Secondary 3 Physics Thermal Physics Quiz

Secondary 3 Physics Quiz - Thermal Physics - ANSWER KEY

Total Marks: 40

Section A: Multiple Choice (5 × 1 mark = 5 marks)

1. C — Particles in a gas move randomly at high speeds and are far apart.

• A is incorrect: particles in a solid vibrate about fixed positions, they are not stationary.
• B is incorrect: at the same temperature, particles in all states have the same average kinetic energy.
• D is incorrect: liquids have no fixed arrangement; particles can slide past each other.

2. B — Conduction

• Conduction is the primary mechanism of heat transfer in solids through particle vibration and free electron movement.

3. B — Air molecules are in constant random motion.

• Brownian motion is the random movement of visible particles caused by collisions with invisible, randomly moving air molecules.

4. B — The heat supplied is used to overcome attractive forces between particles.

• During a change of state, energy is used to break bonds between particles rather than increase kinetic energy, so temperature remains constant.

5. D — The dull black surface is a better absorber and a better emitter of radiation.

• Dull, dark surfaces are good absorbers and good emitters; shiny, light surfaces are poor absorbers and poor emitters.

Section B: Structured Questions (25 marks)

6. (a) Heat gained by water

• Q = mcΔθ
• Q = 0.25 × 4200 × (45 − 25) [1 mark for correct substitution]
• Q = 0.25 × 4200 × 20
• Q = 21,000 J [1 mark for correct answer with units]
• Answer: 21,000 J (or 21 kJ) [2 marks]

(b) Specific heat capacity of copper

• Heat lost by copper = Heat gained by water = 21,000 J
• Q = mcΔθ → 21,000 = 0.50 × c × (90 − 45) [1 mark for correct setup]
• 21,000 = 0.50 × c × 45
• c = 21,000 / (0.50 × 45) = 21,000 / 22.5
• c = 933.3... ≈ 933 J kg⁻¹ °C⁻¹ [1 mark for correct answer]
• Answer: 933 J kg⁻¹ °C⁻¹ (accept 930–940 J kg⁻¹ °C⁻¹) [2 marks]

(c) Explanation

• The water and copper reach thermal equilibrium / the same temperature. [1 mark]
• At this point, there is no net heat transfer between them, so the water temperature stops rising.
• Accept: "Heat transfer occurs from hotter to cooler objects until they reach the same temperature." [1 mark]

7. (a) Thermal energy to heat water to boiling point

• Q = mcΔθ
• Q = 1.2 × 4200 × (100 − 28) [1 mark for correct substitution]
• Q = 1.2 × 4200 × 72
• Q = 362,880 J [1 mark for correct answer]
• Answer: 362,880 J (or 363 kJ) [2 marks]

(b) Minimum time required

• P = E / t → t = E / P
• t = 362,880 / 2400 [1 mark for correct formula and substitution]
• t = 151.2 s
• Answer: 151 s (or 2 min 31 s) [1 mark for correct answer]
• [2 marks total]

(c) Reason for longer actual time

• Energy is lost to the surroundings (as heat). [1 mark]
• Accept: "The kettle itself absorbs some energy" or "Not all electrical energy is converted to thermal energy in the water" or "Heat loss to the environment." [1 mark]

8. (a) Temperature rise

• Δθ = 80 − 20 = 60 °C
• Answer: 60 °C [1 mark]

(b) Total energy supplied

• E = P × t
• E = 50 × 300 [1 mark for correct working]
• E = 15,000 J
• Answer: 15,000 J [1 mark]

(c) Specific heat capacity of aluminium

• Q = mcΔθ
• 15,000 = 0.80 × c × 60 [1 mark for correct setup]
• c = 15,000 / (0.80 × 60)
• c = 15,000 / 48
• c = 312.5 J kg⁻¹ °C⁻¹
• Answer: 313 J kg⁻¹ °C⁻¹ (accept 310–315 J kg⁻¹ °C⁻¹) [1 mark for correct answer]
• [2 marks total]

9. Convection current description

• Water at the bottom is heated and expands. [1 mark]
• The heated water becomes less dense and rises. [1 mark]
• Cooler, denser water at the top sinks to replace it, setting up a circulation current. [1 mark]
• Accept any clear description of the density-driven circulation pattern. [3 marks]

10. (a) Energy to melt ice at 0 °C

• Q = mL
• Q = 0.15 × 3.34 × 10⁵ [1 mark for correct substitution]
• Q = 50,100 J
• Answer: 50,100 J (or 5.01 × 10⁴ J) [1 mark for correct answer]
• [2 marks]

(b) Energy to warm melted ice from 0 °C to 10 °C

• Q = mcΔθ
• Q = 0.15 × 4200 × (10 − 0) [1 mark for correct substitution]
• Q = 0.15 × 4200 × 10
• Q = 6300 J
• Answer: 6300 J [1 mark for correct answer]
• [2 marks]

(c) Total energy absorbed

• Total = 50,100 + 6300 = 56,400 J
• Answer: 56,400 J (or 5.64 × 10⁴ J) [1 mark]

Section C: Data-Based and Application Questions (10 marks)

11. (a) Material with greatest temperature rise

• Lead will show the greatest temperature rise. [1 mark]
• Explanation: Q = mcΔθ, so Δθ = Q / (mc). For the same mass and same heat energy supplied, the material with the smallest specific heat capacity will have the largest temperature rise. Lead has the smallest specific heat capacity (130 J kg⁻¹ °C⁻¹). [1 mark]
• [2 marks]

(b) Water as a coolant

• Water has a very high specific heat capacity (4200 J kg⁻¹ °C⁻¹). [1 mark]
• This means water can absorb a large amount of heat energy with only a small rise in its own temperature, making it effective at removing heat from the engine without boiling quickly. [1 mark]
• Accept any clear explanation linking high specific heat capacity to effective heat absorption. [2 marks]

12. (a) Prediction and explanation

• Beaker B (dull black) will cool faster. [1 mark]
• Explanation: A dull black surface is a better emitter of thermal radiation than a shiny surface. Therefore, Beaker B radiates heat energy to the surroundings at a faster rate, causing it to cool more quickly. [1 mark]
• [2 marks]

(b) Sketch of temperature-time graphs

• Both curves should start at 80 °C and decrease over time. [1 mark for correct starting point and decreasing trend]
• The curve for Beaker B (dull black) should fall more steeply / be lower than the curve for Beaker A (shiny) at any given time. [1 mark for correct relative positions]
• Both curves should be clearly labelled.
• [2 marks]

(c) Precaution for fair comparison

• Use the same initial mass/volume of water in both beakers. [1 mark]
• Accept: "Place both beakers in the same location/draught-free environment" or "Use the same initial temperature" or "Use identical beakers (same size and shape)." [1 mark]

13. Explanation of gas pressure increase with heating

• When a gas is heated, the average kinetic energy of its particles increases. [0.5 marks]
• The particles move faster and collide with the container walls more frequently and with greater force. [0.5 marks]
• This results in a greater force per unit area, i.e., increased pressure.
• Accept any answer linking increased particle kinetic energy to more frequent/harder collisions with walls. [1 mark]

14. Specific heat capacity of the metal

• Heat lost by metal = Heat gained by water
• Heat gained by water: Q = 0.50 × 4200 × (30 − 20) = 21,000 J [1 mark for correct water energy]
• Heat lost by metal: 21,000 = 2.0 × c × (100 − 30)
• 21,000 = 2.0 × c × 70
• c = 21,000 / 140 = 150 J kg⁻¹ °C⁻¹ [1 mark for correct answer]
• Answer: 150 J kg⁻¹ °C⁻¹ [2 marks]

15. Differences between boiling and evaporation

• Boiling occurs at a specific temperature (boiling point); evaporation occurs at any temperature. [1 mark]
• Boiling occurs throughout the liquid; evaporation occurs only at the surface. [1 mark]
• Accept any two valid differences. [2 marks]

Section D: Extended Questions (10 marks)

16. Specific heat capacity of the liquid

• Energy supplied: E = P × t = 100 × 200 = 20,000 J [1 mark]
• Q = mcΔθ → 20,000 = 0.40 × c × (55 − 25) [1 mark for correct substitution]
• 20,000 = 0.40 × c × 30
• c = 20,000 / 12 = 1666.7 J kg⁻¹ °C⁻¹
• Answer: 1670 J kg⁻¹ °C⁻¹ (accept 1667 J kg⁻¹ °C⁻¹) [1 mark for correct answer]
• [3 marks]

17. Expansion of a solid when heated

• When heated, the particles in the solid gain kinetic energy and vibrate more vigorously. [1 mark]
• The increased vibration causes the particles to move slightly further apart on average, leading to an increase in the volume of the solid (expansion). [1 mark]
• [2 marks]

18. (a) Energy released during condensation

• Q = mL = 0.050 × 2.26 × 10⁶ = 113,000 J [1 mark]
• Answer: 113,000 J (or 1.13 × 10⁵ J) [1 mark]

(b) Additional energy released on cooling

• Q = mcΔθ = 0.050 × 4200 × (100 − 40) [1 mark for correct substitution]
• Q = 0.050 × 4200 × 60 = 12,600 J [1 mark for correct answer]
• Answer: 12,600 J [2 marks]

19. Vacuum flask design features

• Conduction and convection are minimised by the vacuum between the double walls, as there are no particles to transfer heat by these processes. [1 mark]
• Radiation is minimised by the silvered surfaces, which are poor emitters and poor absorbers of thermal radiation, reflecting heat back. [1 mark]
• [2 marks]

20. Solar panel design

• The panel is painted black because black surfaces are good absorbers of radiation, maximising the absorption of the Sun's thermal energy. [1 mark]
• The glass sheet traps heat by allowing short-wavelength solar radiation in but preventing long-wavelength thermal radiation from escaping (greenhouse effect), and also reduces heat loss by convection. [1 mark]
• [2 marks]

END OF ANSWER KEY