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## ERQ · 13 marks · Topics: B.1 Thermal energy transfers + A.3 Work, energy and power · Archetype: experimental_analysis
**Integration:** primary=B.1 Thermal energy transfers, secondary=A.3 Work, energy and power (strength: supporting)
**Stem.** A flat-plate solar thermal collector is mounted on a rooftop to heat domestic water. The absorber plate has area A = 2.0 m² and receives solar irradiance I = 800 W m⁻² normal to its surface. Cold water enters the copper pipes at 18 °C and leaves at 33 °C; a small electric pump circulates the water at a constant mass flow rate of 1.2 × 10⁻² kg s⁻¹. The pump is rated at 25 W of electrical input. The specific heat capacity of water is 4200 J kg⁻¹ K⁻¹. A student records the inlet and outlet temperatures with digital probes of uncertainty ±0.5 °C and measures the mass flow rate by collecting outflow in a beaker on a balance over 60 s, giving an uncertainty of ±5 % in the flow rate. The student claims that "the efficiency of the panel is just the thermal power delivered to the water divided by the solar power incident on the absorber".
### Part (a) State [2 marks] · AO1 · Topic: B.1
State the dominant mechanism of thermal energy transfer (i) from the absorber plate to the water flowing in the copper pipes, and (ii) from the absorber plate to the surrounding air on a windy day.
### Part (b) Calculate [3 marks] · AO2 · Topic: B.1
Calculate the rate at which thermal energy is delivered to the water.
### Part (c) Show that [4 marks] · AO2 · Topic: B.1
Using your value from (b), show that the thermal efficiency of the collector, defined as the rate of thermal energy delivered to the water divided by the rate of solar energy incident on the absorber, is approximately 0.47. Then determine the fractional uncertainty in this efficiency, given that the temperature-rise uncertainty contributes ±6.7 % and the flow-rate uncertainty contributes ±5 %.
### Part (d) Evaluate [3 marks] · AO3 · Topic: B.1 + A.3 · ASSUMPTIONS DISCRIMINATOR
The student then revises the claim, proposing instead that "because the pump does 25 W of electrical work on the water, the true useful thermal output is (thermal power − 25 W), so the efficiency should be lowered accordingly". Evaluate this revised claim by reference to the first law of thermodynamics and the fate of the pump's work input.
### Part (e) Suggest [1 mark] · AO3 · Topic: B.1
Suggest one modification to the experimental procedure that would reduce the dominant contribution to the uncertainty calculated in part (c).
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## Mark Scheme
### Part (a) [2 marks]
- M1: (i) forced convection (of water through the pipes) / conduction from plate to pipe wall then convection accepted [no ECF]
- M2: (ii) (forced) convection — accept "convection by the wind" / do NOT accept "radiation" alone as dominant on a windy day [no ECF]
### Part (b) [3 marks] — Calculate
- M1: use of P = ṁ c ΔT with ΔT = 15 K [no ECF]
- M2: substitution: P = (1.2 × 10⁻²)(4200)(15) [no ECF]
- M3: P = 756 W ≈ 7.6 × 10² W [accept 750–760 W]
### Part (c) [4 marks] — Show that
- M1: incident solar power = I A = 800 × 2.0 = 1600 W [no ECF]
- M2: η = 756 / 1600 = 0.4725 ≈ 0.47 [ECF from (b)]
- M3: fractional uncertainties in η combine as Δη/η = ΔΔT/ΔT + Δṁ/ṁ = 6.7 % + 5 % = 11.7 % (≈ 12 %) [no ECF; accept addition in quadrature giving ≈ 8.4 % as alternative]
- M4: hence Δη ≈ 0.12 × 0.47 ≈ 0.06 so η = 0.47 ± 0.06 [ECF from M3]
### Part (d) [3 marks] — Evaluate (position + warrant + quantitative limitation per §4.4.1)
- M1 (position + warrant): claim is invalid / oversimplified — subtracting 25 W of electrical work from a thermal power conflates two different categories of energy input and double-counts losses; efficiency must be defined as useful output over input of the same kind [no ECF]
- M2 (thermodynamic reasoning): by the first law, essentially all of the 25 W pump work is dissipated by viscous friction into internal energy of the water itself, so it is already included in the measured outlet temperature rise ΔT = 15 K and hence in the 756 W figure [no ECF]
- M3 (quantitative correction): a defensible alternative metric is a coefficient including pump cost, η* = thermal gain / (solar + pump) = 756 / (1600 + 25) ≈ 0.465, showing the correct adjustment is to the denominator (≈1 % effect), not by subtracting 25 W from the numerator [ECF from (b), (c)]
### Part (e) [1 mark] — Suggest
- M1: any one valid proposal targeting the dominant (temperature) uncertainty, e.g. use temperature probes of finer resolution (±0.1 °C) / increase ΔT by reducing flow rate so the fractional uncertainty in ΔT falls / take repeated readings and average inlet and outlet temperatures [accept any plausible procedural improvement linked to the larger of the two contributions]
### Marker notes
- Show-that target in (c): η = 0.4725 given to 4 sig figs; student-derived 0.46–0.48 acceptable. Subsequent uncertainty arithmetic must use the given 0.47, not the student's (b) value, to preserve ECF integrity.
- Alternative method accepted for (c) uncertainty: addition in quadrature √(6.7² + 5²) ≈ 8.4 % giving η = 0.47 ± 0.04.
- (d) accept any of: pump work is internal to the control volume of the water; pump dissipation appears as part of ΔT; correct treatment places pump energy in denominator not numerator; subtraction would double-count losses already in measured outlet T.
- (a)(ii): if candidate writes "radiation and convection", award only if convection is identified as dominant for the windy condition.
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