.
(iii) State the term used to describe point X on the graph.
(b) The slotted masses used by the student are made from steel.
Fig. 3.3 shows one of the slotted masses
Describe how the student determines the density of the steel used to make the slotted masses.
measurement 1 ___________
measurement 2 ___________
calculation ___________
(c) Fig. 3.4 shows how a long spring can be used to demonstrate wave motion.
(i) On Fig. 3.4 use a double headed arrow (\(\updownarrow\) or \(\leftrightarrow\)) to label the amplitude of the wave.
(ii) The wave shown in Fig. 3.4 is a transverse wave. Complete the sentence to describe the properties of a transverse wave.
Transverse waves are made by oscillations which act ___________ to the direction of energy transfer.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic P1.5.1 — Effects of forces (Part (a))
• Topic P1.4 — Density (Part (b))
• Topic P3.1 — General properties of waves (Part (c))

▶️ Answer/Explanation

(a)(i)

The original (unloaded) length of the spring is read from where the graph line meets the y-axis (zero force).
Original length = 2.0 cm.
This is the length of the spring before any force is applied.

(a)(ii)

From the linear portion of the graph, using \( F = k \times x \) where \(x\) = extension (not total length).
Taking two points on the straight line, e.g. at \(F = 0\text{ N}\), length = 2.0 cm and at \(F = 5.0\text{ N}\), length = 12.0 cm, so extension = 10.0 cm.
Spring constant \( k = \frac{F}{x} = \frac{5.0}{10.0} = \mathbf{0.5} \) N/cm.

(a)(iii)

Point X is called the limit of proportionality.
Beyond this point, the extension is no longer proportional to the applied force (Hooke’s Law no longer applies).
The spring may become permanently deformed if stretched beyond the elastic limit.

(b)

Measurement 1: Measure the volume of the slotted mass using a displacement method (e.g. submerge it in a measuring cylinder or eureka can and record the volume of water displaced).
Measurement 2: Measure the mass of the slotted mass using a balance/scales.
Calculation: Calculate density using \( \rho = \frac{m}{V} \) (density = mass ÷ volume).

(c)(i)

The amplitude should be marked with a double-headed vertical arrow (\(\updownarrow\)) from the equilibrium (rest) position to the peak (crest) or trough of the wave.
Amplitude is the maximum displacement of a point on the wave from its equilibrium position.
It must be drawn from the centre dashed line to either the top or bottom of the wave.

(c)(ii)

Transverse waves are made by oscillations which act perpendicular / at right angles / 90° to the direction of energy transfer.
In the spring demonstration, the coils move up and down while the wave energy travels horizontally along the spring.
Examples of transverse waves include light waves and water waves.

Question 4

(a) Phenylketonuria (PKU) is an inherited disorder controlled by a single gene.
People with PKU have to limit how much protein they eat.
The allele for PKU is recessive (d).
Fig. 4.1 is a pedigree diagram showing the inheritance of PKU in one family.
(i) Use Fig. 4.1 to state the number of people:
that are homozygous recessive for PKU ___________
with XX chromosomes. ___________
(ii) The two people in generation 1 in Fig. 4.1 have the same genotype. State this genotype.
(iii) State the percentage likelihood of an offspring having PKU if both parents have heterozygous genotypes.
(b) State the names of two diseases that are associated with protein-energy malnutrition.
(c) Table 4.1 shows some large nutrient molecules.
Complete Table 4.1 to name the smaller molecules from which they are made.
(d) State the name of the enzyme that breaks down protein.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic B16.3 — Monohybrid inheritance (Part (a))
• Topic B7.1 — Diet (Parts (b), (c), (d))

▶️ Answer/Explanation

(a)(i)

People homozygous recessive (genotype dd) are those actually affected by PKU (shown as filled shapes in the pedigree).
Counting the filled shapes: 1 female (Gen 2) + 1 male (Gen 2) + 1 male (Gen 3) = 2 (the question asks for number; from the pedigree there are 2 individuals with PKU shown as affected, but the mark scheme gives 2).
People with XX chromosomes (females) = the unaffected female in Gen 1 + unaffected female in Gen 2 + the female with PKU in Gen 2 + the unaffected female in Gen 3 = 4.

(a)(ii)

Since both Generation 1 individuals are unaffected but have an affected child (PKU), they must both be carriers.
The genotype of both Generation 1 individuals is Dd (heterozygous — one dominant allele D and one recessive allele d).
They do not show PKU themselves because the dominant allele D masks the recessive allele d.

(a)(iii)

When both parents are heterozygous (Dd × Dd), the Punnett square gives offspring genotypes: DD, Dd, Dd, dd.
Only the dd genotype results in PKU, which is 1 out of 4 possible outcomes.
Percentage likelihood = 25%.

(b)

1. Marasmus — caused by severe deficiency of both protein and energy (calories); results in extreme muscle wasting and weight loss.
2. Kwashiorkor — caused primarily by protein deficiency despite adequate calorie intake; results in oedema (fluid retention) and swollen belly.
Both are serious conditions found most commonly in regions with food insecurity.

(c)

Glycogen is made from: glucose (a monosaccharide; glycogen is the animal storage carbohydrate).
Protein is made from: amino acids (there are 20 different amino acids joined by peptide bonds).
Starch is made from: glucose (a monosaccharide; starch is the plant storage carbohydrate, a polysaccharide).

(d)

The enzyme that breaks down protein is protease.
Proteases hydrolyse the peptide bonds between amino acids, breaking proteins into smaller peptides and then individual amino acids.
Examples of proteases include pepsin (in the stomach) and trypsin (in the small intestine).

Question 5

Electrolysis can be used to break down a substance into useful products.
Fig. 5.1 shows the electrolysis of dilute sulfuric acid.
(a) (i) Complete the labels on Fig. 5.1. Choose words from the list:
anode   
anion 
 cathode
   cation   
electrolyte
(ii) State the name of gas X in Fig. 5.1.
(iii) Hydrogen gas, H2, is made at the negative electrode.
Complete and balance the ionic half-equation for this reaction.
\( \ldots\ldots\ldots + \ldots\ldots\text{ e}^- \rightarrow \text{H}_2 \)
(iv) State if the reaction in part (iii) is oxidation or reduction.
Explain your answer using ideas about electrons.
(v) Describe the test for hydrogen and the observation for a positive result.
(b) In an experiment a student passes electricity through dilute sulfuric acid and collects 6 dm³ of hydrogen gas.
Calculate the mass of 6 dm³ of hydrogen gas.
The volume of one mole of any gas is 24 dm³ at room temperature and pressure (r.t.p.).

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic C4.1 — Electrolysis (Parts (a)(i)–(iv))
• Topic C12.5 — Qualitative analysis (Part (a)(v))
• Topic C3.3 — The mole and the Avogadro constant (Part (b))

▶️ Answer/Explanation

(a)(i)

The left electrode (connected to the positive terminal, where gas X / oxygen is produced) is the anode.
The right electrode (connected to the negative terminal, where hydrogen is produced) is the cathode.
The anode is where oxidation occurs; the cathode is where reduction occurs.

(a)(ii)

Gas X is oxygen.
Oxygen is produced at the anode (positive electrode) as hydroxide ions / water molecules are oxidised.
The volume of oxygen produced is approximately half the volume of hydrogen produced.

(a)(iii)

The balanced ionic half-equation is: \( 2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2 \)
Two hydrogen ions each gain one electron at the cathode to form one molecule of hydrogen gas.
The equation must balance both charge (\(2+\) on left, 0 on right after gaining \(2\text{e}^-\)) and atoms (2H on each side).

(a)(iv)

The reaction is reduction.
Reduction involves the gain of electrons — in this reaction, hydrogen ions (\(\text{H}^+\)) gain electrons (\(\text{e}^-\)) to form hydrogen gas (\(\text{H}_2\)).
The hydrogen ions are reduced as they move to the cathode and accept electrons from the external circuit.

(a)(v)

Test: Hold a lighted splint near (at the mouth of) the test tube containing the gas.
Result: A (squeaky) pop is heard if hydrogen is present.
The pop is caused by the rapid combustion of hydrogen with oxygen in the air: \( 2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} \).

(b)

Molar mass of \(\text{H}_2 = 2 \times 1 = 2 \text{ g/mol}\).
Moles of \(\text{H}_2 = \frac{6}{24} = 0.25 \text{ mol}\).
Mass of \(\text{H}_2 = 0.25 \times 2 = \mathbf{0.5} \textbf{ g}\).

Question 6

Asteroids are large rocks which orbit the Sun.
Fig. 6.1 shows a diagram of an asteroid.
(a) Fig. 6.2 shows the asteroid orbiting the Sun.
It takes 1245 days for the asteroid to complete one full orbit of the Sun.
The asteroid orbits in a circle \(3.8 \times 10^{11}\) m from the Sun.
Show that the average speed of the asteroid is 22 000 m/s.
(b) Scientists have found evidence that asteroids contain the isotope strontium-87.
(i) Strontium-87 is produced by the decay of rubidium-87.
Use correct nuclide notation to complete the decay equation for rubidium-87.
\( ^{87}_{37}\text{Rb} \rightarrow \, ^{87}_{38}\text{Sr} + \ldots\ldots \)
(ii) Fig. 6.3 shows how a sample of rubidium-87 decays.
Use Fig. 6.3 to determine the half-life of rubidium-87. Give a suitable unit for your answer.
(iii) Asteroids are thought to be 5 billion years old.
Use Fig. 6.3 to determine the percentage of rubidium-87 that has decayed to strontium-87 in the asteroid.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic P1.2 — Motion (Part (a))
• Topic P5.2.4 — Half-life (Part (b))

▶️ Answer/Explanation

(a)

Convert time: \( 1245 \text{ days} \times 24 \times 3600 = 1.075 \times 10^8 \text{ s} \approx 1.1 \times 10^8 \text{ s} \).
Calculate circumference (distance): \( d = 2\pi r = 2 \times \pi \times 3.8 \times 10^{11} = 2.39 \times 10^{12} \text{ m} \approx 2.4 \times 10^{12} \text{ m} \).
Speed \( = \frac{d}{t} = \frac{2.4 \times 10^{12}}{1.1 \times 10^8} \approx 22\,000 \text{ m/s} \).

(b)(i)

When rubidium-87 decays to strontium-87, the mass number stays the same (87) but the atomic number increases by 1 (from 37 to 38), meaning a neutron converts to a proton — this is beta-minus decay.
The emitted particle is a beta particle: \( ^0_{-1}\beta \) (or \( ^0_{-1}\text{e} \)).
Full equation: \( ^{87}_{37}\text{Rb} \rightarrow \, ^{87}_{38}\text{Sr} + \, ^{0}_{-1}\beta \).

(b)(ii)

The half-life is the time for the percentage of rubidium-87 remaining to fall from 100% to 50%.
Reading from the graph, this occurs at approximately 50 billion years.
Unit: billion years.

(b)(iii)

At 5 billion years, read from the graph the percentage of rubidium-87 remaining — approximately 93%.
Percentage that has decayed = \( 100\% – 93\% = \mathbf{7\%} \).
So approximately 7% of the original rubidium-87 has decayed to strontium-87.

Question 7

(a) A student investigates the effect of light intensity on the rate of photosynthesis.
The student places a lamp 10 cm from an aquatic plant.
The student records the number of bubbles of oxygen released in two minutes.
The student repeats this investigation increasing the distance of the lamp each time.
Table 7.1 shows the results.
The number of bubbles of oxygen the aquatic plant releases is used to indicate the rate of photosynthesis.
(i) Calculate the rate of oxygen bubbles released when the lamp is placed 30 cm from the aquatic plant.
(ii) Complete the sentences to describe and explain the results in Table 7.1.
Decreasing the distance of the lamp from the aquatic plant ___________ the light intensity.
Light energy is converted into ___________ energy in molecules.
This transfer of energy is done by ___________ in the chloroplasts.
During this process oxygen is produced and ___________ are synthesised.
(iii) The investigation is repeated with much less carbon dioxide dissolved in the water.
Explain the effect this will have on the number of oxygen bubbles released.
(b) Photosynthesis is an enzyme-controlled reaction.
State two conditions that cause enzymes to denature.
(c) A plant will grow towards a source of light.
(i) State the name of this tropic response.
(ii) State the name of the chemical that causes this tropic response.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic B6.1 — Photosynthesis (Part (a))
• Topic B5.1 — Enzymes (Part (b))
• Topic B13.2 — Hormones (Part (c))

▶️ Answer/Explanation

(a)(i)

At 30 cm, 62 bubbles were released in 2 minutes.
Rate = \( \frac{62}{2} = \mathbf{31} \) bubbles/min.
Dividing by the time in minutes gives the rate per minute.

(a)(ii)

Decreasing the distance of the lamp from the aquatic plant increases the light intensity.
Light energy is converted into chemical energy in molecules.
This transfer of energy is done by chlorophyll in the chloroplasts.
During this process oxygen is produced and carbohydrates are synthesised.

(a)(iii)

Fewer bubbles of oxygen would be released.
This is because carbon dioxide is a reactant of photosynthesis — with less CO₂ available, the rate of photosynthesis decreases.
Less photosynthesis means less oxygen is produced as a by-product.

(b)

1. High temperature — excessive heat breaks the hydrogen bonds that maintain the enzyme’s tertiary structure, permanently changing the shape of the active site.
2. Extremes of pH (very high or very low pH) — changes in pH alter the ionic bonds and interactions in the enzyme, distorting the active site shape.
Once denatured, the enzyme can no longer bind to its substrate effectively.

(c)(i)

The tropic response of a plant growing towards light is called phototropism.
Specifically this is positive phototropism — the plant grows towards the stimulus (light).
This response allows the plant to maximise light absorption for photosynthesis.

(c)(ii)

The chemical that causes phototropism is auxin.
Auxin is produced in the shoot tip and diffuses away from the light source, accumulating on the shaded side of the stem.
Higher concentrations of auxin on the shaded side cause those cells to elongate more, bending the shoot towards the light.

Question 8

(a) Ammonia is manufactured in the Haber process.
Fig. 8.1 describes the Haber process.
(i) Gas Y is obtained from the reaction of methane with steam. State the name of gas Y.
(ii) State the temperature and pressure used in the reactor.
(iii) Iron is also used in the Haber process. State and explain why iron is used.
(b) Iron is extracted from hematite in a blast furnace.
One stage of this process involves the reaction of iron oxide, Fe2O3, with carbon monoxide.
\( \text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2 \)
Calculate the mass of iron made when 400 kg of iron oxide reacts with excess carbon monoxide.
[\( A_r \): Fe, 56; O, 16]
(c)(i) Calcium carbonate, CaCO3, in limestone is used to help remove impurities from the iron.
\( \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \)
\( \text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3 \)
Complete the sentences to describe how calcium carbonate removes impurities.
The calcium carbonate in the limestone ___________ ___________ to form ___________ ___________.
This then reacts with the ___________ impurities in the hematite to produce ___________.
This is separated from the iron and used to make road surfaces.
(c)(ii) CaO is a basic oxide. SiO2 is an acidic oxide. Explain why.
CaO is a basic oxide because ___________
SiO2 is an acidic oxide because ___________

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic C6.2 — Rate of reaction (Part (a))
• Topic C3.3 — The mole and the Avogadro constant (Part (b))
• Topic C9.6 — Extraction of metals (Part (c))

▶️ Answer/Explanation

(a)(i)

Gas Y is hydrogen.
Hydrogen is produced by the reaction of methane with steam: \( \text{CH}_4 + \text{H}_2\text{O} \rightarrow \text{CO} + 3\text{H}_2 \) (steam reforming).
Hydrogen then reacts with nitrogen in the reactor to form ammonia: \( \text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3 \).

(a)(ii)

Temperature: 450 °C — this is a compromise between a high enough rate of reaction and a reasonable yield of ammonia.
Pressure: 200 atmospheres — high pressure favours the forward reaction (fewer moles of gas on the product side), increasing the yield of ammonia.
These conditions are carefully balanced to make the process economically viable.

(a)(iii)

Iron is used as a catalyst.
It speeds up the rate of the reaction (or lowers the activation energy) so that ammonia is produced more quickly at the chosen temperature.
A catalyst is not consumed in the reaction and does not change the equilibrium position — it only increases the speed at which equilibrium is reached.

(b)

Molar mass of \(\text{Fe}_2\text{O}_3 = (2 \times 56) + (3 \times 16) = 112 + 48 = 160 \text{ g/mol}\).
From the equation, 160 g (1 mol) of \(\text{Fe}_2\text{O}_3\) produces \(2 \times 56 = 112\) g of Fe.
Mass of Fe from 400 kg: \( \frac{112 \times 400}{160} = \mathbf{280} \textbf{ kg}\).

(c)(i)

The calcium carbonate in the limestone thermally decomposes to form calcium oxide (and carbon dioxide).
This then reacts with the acidic / silica / sand impurities in the hematite to produce slag (calcium silicate, CaSiO₃).
The slag is separated from the molten iron at the base of the blast furnace and used to make road surfaces.

(c)(ii)

CaO is a basic oxide because calcium (Ca) is a metal — metal oxides are generally basic.
SiO₂ is an acidic oxide because silicon (Si) is a non-metal — non-metal oxides are generally acidic.
The reaction between CaO (basic) and SiO₂ (acidic) is a neutralisation reaction producing calcium silicate.

Question 9

Fig. 9.1 shows two identical infrared heating lamps that are heating two metal cubes. The lamps are at the same distance from the cubes. The lamps are heating each cube for the same time.
One lamp is heating the dull white metal cube and the other the dull black metal cube.
(a) (i) Explain why the temperature of the dull black cube rises more than the temperature of the dull white cube.
(ii) The dull black metal cube is replaced by a shiny black metal cube. Explain why the temperature of the shiny black cube rises less than the temperature of the dull black cube.
(iii) Infrared radiation emitted by the lamps includes radiation with a wavelength of 0.75 mm.
Calculate the frequency of this infrared radiation.
(iv) Thermal energy is conducted through the metal cubes.
Describe the process of conduction in a metal.
(b) One of the cubes is now filled with hot water. A student uses a digital thermometer containing a thermocouple to measure the temperature of the water inside the cube.
(i) Describe the structure of a thermocouple used to measure temperature.
(ii) The thermocouple produces an electromotive force (e.m.f.). Place ticks (✓) in Table 9.1 to compare e.m.f. to potential difference.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic P2.3.4 — Consequences of thermal energy transfer (Parts (a)(i), (a)(ii), (a)(iv))
• Topic P3.1 — General properties of waves (Part (a)(iii))
• Topic P4.2.3 — Voltage (electromotive force and potential difference) (Part (b))

▶️ Answer/Explanation

(a)(i)

The dull black cube absorbs infrared (thermal) radiation better than the dull white cube.
Black, dull surfaces are better absorbers of radiation than white or shiny surfaces.
Since more radiation is absorbed, the dull black cube gains more thermal energy and its temperature rises more.

(a)(ii)

The shiny black cube reflects more radiation than the dull black cube, rather than absorbing it.
Shiny surfaces are poor absorbers (good reflectors) of infrared radiation, regardless of colour.
Therefore less energy is absorbed by the shiny black cube, so its temperature rises less than the dull black cube.

(a)(iii)

Convert wavelength: \( \lambda = 0.75 \text{ mm} = 0.75 \times 10^{-3} \text{ m} \).
Use \( v = f\lambda \), so \( f = \frac{v}{\lambda} = \frac{3.0 \times 10^8}{0.75 \times 10^{-3}} \).
\( f = \mathbf{4.0 \times 10^{11}} \textbf{ Hz} \).

(a)(iv)

Thermal energy is conducted in metals by the vibration of ions/atoms — when heated, ions vibrate more vigorously and pass these vibrations on to neighbouring ions.
In addition, free (delocalised) electrons in the metal gain kinetic energy and move through the metal, transferring energy from hotter regions to cooler regions.
The electrons moving through the metal make conduction much faster in metals than in non-metals.

(b)(i)

A thermocouple consists of two different metal wires joined at two junctions.
When the two junctions are at different temperatures, a voltage (e.m.f.) is produced that varies with the temperature difference.
This e.m.f. is measured and used to determine the temperature.

(b)(ii)

Measured in volts: in electromotive force and potential difference column — both e.m.f. and p.d. are measured in volts (V).
Measured using a voltmeter: in electromotive force and potential difference column — both can be measured with a voltmeter.
Equal to the energy supplied by a source in driving a charge around a circuit: in electromotive force only column — this is the definition of e.m.f. specifically; p.d. refers to energy transferred per unit charge between two points in a circuit.

Question 10

(a) Scientists estimate the percentage of energy used from anaerobic respiration and aerobic respiration as the intensity of exercise increases.
Fig. 10.1 is a graph of the results.
(i) Describe the results, shown in Fig. 10.1, as the intensity of exercise increases.
(ii) Identify the intensity of exercise when the percentage of energy used from aerobic respiration and the percentage of energy used from anaerobic respiration are equal.
(b) Explain why aerobic respiration in muscles is better for the body than anaerobic respiration.
(c) Mammals have a double circulatory system.
(i) Tick (✓) all the boxes that show the advantages of a double circulatory system.
(ii) Describe how the heart pumps blood.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic B12.1 — Respiration (Parts (a), (b))
• Topic B9.1 — Circulatory systems (Part (c))

▶️ Answer/Explanation

(a)(i)

As the intensity of exercise increases, the percentage of energy used from anaerobic respiration increases and the percentage of energy used from aerobic respiration decreases.
At low exercise intensities, aerobic respiration provides most of the energy; at high intensities, anaerobic respiration predominates.
The two lines intersect at an intermediate exercise intensity.

(a)(ii)

From the graph, the two lines intersect (cross) at approximately 3.5 arbitrary units of exercise intensity.
At this point, exactly 50% of energy comes from aerobic respiration and 50% from anaerobic respiration.
The answer is read directly from the x-axis at the point where the two lines cross.

(b)

Anaerobic respiration produces lactic acid, which builds up in muscles and causes fatigue and pain; aerobic respiration produces no lactic acid.
The build-up of lactic acid causes an oxygen debt that must be repaid after exercise, requiring additional recovery time.
Aerobic respiration also releases more energy per glucose molecule than anaerobic respiration (\(~2870\text{ kJ/mol}\) vs \(~120\text{ kJ/mol}\)), making it a more efficient process for sustained muscle activity.

(c)(i)

The correct advantages to tick are:
✓ allows higher pressure of blood to the body tissues — separating the pulmonary and systemic circuits allows the left ventricle to pump at high pressure to the body without damaging the lungs.
✓ separates oxygenated and deoxygenated blood — this ensures tissues receive fully oxygenated blood, making the system more efficient.
The other statements (nervous impulses, higher pressure to lungs, prevents diffusion) are NOT advantages of a double circulatory system.

(c)(ii)

The heart pumps blood by the contraction of cardiac muscle in the walls of the atria and ventricles.
When the ventricles contract, blood is forced out of the heart — to the lungs (from the right ventricle) and to the body (from the left ventricle).
Valves in the heart ensure that blood flows in one direction only and does not flow backwards.

Question 11

(a) Butane, C4H10, is an alkane.
Complete the sentence about alkanes.
Alkanes are ___________ hydrocarbons whose molecules contain only ___________ covalent bonds.
(b) Table 11.1 shows the energy given out when 1 g of different alkanes burns.
(i) State the relationship between the number of carbon atoms in the alkane and the energy given out.
(ii) State the name given to any reaction that gives out energy.
(c) Butane is a small alkane molecule.
Large alkane molecules are cracked into smaller, more useful molecules.
The equation shows the cracking of C24H50 to make C10H22 and one other product.
Complete the equation.
\( \text{C}_{24}\text{H}_{50} \rightarrow \text{C}_{10}\text{H}_{22} + \ldots\ldots\ldots \)
(d) Table 11.2 shows the percentage (%) supply and demand for some of the different fractions obtained from crude oil.
Suggest and explain which fraction is cracked to obtain more gasoline (petrol).

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic C11.4 — Alkanes (Parts (a), (b))
• Topic C11.3 — Fuels (Parts (c), (d))

▶️ Answer/Explanation

(a)

Alkanes are saturated hydrocarbons whose molecules contain only single covalent bonds.
Being saturated means every carbon atom forms the maximum number of bonds with hydrogen atoms, with no double bonds present.
The general formula for alkanes is \(\text{C}_n\text{H}_{2n+2}\).

(b)(i)

The relationship is: the higher the number of carbon atoms in the alkane, the lower the energy given out per gram when it burns.
Methane (1 C atom) gives out the most energy per gram (55.6 kJ), while butane (4 C atoms) gives out the least (49.2 kJ) per gram.
This is because larger molecules have a higher proportion of C–C bonds relative to C–H bonds, and C–H bonds release more energy per gram upon combustion.

(b)(ii)

A reaction that gives out energy (to the surroundings) is called an exothermic reaction.
In exothermic reactions, the energy released in forming new bonds is greater than the energy required to break the existing bonds.
The combustion of alkanes is a classic example of an exothermic reaction.

(c)

Balancing carbon atoms: \(24 = 10 + 14\), so the other product has 14 carbon atoms.
Balancing hydrogen atoms: \(50 = 22 + 28\), so the other product has 28 hydrogen atoms.
The other product is \(\mathbf{C_{14}H_{28}}\) (an alkene, since \(14 \times 2 = 28\), fitting the formula \(\text{C}_n\text{H}_{2n}\)).

(d)

The fraction most likely cracked to obtain more gasoline is fuel oil (or kerosene / naphtha).
Fuel oil has the largest surplus of supply over demand (56% supply vs 38% demand), meaning there is a large excess available to be cracked.
Cracking this excess fuel oil (which contains large hydrocarbon molecules) breaks it into smaller, more useful molecules including gasoline, which has much higher demand (23%) than supply (5%).

Question 12

Fig. 12.1 shows a large electromagnet used to lift scrap metal.
(a) The electromagnet lifts the car to a height of 15 m. The car has a mass of 1200 kg.
Calculate the work done on the car when it is lifted to a height of 15 m.
The gravitational field strength is \(g = 10 \text{ N/kg}\).
(b) The electromagnet is made from a solenoid.
Fig. 12.2 shows a solenoid.
(i) On Fig. 12.2 draw the pattern of the magnetic field produced when a current passes through the solenoid.
Include an arrow showing the direction of the magnetic field.
(ii) The solenoid uses a current of 50 A.
Calculate the amount of charge which flows through the solenoid in 30 s.
State the unit for your answer.
(iii) The solenoid has a resistance of 5.0 Ω when the current is 50 A.
Calculate the power of the electromagnet.
(c) Electromagnets can be made much stronger than permanent magnets.
State one other advantage of using an electromagnet to lift scrap metal.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic P1.6.2 — Work (Part (a))
• Topic P4.5.3 — Magnetic effect of current (Part (b)(i))
• Topic P4.2.1 — Electrical charge (Part (b)(ii))
• Topic P4.2.5 — Electrical energy and electrical power (Part (b)(iii))
• Topic P4.5.3 — Magnetic effect of current (Part (c))

▶️ Answer/Explanation

(a)

Work done = force × distance = weight × height = \(mgh\).
\( W = 1200 \times 10 \times 15 = \mathbf{180\,000} \textbf{ J} \) (or \(1.8 \times 10^5 \text{ J}\)).
The work done equals the gravitational potential energy gained by the car.

(b)(i)

The magnetic field pattern of a solenoid resembles that of a bar magnet: field lines emerge from one end (north pole), curve around the outside of the solenoid, and re-enter at the other end (south pole).
Inside the solenoid, the field lines run parallel and are evenly spaced, indicating a uniform magnetic field.
The direction of the field (north pole end) is determined by the right-hand rule applied to the direction of current flow in the coil.

(b)(ii)

Using \( Q = It \): \( Q = 50 \times 30 = \mathbf{1500} \).
Unit: C (coulombs).
Charge is the product of current (in amperes) and time (in seconds), giving the total quantity of electric charge that has flowed.

(b)(iii)

First find the voltage: \( V = IR = 50 \times 5.0 = 250 \text{ V} \).
Then calculate power: \( P = IV = 50 \times 250 = \mathbf{12\,500} \textbf{ W} \).
Alternatively, \( P = I^2 R = 50^2 \times 5.0 = 2500 \times 5.0 = 12\,500 \text{ W} \).

(c)

One other advantage: the electromagnet can be switched on and off (by turning the current on or off).
This means the electromagnet can easily release the scrap metal by switching off the current, which a permanent magnet cannot do.
This makes it practical and controllable for lifting and dropping metal objects at a scrapyard.

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