2025 Cambridge IGCSE Science - Combined (0653) Practice Paper with Answers

(a) Double circulation refers to a system where blood passes through the heart twice for each complete circuit around the body. It consists of two distinct circuits: the pulmonary circulation (heart to lungs and back) and the systemic circulation (heart to body tissues and back).

(b) The left ventricle has a thicker muscular wall because it is responsible for pumping blood to all the systemic organs of the body, which requires a much higher pressure to overcome the friction and resistance of a long network of blood vessels. In contrast, the right ventricle only pumps blood to the lungs, which are nearby and have a delicate capillary network that would be damaged by high pressures.

(c) (i) The coronary artery supplies the heart tissue itself with oxygen and nutrients.
(ii) The pulmonary artery carries deoxygenated blood from the right ventricle of the heart to the lungs.

Part (a) [3 marks total]:
- Mention that blood passes through the heart twice [1 mark]
- for one complete circuit of the body [1 mark]
- identifies the two separate circuits: pulmonary and systemic [1 mark]

Part (b) [3.88 marks total]:
- Left ventricle pumps blood to the rest of the body / systemic circulation [1 mark]
- Right ventricle pumps blood only to the lungs / pulmonary circulation [1 mark]
- Left ventricle needs to generate much higher blood pressure [1 mark]
- To push blood over a greater distance / overcome high resistance [0.88 marks]

Part (c) [2 marks total]:
- (i) Coronary artery [1 mark]
- (ii) Pulmonary artery [1 mark]

(a) (i) The process is exothermic because the temperature of the water increases, indicating that thermal energy is released from the chemical system into the surroundings.
(ii) First, calculate the temperature change:
\(\Delta T = 27.2\ ^\circ\text{C} - 19.5\ ^\circ\text{C} = 7.7\ ^\circ\text{C}\)
Next, calculate the energy transferred using the given values:
\(q = m \times c \times \Delta T\)
\(q = 50.0\text{ g} \times 4.2\text{ J}/(\text{g}\ ^\circ\text{C}) \times 7.7\ ^\circ\text{C} = 1617\text{ J}\) (or \(1.617\text{ kJ}\), which rounds to \(1.62\text{ kJ}\)).

(b) Activation energy is defined as the minimum energy that colliding particles must possess to overcome the energy barrier and initiate a chemical reaction.

(c) In an energy level diagram for an exothermic reaction, the reactants are placed at a higher potential energy level than the products. The curve rises from the reactants to a peak (representing the transition state, with the height from the reactants level being the activation energy, \(E_a\)) and then drops down to the products level. The overall enthalpy change (\(\Delta H\)) is negative, indicated by a downward-pointing arrow from the reactants line to the products line.

Part (a)(i) [2 marks total]:
- Exothermic [1 mark]
- Reason: temperature increases / heat energy is released to the surroundings [1 mark]

Part (a)(ii) [3 marks total]:
- Temperature difference calculation: \(7.7\ ^\circ\text{C}\) [1 mark]
- Correct substitution: \(50.0 \times 4.2 \times 7.7\) [1 mark]
- Correct final answer: \(1617\text{ J}\) (accept \(1620\text{ J}\) or \(1.62\text{ kJ}\)) [1 mark]

Part (b) [1.88 marks total]:
- Minimum energy [1 mark]
- Required by colliding particles to react [0.88 marks]

Part (c) [2 marks total]:
- Energy level of reactants shown higher than energy level of products [1 mark]
- Activation energy (peak) and negative enthalpy change (downward step) correctly identified/described [1 mark]

(a) To find the acceleration (\(a\)), use the formula:
\(a = \frac{v - u}{t}\)
where \(u = 0\text{ m/s}\), \(v = 3.0\text{ m/s}\), and \(t = 4.0\text{ s}\).
\(a = \frac{3.0 - 0}{4.0} = 0.75\text{ m/s}^2\)

(b) Use the kinetic energy formula:
\(E_k = \frac{1}{2} m v^2\)
\(E_k = 0.5 \times 1.2\text{ kg} \times (3.0\text{ m/s})^2\)
\(E_k = 0.5 \times 1.2 \times 9.0 = 5.4\text{ J}\)

(c) (i) Use the relation between energy, power, and time:
\(E = P \times t\)
\(E = 2.5\text{ W} \times 4.0\text{ s} = 10\text{ J}\)

(ii) The useful kinetic energy (\(5.4\text{ J}\)) is less than the electrical energy input (\(10\text{ J}\)) because some energy is dissipated as heat/thermal energy due to friction between moving parts of the car or air resistance, and some is lost as sound energy.

Part (a) [3 marks total]:
- Correct formula: \(a = \frac{\Delta v}{t}\) or substitution seen [1 mark]
- Correct numerical value: \(0.75\) [1 mark]
- Correct unit: \(\text{m/s}^2\) (or \(\text{m}\ \text{s}^{-2}\)) [1 mark]

Part (b) [2.88 marks total]:
- Correct formula: \(E_k = \frac{1}{2} m v^2\) [1 mark]
- Correct substitution: \(0.5 \times 1.2 \times 3.0^2\) [1 mark]
- Correct final calculation: \(5.4\text{ J}\) [0.88 marks]

Part (c)(i) [2 marks total]:
- Correct formula: \(E = P \times t\) or substitution seen [1 mark]
- Correct final calculation: \(10\text{ J}\) [1 mark]

Part (c)(ii) [1 mark total]:
- Energy is lost/dissipated as thermal energy (due to friction/air resistance) or sound energy [1 mark]

(a) Acceleration is defined as the change in velocity per unit time (or rate of change of velocity).

(b) Using the formula for acceleration:
\(a = \frac{v - u}{t}\)
Given:
Initial velocity, \(u = 0\text{ m/s}\) (starts from rest)
Final velocity, \(v = 4.0\text{ m/s}\)
Time, \(t = 2.0\text{ s}\)
\(a = \frac{4.0 - 0}{2.0} = 2.0\text{ m/s}^2\)

(c) Using the formula for kinetic energy:
\(E_k = \frac{1}{2} m v^2\)
\(E_k = \frac{1}{2} \times 1.5\text{ kg} \times (4.0\text{ m/s})^2\)
\(E_k = 0.75 \times 16 = 12\text{ J}\)

(d) (i) Using Newton's second law for the resultant force:
\(F_{\text{resultant}} = m \times a\)
\(F_{\text{resultant}} = 1.5\text{ kg} \times 2.0\text{ m/s}^2 = 3.0\text{ N}\)

(ii) The resultant force is the driving force minus the resistive forces:
\(F_{\text{resultant}} = F_{\text{driving}} - F_{\text{friction}}\)
\(3.0\text{ N} = F_{\text{driving}} - 0.50\text{ N}\)
\(F_{\text{driving}} = 3.0 + 0.50 = 3.5\text{ N}\)

(a) Rate of change of velocity / change in velocity per unit time [1 mark].

(b)
- Correct formula or substitution: \(a = \frac{4.0}{2.0}\) [1 mark].
- Correct answer with units: \(2.0\text{ m/s}^2\) [1 mark].

(c)
- Correct formula or substitution: \(\frac{1}{2} \times 1.5 \times 4.0^2\) [1 mark].
- Correct calculation: \(12\text{ J}\) [1 mark].

(d)
(i)
- Correct formula or substitution: \(F = 1.5 \times 2.0\) [1 mark].
- Correct resultant force: \(3.0\text{ N}\) [1 mark].
(ii)
- Recognising that driving force = resultant force + resistive force: \(F_{\text{driving}} = 3.0 + 0.50\) [1 mark].
- Correct value: \(3.5\text{ N}\) [0.88 marks].

(a) The reaction is exothermic. This is because the temperature of the mixture increases, meaning thermal energy is released to the surroundings.

(b) Chemical reactions involve breaking existing bonds (which is an endothermic process as it requires energy) and making new bonds (which is an exothermic process as it releases energy). In an exothermic reaction, the quantity of energy released when new bonds form in the products is greater than the quantity of energy absorbed to break bonds in the reactants.

(c) First, determine the temperature change:
\(\Delta T = 38.5^\circ\text{C} - 21.5^\circ\text{C} = 17.0^\circ\text{C}\)

Next, calculate the mass of the solution:
\(m = 50.0\text{ cm}^3 \times 1.00\text{ g/cm}^3 = 50.0\text{ g}\)

Now use the heat transfer equation:
\(Q = m c \Delta T\)
\(Q = 50.0\text{ g} \times 4.18\text{ J/(g}\cdot^\circ\text{C)} \times 17.0^\circ\text{C}\)
\(Q = 3553\text{ J}\) (or \(3.55\text{ kJ}\))

(a)
- Exothermic [1 mark].
- Because the temperature of the surroundings/solution increases [1 mark].

(b)
- States that bond breaking is endothermic / requires energy [1 mark].
- States that bond making is exothermic / releases energy [1 mark].
- States that more energy is released when bonds are made than is absorbed to break bonds [1 mark].

(c)
- Calculates the temperature change correctly: \(17.0^\circ\text{C}\) [1 mark].
- Determines mass of the solution: \(50.0\text{ g}\) [1 mark].
- Correctly substitutes values into \(Q = m c \Delta T\): \(50.0 \times 4.18 \times 17.0\) [1 mark].
- Final correct calculated energy value: \(3553\text{ J}\) (accept \(3550\text{ J}\) or \(3.55\text{ kJ}\)) [0.88 marks].

(a) A double circulatory system means that for every complete circuit of the body, blood passes through the heart twice. It consists of the pulmonary circulation (heart to lungs and back) and systemic circulation (heart to body tissues and back). The main advantage is that blood pressure can be boosted after passing through the lungs, allowing oxygenated blood to be pumped around the body rapidly and at higher pressure.

(b) Arteries have thick, highly muscular and elastic walls to withstand and maintain high pressure of blood delivered directly from the heart. In contrast, veins have thinner walls with less muscle and elastic fibers because they carry blood under much lower pressure. Veins contain internal valves to prevent the backflow of blood, ensuring it flows in one direction back to the heart, whereas arteries do not contain valves (except semilunar valves at the exit of the heart).

(c) (i) The pulmonary artery carries deoxygenated blood from the right ventricle of the heart to the lungs.
(ii) The pulmonary vein carries oxygenated blood from the lungs back to the left atrium of the heart.

(a)
- Idea that blood passes through the heart twice for each complete circuit of the body [1 mark].
- Reference to two loops/circuits (pulmonary and systemic) [1 mark].
- Advantage: blood can be pumped to the tissues at a higher pressure / increases speed of delivery of oxygen or glucose [1 mark].

(b)
- Artery has thicker wall / thicker muscle layer / more elastic tissue than a vein [1 mark].
- Artery structure is adapted to withstand or maintain high blood pressure [1 mark].
- Vein has valves (to prevent backflow) but artery does not [1 mark].
- Vein structure is adapted to prevent backflow under low blood pressure [0.88 marks].

(c)
- (i) Pulmonary artery [1 mark].
- (ii) Pulmonary vein [1 mark].

(a) Use the acceleration formula: \(a = \frac{v - u}{t} = \frac{6.0\text{ m/s} - 0\text{ m/s}}{4.0\text{ s}} = 1.5\text{ m/s}^2\).

(b) Use the kinetic energy formula: \(E_k = \frac{1}{2}mv^2 = 0.5 \times 1.2\text{ kg} \times (6.0\text{ m/s})^2 = 0.6 \times 36 = 21.6\text{ J}\).

(c) (i) Since the car travels at a constant speed, the acceleration is zero, meaning there is no resultant force. Therefore, the driving force must be equal and opposite to the resistive force. Driving force = \(3.6\text{ N}\).
(ii) Use the power formula: \(P = F \times v = 3.6\text{ N} \times 6.0\text{ m/s} = 21.6\text{ W}\) (or \(\text{J/s}\)).

(a) [3 marks total]:
- 1 mark for correct working or formula: \((6.0 - 0) / 4\)
- 1 mark for value: \(1.5\)
- 1 mark for correct unit: \(\text{m/s}^2\) or \(\text{m s}^{-2}\)

(b) [2 marks total]:
- 1 mark for correct substitution: \(0.5 \times 1.2 \times 6.0^2\)
- 1 mark for correct value and unit: \(21.6\text{ J}\)

(c) (i) [2 marks total]:
- 1 mark for stating \(3.6\text{ N}\)
- 1 mark for explaining that the forces are balanced / net force is zero because it is travelling at a constant speed
(ii) [2 marks total]:
- 1 mark for substitution: \(3.6 \times 6.0\)
- 1 mark for correct value and unit: \(21.6\text{ W}\) (accept \(\text{J/s}\))

(a) (i) Balancing the carbon atoms gives \(3\text{CO}_2\), and balancing the hydrogen atoms gives \(4\text{H}_2\text{O}\). This gives a total of \((3 \times 2) + 4 = 10\) oxygen atoms on the right, which requires \(5\text{O}_2\) on the left.
(ii) Carbon monoxide, \(\text{CO}\), is formed during incomplete combustion.

(b) (i) Catalytic cracking requires high temperature (around \(450\,^\circ\text{C}\) to \(800\,^\circ\text{C}\)) and a catalyst (e.g., zeolite, alumina, or silica).
(ii) Decane (\(\text{C}_{10}\text{H}_{22}\)) cracking equation: \(\text{C}_{10}\text{H}_{22} \rightarrow \text{C}_2\text{H}_4 + X\). Subtracting the atoms of ethene from decane gives \(X = \text{C}_8\text{H}_{18}\). This fits the general formula of alkanes, \(\text{C}_n\text{H}_{2n+2}\).
(iii) Test: Bromine water (aqueous bromine) test. Ethene (unsaturated alkene) decolourises bromine water from orange-brown to colourless. Hydrocarbon \(X\) (saturated alkane) does not react, so the solution remains orange-brown.

(a) (i) [2 marks total]:
- 1 mark for balancing oxygen: \(5\text{O}_2\)
- 1 mark for balancing carbon dioxide and water: \(3\text{CO}_2 + 4\text{H}_2\text{O}\)
(ii) [1 mark total]:
- 1 mark for carbon monoxide (reject carbon dioxide / soot / carbon)

(b) (i) [2 marks total]:
- 1 mark for high temperature / heat
- 1 mark for a catalyst (accept specific examples like alumina / silica / zeolite)
(ii) [2 marks total]:
- 1 mark for formula: \(\text{C}_8\text{H}_{18}\)
- 1 mark for stating it is an alkane
(iii) [2 marks total]:
- 1 mark for identifying the reagent: bromine water / aqueous bromine
- 1 mark for correct results: decolourises with ethene AND remains orange-brown (no change) with hydrocarbon \(X\)

(a) In photosynthesis, carbon dioxide reacts with water (\(A\)) in the presence of light and chlorophyll to produce glucose (\(B\)) and oxygen.

(b) (i) Oxygen is released as a byproduct of photosynthesis.
(ii) Relationship: As the distance of the light source increases, the rate of bubble production decreases (inverse relationship). Explanation: Increasing the distance decreases the light intensity. Since light intensity is a limiting factor for photosynthesis under these conditions, less light energy is absorbed by chlorophyll, slowing down the rate of the reaction.
(iii) Reasons for inaccuracy:
1. Bubbles can be of different sizes, so counting bubbles does not give an accurate measurement of gas volume.
2. Some oxygen dissolves in the water or is used up in cellular respiration by the plant itself.

Improvement: Collect the gas using an inverted measuring cylinder over a funnel, or a capillary gas syringe, to measure the exact volume of gas produced over a given time.

(a) [2 marks total]:
- 1 mark for \(A\) = water
- 1 mark for \(B\) = glucose (accept starch / sugar / carbohydrate)

(b) (i) [1 mark total]:
- 1 mark for oxygen
(ii) [3 marks total]:
- 1 mark for describing the relationship (as distance increases, bubbling rate decreases)
- 1 mark for linking distance to light intensity (increasing distance decreases light intensity)
- 1 mark for explaining that light is a limiting factor / less light energy absorbed by chlorophyll
(iii) [3 marks total]:
- 2 marks for any two reasons: bubbles are different sizes / some gas dissolves in water / some gas is used in respiration / bubbles are released too fast to count accurately
- 1 mark for a valid improvement: collect gas in a gas syringe / inverted measuring cylinder to measure actual volume (or use a dissolved oxygen sensor with a data logger)