(a) State what is meant by specific latent heat. (b) A student determines the specific latent heat of vaporisation of a liquid using the apparatus illustrated in Fig. 3.1. [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.
- (a)State what is meant by specific latent heat.[2]
- (b)(i)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.[1]
- (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]
- (b)(iii)Determine a value for the specific latent heat of vaporisation L of the liquid.[4]
- (b)(iv)Calculate the rate at which thermal energy is transferred to the surroundings.[2]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.
- [B1] (a) (thermal) energy per (unit) mass (to change state)
- [B1] (a) (heat transfer during) change of state at constant temperature
- [A1] (b)(i) 32 g
- [B1] (b)(ii) temperature difference (between liquid and surroundings) does not change
- [C1] (b)(iii) VIt = mL
- [C1] (b)(iii) 230 x 1.2 × 60 × 10 = (56 × L) + H
- [C1] (b)(iii) 190 x 1.0 × 60 × 10 = (32 x L) + H
- [C1] (b)(iii) 86 × 600 = (56 – 32) × L
- [A1] (b)(iii) L = 2200 J g⁻¹
- [C1] (b)(iv) 230 × 1.2 × 600 = (56 × 2150) + H or 190 x 1.0 × 600 = (32 × 2150) + H (accept both or either for C1)
- [A1] (b)(iv) = 75 W
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