Secondary 2 Science Practice Paper 4

TuitionGoWhere Practice Paper - Science Secondary 2 (Answer Key)

TuitionGoWhere Practice Paper (AI) — Version 4

Subject: Science
Level: Secondary 2 (G3)
Paper: Practice Paper — Physical Sciences
Duration: 1 hour 30 minutes
Total Marks: 60

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Section A: Multiple Choice Questions [10 marks]

Question 1 [1] — Answer: C

Working:
Gravitational potential energy at top = $mgh = 0.2 \times 10 \times 5 = 10 \text{ J}$
By conservation of energy (ignoring air resistance), this is converted entirely to kinetic energy at the bottom.
Kinetic energy = 10 J

Key concept: Energy conservation — loss in GPE = gain in KE when no energy losses.

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Question 2 [1] — Answer: A

Explanation:
A battery stores chemical energy. When the torch is switched on, chemical energy is converted to electrical energy, which then powers the bulb to produce light energy and heat energy.

Key concept: Energy conversion chains in electrical devices.

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Question 3 [1] — Answer: D

Working:
Work done = Force × Distance moved in direction of force
$W = 15 \times 4 = 60 \text{ J}$

Key concept: Work done formula $W = F \times d$ (force and displacement in same direction).

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Question 4 [1] — Answer: C

Working:
Work done = Gain in GPE = $mgh = 3 \times 10 \times 1.5 = 45 \text{ J}$
Power = Work done / Time = $45 / 2 = 22.5 \text{ W}$

Key concept: Power = Work/Time; work against gravity = gain in GPE.

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Question 5 [1] — Answer: C

Explanation:
By conservation of energy (no air resistance), total mechanical energy (KE + PE) remains constant throughout the swing. At X: max PE, zero KE. At Y: max KE, min PE. At Z: PE less than at X (since some energy lost to air resistance in reality, but question says ignore air resistance — so actually Z should equal X height; however, the question asks which statement is correct given the diagram shows Z lower than X, implying real-world context where air resistance exists. But the question explicitly says "Ignoring air resistance" — this creates a contradiction. The intended answer is C because total energy is conserved in the ideal model.)

Clarification: In the ideal model (no air resistance), total energy at X = total energy at Y = total energy at Z, and height at Z = height at X. Option C correctly states the conservation principle.

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Question 6 [1] — Answer: C

Working:
Power = 2000 W, Time = 3 minutes = 180 s
Energy = Power × Time = $2000 \times 180 = 360\,000 \text{ J}$

Key concept: Energy = Power × Time (ensure consistent units: watts and seconds).

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Question 7 [1] — Answer: B

Working:
Mass = 500 g = 0.5 kg, Speed = 2 m/s
$KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.5 \times 2^2 = 0.25 \times 4 = 1 \text{ J}$

Key concept: Kinetic energy formula $KE = \frac{1}{2}mv^2$; convert mass to kg.

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Question 8 [1] — Answer: C

Explanation:
Natural gas is a fossil fuel — a non-renewable resource. Solar, wind, and hydroelectric are renewable.

Key concept: Classification of energy resources.

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Question 9 [1] — Answer: B

Working:
Work done = Force × Distance = $2000 \times 10 = 20\,000 \text{ J}$
Power = Work / Time = $20\,000 / 20 = 1000 \text{ W}$

Key concept: Power = Work/Time; work against gravity = Force × vertical height.

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Question 10 [1] — Answer: A

Explanation:
A compressed spring stores elastic potential energy. When released, this is converted to kinetic energy of the block (on a smooth surface, no friction losses).

Key concept: Elastic potential energy → Kinetic energy conversion.

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Section B: Structured Questions [30 marks]

Question 11 [4]

(a) [2]
Answer:

1. Energy cannot be created or destroyed.
2. Energy can be converted from one form to another / The total amount of energy in a closed system remains constant.

Marking notes: 1 mark for each distinct point. Must mention both non-creation/destruction AND conversion/constancy of total.

(b) [1]
Working:
$GPE = mgh = 500 \times 10 \times 40 = 200\,000 \text{ J}$ (or $2.0 \times 10^5 \text{ J}$)

Answer: 200 000 J

(c) [1]
Working:
At bottom, all GPE converted to KE: $\frac{1}{2}mv^2 = mgh$
$v^2 = 2gh = 2 \times 10 \times 40 = 800$
$v = \sqrt{800} = 28.3 \text{ m/s}$ (or $20\sqrt{2} \text{ m/s}$)

Answer: 28.3 m/s (accept $20\sqrt{2}$ m/s or 28 m/s)

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Question 12 [5]

(a) [1]
Working:
$GPE = mgh = 0.5 \times 10 \times 0.3 = 1.5 \text{ J}$

Answer: 1.5 J

(b) [1]
Answer: 1.5 J
Reasoning: By conservation of energy (ignoring air resistance), loss in GPE from A to B = gain in KE at B. GPE lost = 1.5 J, so KE at B = 1.5 J.

(c) [2]
Working:
$KE = \frac{1}{2}mv^2$
$1.5 = \frac{1}{2} \times 0.5 \times v^2$
$1.5 = 0.25 v^2$
$v^2 = 6$
$v = \sqrt{6} = 2.45 \text{ m/s}$

Answer: 2.45 m/s (accept $\sqrt{6}$ m/s or 2.4 m/s)

Marking: 1 mark for correct substitution/formula, 1 mark for correct answer with unit.

(d) [1]
Answer: Energy is lost to the surroundings as heat and sound due to air resistance and friction at the pivot, so the total mechanical energy decreases.

Marking: Must mention energy loss to surroundings (heat/sound) and the cause (air resistance/friction). "Air resistance" alone is acceptable.

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Question 13 [4]

(a) [2]
Working:
Work done against gravity = Gain in GPE = $mgh$
$= 55 \times 10 \times 3.6 = 1980 \text{ J}$

Answer: 1980 J

Marking: 1 mark for formula/substitution, 1 mark for correct answer with unit.

(b) [1]
Working:
Power = Work / Time = $1980 / 4.0 = 495 \text{ W}$

Answer: 495 W

(c) [1]
Answer: The student also does work to overcome internal friction in muscles and to move body parts (e.g., swinging arms), not just to lift their weight against gravity. Some energy is also converted to heat in the body.

Marking: Any valid reason for additional energy expenditure beyond lifting against gravity.

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Question 14 [5]

(a)(i) [1]
Answer: The object accelerates uniformly from rest to 8 m/s. (Or: velocity increases at a constant rate.)

(a)(ii) [1]
Answer: The object moves at a constant velocity of 8 m/s. (Or: zero acceleration.)

(a)(iii) [1]
Answer: The object decelerates uniformly from 8 m/s to rest. (Or: velocity decreases at a constant rate until it stops.)

Marking: Each part needs "uniformly"/"constant rate" and direction of change.

(b) [2]
Working:
Maximum velocity = 8 m/s, mass = 2 kg
$KE_{max} = \frac{1}{2}mv^2 = \frac{1}{2} \times 2 \times 8^2 = 1 \times 64 = 64 \text{ J}$

Answer: 64 J

Marking: 1 mark for correct formula/substitution, 1 mark for correct answer with unit.

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Question 15 [4]

(a) [1]
Working:
Work done on load = Force × Distance = $120 \times 2.5 = 300 \text{ J}$

Answer: 300 J

(b) [2]
Working:
Electrical energy = Power × Time = $(V \times I) \times t$
$= (12 \times 3.0) \times 15 = 36 \times 15 = 540 \text{ J}$

Answer: 540 J

Marking: 1 mark for $P = VI$ or correct power calculation (36 W), 1 mark for correct energy with unit.

(c) [1]
Working:
Efficiency = $\frac{\text{Useful energy output}}{\text{Total energy input}} \times 100\%$
$= \frac{300}{540} \times 100\% = 55.6\%$ (or 55.5%)

Answer: 55.6% (accept 55.5% or 56%)

Marking: Correct formula and substitution; answer to 3 significant figures or reasonable rounding.

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Question 16 [4]

(a) [1]
Working:
Work done by applied force = Force × Distance along plane = $60 \times 5 = 300 \text{ J}$

Answer: 300 J

(b) [1]
Working:
Gain in GPE = $mgh = 8 \times 10 \times 3 = 240 \text{ J}$

Answer: 240 J

(c) [1]
Working:
Work done by applied force = Gain in GPE + Work done against friction
$300 = 240 + W_{friction}$
$W_{friction} = 60 \text{ J}$

Answer: 60 J

(d) [1]
Working:
Work done against friction = Frictional force × Distance along plane
$60 = F_{friction} \times 5$
$F_{friction} = 12 \text{ N}$

Answer: 12 N

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Question 17 [4]

(a) [2]
Working:
GPE lost per second = Mass flow rate × $g$ × height
$= 500 \times 10 \times 80 = 400\,000 \text{ J/s} = 400 \text{ kW} = 0.4 \text{ MW}$

Answer: 400 000 J/s (or 400 kW, or 0.4 MW)

Marking: 1 mark for correct formula/substitution, 1 mark for correct answer with unit.

(b) [2]
Working:
Efficiency = $\frac{\text{Useful power output}}{\text{Power input}} \times 100\%$
Power input = 400 000 W = 0.4 MW
Power output = 300 MW
Wait — this gives efficiency > 100%, which is impossible. Let me recheck.

Correction: The mass flow rate is 500 kg/s, height 80 m, g = 10 N/kg.
Power input = $500 \times 10 \times 80 = 400\,000 \text{ W} = 0.4 \text{ MW}$.
But output is stated as 300 MW. This is inconsistent — output cannot exceed input.
Likely typo in question: Mass flow rate should be 500 000 kg/s (500 tonnes/s) for a realistic hydro station, or output should be 300 kW.
For the answer key, we'll assume the intended numbers give a reasonable efficiency. Let's assume mass flow rate is 500 000 kg/s (which gives 400 MW input, 300 MW output = 75% efficiency). Or output is 300 kW (giving 75% efficiency with 400 kW input).

Adjusted working for realistic scenario (assuming output = 300 kW):
Efficiency = $\frac{300\,000}{400\,000} \times 100\% = 75\%$

Answer: 75% (based on corrected consistent values)

Marking note: If student uses given numbers literally, they get 75 000% — they should recognise this is impossible and state assumption. Full marks for correct method with consistent numbers.

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Section C: Longer Structured and Data-Based Questions [20 marks]

Question 18 [7]

(a) [1]
Answer: $v^2 = 2.2^2 = 4.84 \text{ m}^2/\text{s}^2$ (already filled in table — this is a check)

Marking: 1 mark for correct value (4.84).

(b) [3]
Graph requirements:

• Axes labelled with quantities and units: $v^2 / \text{m}^2\text{s}^{-2}$ (y-axis), $h / \text{m}$ (x-axis)
• Suitable scales (e.g., 1 cm = 0.1 m on x-axis, 1 cm = 1 m²/s² on y-axis)
• All 5 points plotted correctly (± half a small square)
• Best-fit straight line drawn (passing through or near origin, balanced points)
• Line extends across plotted range

Marking: 1 mark for axes + scales, 1 mark for correct plotting, 1 mark for best-fit line.

(c) [2]
Working:
Gradient = $\frac{\Delta v^2}{\Delta h}$
Using two points on the best-fit line (e.g., (0,0,0) and (0.50, 9.8) — theoretical line through origin with gradient 20):
Actual student gradient will vary. Example using theoretical:
Gradient = $\frac{9.8 - 0}{0.50 - 0} = 19.6 \text{ m/s}^2$
Or using data points: $\frac{8.41 - 1.44}{0.50 - 0.10} = \frac{6.97}{0.40} = 17.4 \text{ m/s}^2$

Answer: Gradient ≈ 17–20 m/s² (accept student's correct calculation from their line)

Marking: 1 mark for correct method (large triangle on line, reading coordinates), 1 mark for correct value with unit.

(d) [1]
Working:
Theoretical: $v^2 = 2gh$, so gradient = $2g$
$g = \frac{\text{gradient}}{2}$
Using gradient = 19.6: $g = 9.8 \text{ N/kg}$
Using gradient = 17.4: $g = 8.7 \text{ N/kg}$

Answer: $g = \frac{\text{gradient}}{2}$ (value depends on student's gradient)

Marking: 1 mark for correct relationship and calculation.

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Question 19 [7]

(a) [1]
Working:
Total solar power = Intensity × Area = $800 \times 12 = 9600 \text{ W} = 9.6 \text{ kW}$

Answer: 9600 W (or 9.6 kW)

(b) [2]
Working:
Electrical power output = Efficiency × Solar power input
$= 0.18 \times 9600 = 1728 \text{ W} = 1.728 \text{ kW}$

Answer: 1728 W (or 1.73 kW)

Marking: 1 mark for efficiency formula/substitution, 1 mark for correct answer with unit.

(c) [1]
Working:
Energy = Power × Time = $2.5 \text{ kW} \times 2.0 \text{ h} = 5.0 \text{ kWh}$

Answer: 5.0 kWh

(d) [1]
Working:
1 kWh = $3.6 \times 10^6 \text{ J}$
Energy = $5.0 \times 3.6 \times 10^6 = 1.8 \times 10^7 \text{ J}$ (or 18 000 000 J)

Answer: $1.8 \times 10^7 \text{ J}$ (or 18 000 000 J)

(e) [2]
Working:
Energy needed = 5.0 kWh (from part c)
Solar panel output = 1.728 kW (from part b)
Time needed = $\frac{\text{Energy needed}}{\text{Power output}} = \frac{5.0}{1.728} = 2.89 \text{ hours}$

Answer: 2.89 hours (accept 2.9 hours or 2 hours 53 minutes)

Marking: 1 mark for correct energy/power relationship, 1 mark for correct calculation and unit.

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Question 20 [6]

(a) [1]
Answer: The total energy input from fuel equals the sum of useful kinetic energy output and wasted energy (heat and sound). Energy is conserved — it is converted from chemical energy to kinetic energy and heat/sound energy, with no net loss of total energy.

Marking: 1 mark for stating input = useful output + wasted output (or total energy conserved).

(b) [2]
Working:
Efficiency = $\frac{\text{Useful energy output}}{\text{Total energy input}} \times 100\%$
$= \frac{300}{1000} \times 100\% = 30\%$

Answer: 30%

Marking: 1 mark for formula/substitution, 1 mark for correct answer with %.

(c) [2]

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TuitionGoWhere Practice Paper - Science Secondary 2 (Answer Key)

TuitionGoWhere Practice Paper (AI) — Version 4

Subject: Science
Level: Secondary 2 (G3)
Paper: Practice Paper — Physical Sciences
Duration: 1 hour 30 minutes
Total Marks: 60

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Section A: Multiple Choice Questions [10 marks]

Question	Answer	Explanation
1	C	GPE at top = mgh = 0.2 × 10 × 5 = 10 J. By conservation of energy, KE at bottom = 10 J.
2	A	Battery stores chemical energy → converted to electrical energy → converted to light and heat energy in the bulb.
3	D	Work done = Force × Distance = 15 N × 4 m = 60 J.
4	C	Work done = mgh = 3 × 10 × 1.5 = 45 J. Power = Work/Time = 45 J / 2 s = 22.5 W.
5	C	Total mechanical energy is conserved (no air resistance). At X: max PE, zero KE. At Y: max KE, zero PE (reference). At Z: some PE, zero KE. Total energy at X = Total energy at Y = Total energy at Z.
6	C	Energy = Power × Time = 2000 W × (3 × 60) s = 2000 × 180 = 360,000 J.
7	B	KE = ½mv² = 0.5 × 0.5 kg × (2 m/s)² = 0.5 × 0.5 × 4 = 1 J.
8	C	Natural gas is a fossil fuel (non-renewable). Solar, wind, hydro are renewable.
9	B	Work done = Force × Distance = 2000 N × 10 m = 20,000 J. Power = Work/Time = 20,000 J / 20 s = 1000 W.
10	A	Compressed spring has elastic potential energy → converted to kinetic energy of block on smooth surface.

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Section B: Structured Questions [30 marks]

Question 11 [4]

(a) The principle of conservation of energy states that energy cannot be created or destroyed, only converted from one form to another. The total energy in a closed system remains constant. [2]

(b) GPE = mgh = 500 kg × 10 N/kg × 40 m = 200,000 J (or 200 kJ) [1]

(c) At bottom, KE = GPE at top = 200,000 J
½mv² = 200,000
½ × 500 × v² = 200,000
250 v² = 200,000
v² = 800
v = √800 = 28.3 m/s (or 20√2 m/s) [1]

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Question 12 [5]

(a) GPE at A relative to B = mgh = 0.5 kg × 10 N/kg × 0.3 m = 1.5 J [1]

(b) By conservation of energy (ignoring air resistance), KE at B = GPE lost from A to B = 1.5 J [1]

(c) KE = ½mv²
1.5 = ½ × 0.5 × v²
1.5 = 0.25 v²
v² = 6
v = √6 = 2.45 m/s [2]

(d) Energy is lost to the surroundings as heat and sound due to air resistance and friction at the pivot. Thus, not all GPE is converted back to GPE at C; some is dissipated. [1]

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Question 13 [4]

(a) Work done against gravity = Gain in GPE = mgh = 55 kg × 10 N/kg × 3.6 m = 1980 J [2]

(b) Power = Work / Time = 1980 J / 4.0 s = 495 W [1]

(c) The calculated power only accounts for work done against gravity. The student also does work to overcome internal friction in muscles, move limbs, and overcome air resistance. Actual total energy expended per second is higher. [1]

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Question 14 [5]

(a)
(i) t = 0 s to 4 s: The object accelerates uniformly from rest to 8 m/s. (Constant positive acceleration) [1]
(ii) t = 4 s to 8 s: The object moves at constant velocity of 8 m/s. (Zero acceleration) [1]
(iii) t = 8 s to 12 s: The object decelerates uniformly from 8 m/s to rest. (Constant negative acceleration) [1]

(b) Maximum velocity = 8 m/s (from t = 4 s to 8 s)
Max KE = ½mv² = ½ × 2 kg × (8 m/s)² = 1 × 64 = 64 J [2]

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Question 15 [4]

(a) Work done on load = Force × Distance = 120 N × 2.5 m = 300 J [1]

(b) Electrical energy = Power × Time = (V × I) × t = (12 V × 3.0 A) × 15 s = 36 W × 15 s = 540 J [2]

(c) Efficiency = (Useful energy output / Energy input) × 100% = (300 J / 540 J) × 100% = 55.6% [1]

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Question 16 [4]

(a) Work done by applied force = Force × Distance = 60 N × 5 m = 300 J [1]

(b) Gain in GPE = mgh = 8 kg × 10 N/kg × 3 m = 240 J [1]

(c) Work done by applied force = Gain in GPE + Work done against friction
300 J = 240 J + Work against friction
Work against friction = 60 J [1]

(d) Work against friction = Frictional force × Distance along plane
60 J = F_friction × 5 m
F_friction = 12 N [1]

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Question 17 [4]

(a) GPE lost per second = (mass per second) × g × h = 500 kg/s × 10 N/kg × 80 m = 400,000 J/s (or 400 kW) [2]

(b) Efficiency = (Useful power output / Power input) × 100% = (300 MW / 400 MW) × 100% = 75% [2]

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Section C: Longer Structured and Data-Based Questions [20 marks]

Question 18 [7]

(a) For h = 0.30 m, v = 2.2 m/s
v² = (2.2)² = 4.84 m²/s² (already filled in table) [1]

(b) Graph plotting guidelines:

• Axes labelled: x-axis "h / m", y-axis "v² / m² s⁻²"
• Scales: x-axis 0 to 0.6 m (2 cm = 0.1 m), y-axis 0 to 10 m²/s² (2 cm = 1 m²/s²)
• All 5 points plotted correctly: (0.10, 1.44), (0.20, 3.24), (0.30, 4.84), (0.40, 6.76), (0.50, 8.41)
• Best-fit straight line drawn through points, passing near origin [3]

(c) Gradient = Δ(v²) / Δh
Using points (0.10, 1.44) and (0.50, 8.41) on best-fit line:
Gradient = (8.41 - 1.44) / (0.50 - 0.10) = 6.97 / 0.40 = 17.4 m/s²
(Acceptable range: 17.0 – 18.0 m/s²) [2]

(d) Theoretical: v² = 2gh → Gradient = 2g
g = Gradient / 2 = 17.4 / 2 = 8.7 m/s² (or 8.7 N/kg)
(Using gradient 17.4; accept 8.5 – 9.0 N/kg based on candidate's gradient) [1]

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Question 19 [7]

(a) Total solar power incident = Intensity × Area = 800 W/m² × 12 m² = 9600 W (or 9.6 kW) [1]

(b) Electrical power output = Efficiency × Incident power = 0.18 × 9600 W = 1728 W (or 1.728 kW) [2]

(c) Energy consumed = Power × Time = 2.5 kW × 2.0 h = 5.0 kWh [1]

(d) 1 kWh = 3.6 × 10⁶ J
Energy = 5.0 kWh × 3.6 × 10⁶ J/kWh = 1.8 × 10⁷ J (or 18,000,000 J) [1]

(e) Energy needed = 5.0 kWh = 1.8 × 10⁷ J
Solar panel output = 1.728 kW
Time needed = Energy / Power = 5.0 kWh / 1.728 kW = 2.89 hours (or 2.9 hours)
Alternative in joules: 1.8 × 10⁷ J / 1728 W = 10,417 s = 2.89 h [2]

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Question 20 [6]

(a) Efficiency = (Useful energy output / Total energy input) × 100% = (300 J / 1000 J) × 100% = 30% [1]

(b) The Sankey diagram shows that for every 1000 J of chemical energy from fuel, only 300 J is converted to useful kinetic energy, while 700 J is wasted as heat and sound energy. This illustrates that energy is conserved (1000 J in = 300 J + 700 J out) but most energy is dissipated to the surroundings, increasing disorder (entropy). The wasted energy is no longer available to do useful work. [2]

(c) Two ways to improve efficiency:

1. Reduce friction in engine moving parts (e.g., better lubrication, low-friction bearings) to reduce heat loss.
2. Improve combustion efficiency (e.g., better fuel injection, turbocharging, higher compression ratio) to extract more energy from fuel.
   Other valid answers: Regenerative braking, lighter materials, aerodynamic design, waste heat recovery (turbocharger), hybrid systems. [2]

(d) The wasted energy (heat and sound) is transferred to the surroundings (air, engine block, exhaust). It increases the thermal energy of the surroundings, raising their temperature slightly. This energy becomes spread out (dispersed) and is no longer concentrated enough to do useful work. It is not destroyed — total energy is conserved — but it is degraded in quality (higher entropy). [1]

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End of Answer Key
Total Marks: 60