O Level Physics Thermal Physics Quiz

O-Level Physics Quiz - Thermal Physics (Answer Key)

Total Marks: 40
Topic: Thermal Physics

Section A: Short Answer (Q1–5)

Q1. [1 mark]
Solid, liquid, gas.
Teaching note: Matter exists in three common states. Marks awarded for naming all three. Common mistake: omitting one state.

Q2. [1 mark]
Evaporation.
Teaching note: Evaporation occurs at the surface below boiling point; boiling is throughout the liquid at boiling point.

Q3. [1 mark]
Thermometer.
Teaching note: A thermometer measures temperature; types include mercury, alcohol, digital.

Q4. [1 mark]
Internal energy (or thermal energy).
Teaching note: Internal energy = sum of kinetic and potential energies of particles. For this level, "internal energy" accepted.

Q5. [1 mark]
J/(kg·°C) or J kg⁻¹ °C⁻¹.
Teaching note: Specific heat capacity unit from E = mcΔT → J / (kg·°C).

Section B: Structured Questions (Q6–10)

Q6. [2 marks]

• Solid: particles closely packed in fixed positions, strong forces, vibrate only → fixed shape. [1]
• Gas: particles far apart, weak forces, move randomly and fill container → no fixed shape. [1]
  Teaching note: Use kinetic model: arrangement and motion explain macroscopic properties.

Q7. [2 marks]
State of matter: Solid. [1]
Reason: Particles are in a regular grid with small equal gaps, showing fixed arrangement. [1]
Image note: Diagram shows 12 circles in regular grid; this is solid lattice.

Q8. [2 marks]

1. Conduction needs matter/contact; convection needs fluid movement. [1]
2. Conduction by particle vibration/electron transfer; convection by bulk fluid circulation. [1]
   Teaching note: Accept any two clear differences.

Q9. [2 marks]
Phenomenon: Diffusion. [1]
Suggestion: Particles of ink and water move randomly and mix. [1]
Teaching note: Brownian/ diffusion shows particle motion.

Q10. [2 marks]
Mercury expands when heated and contracts when cooled, so column length changes with temperature. [1] Calibrated scale gives reading. [1]
Teaching note: Thermal expansion principle.

Section C: Calculation and Explanation (Q11–20)

Q11. [3 marks]
Formula: E = mcΔT [1]
m = 2.0 kg, c = 4200, ΔT = 100−20 = 80 °C [1]
E = 2.0 × 4200 × 80 = 672 000 J [1]
Answer: 6.72 × 10⁵ J
Teaching note: ΔT must be final−initial.

Q12. [3 marks]
E = mcΔT [1]
m = 0.50, c = 900, ΔT = 80−20 = 60 [1]
E = 0.50 × 900 × 60 = 27 000 J lost [1]
Answer: 2.7 × 10⁴ J
Teaching note: Cooling → energy lost.

Q13. [3 marks]
Steam at 100 °C releases latent heat of vaporisation when condensing to water at 100 °C. [1] This extra energy (no temp drop) transfers to skin. [1] Water at 100 °C only gives sensible heat on cooling. [1]
Teaching note: Latent heat = phase change energy without ΔT.

Q14. [3 marks]
Q = mL → m = Q/L [1]
Q = 12 600 J, L = 3.34×10⁵ [1]
m = 12 600 / 334 000 = 0.0377 kg [1]
Answer: 3.77×10⁻² kg
Teaching note: Only mass melted, not all if insufficient.

Q15. [3 marks]
Melting point: 0 °C [1]
Boiling point: 100 °C [1]
Explanation: Flat regions at constant temp are phase changes. [1]
Image note: Graph plateaus at 0 and 100 °C.

Q16. [3 marks]
E = mcΔT = 1.0×4200×(100−25)=315 000 J [1]
P = 1.5 kW = 1500 W, t = E/P [1]
t = 315 000/1500 = 210 s [1]
Answer: 210 s (3.5 min)
Teaching note: Convert kW to W.

Q17. [3 marks]

1. Wrap block in insulation, insert heater and thermometer. [1]
2. Record initial T, supply known V,I for time t, record final T. [1]
3. Use E=VIt=mcΔT to calculate c. [1]
   Teaching note: Reduces heat loss, measures energy input.

Q18. [3 marks]
Internal energy increases as bonds break. [1] Temperature stays constant during boiling. [1] Energy goes to potential not kinetic. [1]
Teaching note: At phase change, KE constant, PE rises.

Q19. [3 marks]
Heat lost by metal = heat gained by water [1]
0.80×c×(90−30) = 0.50×4200×(30−20)
48c = 21 000 → c = 437.5 J/(kg·°C) [2]
Answer: 438 J/(kg·°C)
Teaching note: Equilibrium method.

Q20. [3 marks]
Advantage: cheap, no battery, visual. [1]
Disadvantage: slower, less precise, breakable. [1+1 for two distinct]
Teaching note: Accept any valid pair.