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A-LevelPhysicsTemperature9702/43 · Oct/Nov 2017 · Q110 Marks

State

  1. (a)State
  2. (a)(i)what may be deduced from the difference in the temperatures of two objects,[1]
  3. (a)(ii)the basic principle by which temperature is measured.[1]
  4. (b)By reference to your answer in (a)(ii), explain why two thermometers may not give the same temperature reading for an object.[2]
  5. (c)A block of aluminium of mass 670 g is heated at a constant rate of 95W for 6.0 minutes. The specific heat capacity of aluminium is 910Jkg⁻¹K⁻¹. The initial temperature of the block is 24 °C.
  6. (c)(i)Assuming that no thermal energy is lost to the surroundings, show that the final temperature of the block is 80°C.[3]
  7. (c)(ii)In practice, there are energy losses to the surroundings. The actual variation with time t of the temperature θ of the block is shown in Fig. 1.1. [Figure 1.1]
  8. (c)(ii)1Use the information in (i) to draw, on Fig. 1.1, a line to represent the temperature of the block, assuming no energy losses to the surroundings.[1]
  9. (c)(ii)2Using Fig. 1.1, calculate the total energy loss to the surroundings during the heating process. energy loss = ..................................................... J[2]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2017). The mark or mark type is shown in brackets.

  1. [B1] direction or rate of transfer of (thermal) energy or (if different,) not in thermal equilibrium/energy is transferred
  2. [B1] uses a property (of a substance) that changes with temperature
  3. [B2] ∞ temperature scale assumes linear change of property with temperature ∞ physical properties may not vary linearly with temperature ∞ agrees only at fixed points Any 2 points.
  4. [C1] Pt = mc(Δ)θ
  5. [M1] 95 × 6 × 60 = 0.670 × 910 × Δθ
  6. [A1] Δθ = 56°C so final temperature = 56 + 24 = 80 °C
  7. [(M1)] or 95 × 6 × 60 = 0.67 × 910 × (θ−24)
  8. [(A1)] so final temperature or θ = 80 °C
  9. [B1] sketch: straight line from (0,24) to (6,80)
  10. [C1] temperature drop due to energy loss = (80 – 64) = 16°C
  11. [A1] energy loss = 0.670 × 910 × (80 – 64) = 9800 J
  12. [(C1)] or energy to raise temperature to 64°C = 0.670 × 910 × (64 – 24)
  13. [(A1)] = 24400 J
  14. [M1] loss = (95 × 6 × 60) – 24400 = 9800 J

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About this question

Topic
Temperature, A-Level Physics (syllabus 9702). Structured question, 10 marks.
Source
A-Level Physics, Oct/Nov 2017 Paper 4 Variant 3, question 1.

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