Generated ERQ

✓ passed A.3 Work, energy and power × B.1 Thermal energy transfers 12 marks HL 3 passes 106.65s $0.6572
## ERQ · 12 marks · Topics: A.3 Work, energy and power + B.1 Thermal energy transfers · Archetype: theory_application **Integration:** primary=A.3 Work, energy and power, secondary=B.1 Thermal energy transfers (strength: supporting) **Stem.** A small farm uses an electrically driven pump to lift water from a ground-level reservoir into a storage tank whose inlet is 18.0 m above the water surface of the reservoir. The pump draws 1.50 kW of electrical power from the mains supply and delivers water at a steady volume flow rate of 4.20 × 10⁻³ m³ s⁻¹ through a long pipe of constant cross-section. Frictional losses occur in the pipework and in the motor. The density of water is 1.00 × 10³ kg m⁻³ and its specific heat capacity is 4.18 × 10³ J kg⁻¹ K⁻¹. Assume the kinetic energy of the water leaving the pipe is negligible and that the system has reached steady state. ### Part (a) Define [2 marks] · AO1 · Topic: A.3 Define the *efficiency* of an energy conversion device and state its SI unit. ### Part (b)(i) Show that [3 marks] · AO2 · Topic: A.3 Show that the useful mechanical power delivered to the water as it is raised to the tank inlet is approximately 741 W. ### Part (b)(ii) Determine [3 marks] · AO2 · Topic: A.3 Determine the overall efficiency of the pump-and-pipe system, expressing your answer as a percentage. ### Part (c) Explain [2 marks] · AO3 · Topic: A.3+B.1 The water in the storage tank is found to be at a slightly higher temperature than the water in the reservoir. Explain, with reference to the energy balance in the pipework, why this temperature rise occurs. ### Part (d) Suggest [2 marks] · AO3 · Topic: A.3+B.1 · ASSUMPTIONS DISCRIMINATOR A student predicts the temperature rise of the water using the assumption that *all* the power lost in the system goes into heating the water. Suggest one reason why the actual temperature rise of the water in the tank would be smaller than this prediction. --- ## Mark Scheme ### Part (a) [2 marks] — Define - M1: efficiency = useful (output) energy / total (input) energy ✓ OR equivalent ratio in terms of power [no ECF] - M2: states that efficiency is dimensionless / has no SI unit (accept "ratio" or "%") ✓ [no ECF] ### Part (b)(i) [3 marks] — Show that - M1: mass flow rate Δm/Δt = ρ × (ΔV/Δt) = 1.00 × 10³ × 4.20 × 10⁻³ = 4.20 kg s⁻¹ ✓ [no ECF] - M2: applies P = (Δm/Δt) g h with substitution: P = 4.20 × 9.81 × 18.0 ✓ [ECF from M1] - M3: P = 741.6 W ≈ 741 W (to 3 s.f.) ✓ — answer must show derivation, not assertion [no ECF — target value given] ### Part (b)(ii) [3 marks] — Determine - M1: identifies η = P_useful / P_input ✓ [no ECF] - M2: correct substitution η = 741 / 1500 ✓ [ECF from (b)(i) if alternative derived value used] - M3: η = 0.494 or 49.4 % (accept 49 %–50 %) ✓ [ECF from M2] ### Part (c) [2 marks] — Explain (causal chain per §4.4.1) - M1: identifies that frictional/viscous forces in the pipe (or turbulence) dissipate mechanical energy as thermal energy / internal energy of the water ✓ - M2: therefore (causal link) this thermal energy raises the internal energy of the water, which means its temperature increases as it travels to the tank ✓ [linking marking point] ### Part (d) [2 marks] — Suggest (proposal + justification per §4.4.1) - M1: **Proposal** — identifies a specific pathway by which dissipated energy does NOT enter the water, e.g. heat conduction through pipe walls to surroundings / energy dissipated in motor windings or casing (not in contact with the water stream) / energy radiated/lost as sound and vibration ✓ - M2: **Justification** — explains that because some of the lost power is therefore transferred to the surroundings rather than to the water mass, less energy is available to raise the internal energy of the water, so ΔT = Q/(mc) is smaller than the prediction ✓ ### Marker notes - (b)(i): accept P = ρ(ΔV/Δt) g h done in a single combined step provided all three numerical factors and the final value 741 W appear; award M1+M2 for combined substitution and M3 for the value. - (b)(ii): alternative method — compute P_lost = 1500 − 741 = 759 W and η = 1 − 759/1500; equivalent marks awarded. - (c): do NOT award M2 for merely re-stating "energy is lost as heat" without linking to the temperature rise of the water specifically. - (d): accept any single valid loss pathway in M1 provided M2 explicitly links it to a *reduction* in energy delivered to the water (and hence smaller ΔT). Do not award M2 for vague statements such as "some energy is lost elsewhere". - Show-that target in (b)(i): 741.6 W given to 3 s.f.; student-derived 740–742 W acceptable.