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A-LevelPhysicsThermodynamics9702/41 · Oct/Nov 2019 · Q310 Marks

State what is meant by specific latent heat.

  1. (a)State what is meant by specific latent heat.[2]
  2. (b)(i)A student determines the specific latent heat of vaporisation of a liquid using the apparatus illustrated in [Figure 3.1]. The heater is switched on. When the liquid is boiling at a constant rate, the balance reading is noted at 2.0 minute intervals. After 10 minutes, the current in the heater is reduced and the balance readings are taken for a further 12 minutes. The readings of the ammeter and of the voltmeter are given in [Figure 3.2]. The variation with time of the balance reading is shown in [Figure 3.3]. From time 0 to time 10.0 minutes, the mass of liquid evaporated is 56 g. Use Fig. 3.3 to determine the mass of liquid evaporated from time 12.0 minutes to time 22.0 minutes. mass = ........................................................... g[1]
  3. (b)(ii)Explain why, although the power of the heater is changed, the rate of loss of thermal energy to the surroundings may be assumed to be constant.[1]
  4. (b)(iii)Determine a value for the specific latent heat of vaporisation L of the liquid. L = ........................................................... Jg⁻¹[4]
  5. (b)(iv)Calculate the rate at which thermal energy is transferred to the surroundings. rate = ........................................................... W[2]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.

  1. [B1] (thermal) energy per (unit) mass (to change state)
  2. [B1] (heat transfer during) change of state at constant temperature
  3. [A1] 32 g
  4. [B1] temperature difference (between liquid and surroundings) does not change
  5. [C1] VIt = mL
  6. [C1] 230 x 1.2 × 60 × 10 = (56 × L) + H
  7. [C1] or 190 x 1.0 × 60 × 10 = (32 x L) + H
  8. [C1] 86 × 600 = (56 – 32) × L
  9. [(C1)] or 230 × 1.2 = (56 × L) / (60 × 10) + P
  10. [(C1)] or 190 x 1.0 = (32 × L) / (60 × 10) + P
  11. [(C1)] 276-190 = (24 × L) / 600
  12. [A1] L = 2200 J g-1
  13. [C1] 230 × 1.2 × 600 = (56 × 2150) + Η
  14. [C1] or 190 × 1.0 × 600 = (32 × 2150) + Η
  15. [A1] H = 45200
  16. [(A1)] rate = 45200/600
  17. [(A1)] = 75 W
  18. [(C1)] or 230 × 1.2 = (56 × 2150) / (60 × 10) + P
  19. [(C1)] or 190 × 1.0 = (32 × 2150) / (60 × 10) + P
  20. [(A1)] rate (= P) = 75 W

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About this question

Topic
Thermodynamics, A-Level Physics (syllabus 9702). Structured question, 10 marks.
Source
A-Level Physics, Oct/Nov 2019 Paper 4 Variant 1, question 3.

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