Generated ERQ

✓ passed A.3 Work, energy and power × B.5 Current and circuits 12 marks HL 2 passes 67.55s $0.3708

## ERQ · 12 marks · Topics: A.3 Work, energy and power + B.5 Current and circuits **Stem.** A portable electric heater is rated 2000 W at 230 V rms and is used to heat 1.50 kg of water inside a sealed metal chamber. The heater is connected to the mains socket by a 5.0 m two-core flexible cable; each copper core has cross-sectional area 1.0 × 10⁻⁶ m² and resistivity 1.7 × 10⁻⁸ Ω m. When switched on at a mains voltage of 230 V rms, the water temperature rises from 18.0 °C to 92.0 °C in 240 s. The specific heat capacity of water is 4180 J kg⁻¹ K⁻¹. The student assumes the heater behaves as an ohmic resistor at its rated values and that the cable resistance is the only significant circuit loss outside the heating element. ### Part (a) Define [2 marks] · AO1 · Topic: A.3 Define *power* and state the unit in fundamental SI base units. ### Part (b)(i) Calculate [2 marks] · AO2 · Topic: A.3 Calculate the useful thermal energy transferred to the water during the 240 s. ### Part (b)(ii) Determine [2 marks] · AO2 · Topic: A.3 Determine the efficiency of the heater, taking the rated electrical input power as the total input. ### Part (c) Show that [3 marks] · AO3 · Topic: A.3 + B.5 The student now models the cable + heater as a series circuit on the 230 V supply. Show that the power dissipated in the cable (both cores combined) is approximately 6.7 W. ### Part (d) Suggest [3 marks] · AO3 · ASSUMPTIONS DISCRIMINATOR The efficiency obtained in (b)(ii) is less than 100 %, yet the cable loss calculated in (c) accounts for only a small part of this shortfall. Suggest three further reasons why the experimentally determined efficiency is below 100 %. --- ## Mark Scheme ### Part (a) [2 marks] - M1: Power defined as rate of energy transfer (or rate of doing work) [ECF: no] - M2: Unit kg m² s⁻³ [ECF: no] ### Part (b)(i) [2 marks] - M1: Use of Q = mcΔT with ΔT = 74.0 K [ECF: no] - M2: Q = 1.50 × 4180 × 74.0 = 4.64 × 10⁵ J (accept 4.6 × 10⁵ J) [ECF: yes] ### Part (b)(ii) [2 marks] - M1: Total electrical input energy = 2000 × 240 = 4.80 × 10⁵ J [ECF: no] - M2: Efficiency = 4.64 × 10⁵ / 4.80 × 10⁵ = 0.967 or 96.7 % (accept 96–97 %) [ECF: yes] ### Part (c) [3 marks] - M1: Total cable resistance R_c = ρL/A summed over both cores = 2 × (1.7 × 10⁻⁸ × 5.0)/(1.0 × 10⁻⁶) = 0.17 Ω [ECF: no] - M2: Heater resistance from rated values R_h = V²/P = 230²/2000 = 26.45 Ω, giving series current I = 230/(26.45 + 0.17) = 8.64 A [ECF: yes] - M3: P_cable = I²R_c = 8.64² × 0.17 ≈ 6.7 W (accept 6.6–6.8 W) [ECF: yes] ### Part (d) [3 marks] — award exactly one mark per distinct point below (max 3) - M1: Heat loss from the chamber/water to the surroundings during the 240 s (conduction/convection/radiation) [ECF: no] - M2: Thermal energy absorbed by the metal chamber/heating element itself is not counted in Q = mcΔT for the water [ECF: no] - M3: One of: (i) actual mains voltage fluctuates from 230 V; (ii) resistance of the heating element increases with temperature, so it does not draw the rated 2000 W throughout; (iii) losses in internal components (switch, fuse, contacts) not modelled [ECF: no] ### Marker notes - (b)(ii): Accept alternative input = V²t/R using rated R giving same answer. - (c) alternative: candidates who approximate I ≈ 2000/230 = 8.70 A (ignoring cable in current calculation) and then compute P_cable = 8.70² × 0.17 = 12.9 W do NOT earn M3, but earn M1 and a partial M2 for the heater-resistance step (max 2/3). - (d): Do not award two marks for two phrasings of the same physical loss (e.g. "heat escapes to air" and "heat lost by convection" count once). Reject vague answers such as "energy is wasted" or "the heater is not perfect".

Critic verdicts

Iteration 0 — failed

ao_ladder pass

The ERQ strictly obeys the cognitive ladder. Part (a) is AO1 (Define) asking for a definition and unit. Parts (b)(i)–(b)(ii) and (c) are AO2/AO3 (Calculate, Determine, Show that) with multi-step quantitative reasoning. Part (d) is AO3 (Suggest) requiring reasoning and discrimination among distinct explanations. The progression from lower to higher-order thinking is clear and unbroken.
cross_theme pass

The ERQ genuinely blends A.3 (Work, energy and power) and B.5 (Current and circuits) across both parts (b)–(c). Part (b) uses energy conservation and thermal concepts (A.3); part (c) requires circuit analysis (resistance calculation via ρL/A, Ohm's law, power in resistances) to determine cable losses. Both concepts are essential to solving (c)—it cannot be solved by energy methods alone. The second topic is not merely name-checked or relegated to assumptions; it is central to the quantitative argument.
discriminator pass

Part (d) explicitly asks students to 'Suggest, with reasoning, three distinct reasons why the energy delivered to the water is less than the rated electrical energy supplied.' This is a genuine model-critique question that requires students to identify and justify limitations and assumptions in the experimental setup (perfect insulation, temperature independence of resistance, voltage stability, etc.). It goes well beyond calculation.
rule_of_one fail

Part (c) lists 3 award criteria but the mark scheme conflates M2 and M3 into a single computational step (finding I and then P_cable). The actual distinct operations are: (1) calculate cable resistance, (2) calculate heater resistance, (3) calculate current, (4) calculate cable power — which is 4 operations for a 3-mark part. Additionally, part (d) is marked as 3 marks but the scheme awards '1 mark for each distinct point (max 3)' without explicitly listing 3 separate criteria; this makes the mark-to-bullet relationship ambiguous.

FIX: For part (c): Separate M2 into two distinct marks: M2 for calculating heater resistance and current (R_h = 26.45 Ω, I = 8.67 A), and M3 for calculating cable power (P_cable ≈ 6.8 W or 3.4 W per core). Alternatively, compress to 2 marks by combining resistance calculations into one criterion. For part (d): Explicitly list exactly 3 distinct award criteria as bullet points (e.g. M1: Heat loss to surroundings; M2: Thermal energy absorbed by calorimeter/vessel; M3: Either mains voltage variation OR temperature-dependent resistance OR internal component losses). This removes ambiguity and ensures the 3-mark count matches 3 independent award points.
topic pass

The ERQ authentically integrates both topics. Part (c) explicitly requires **A.3 concepts** (power dissipation as P = I²R, energy conversion) **and B.5 concepts** (circuit resistance, series circuit analysis, ohm's law applied to find current). The cable resistance calculation and power loss in the conductor are intrinsically circuit-based; the power dissipation interpretation is intrinsically energy-based. Neither topic is merely nominal—both are structurally essential to the solution.

Iteration 1 — passed

ao_ladder pass

The ERQ correctly obeys the cognitive ladder: part (a) is AO1 with Define command term; parts (b)(i)–(b)(ii) and (c) are AO2–AO3 with Calculate/Determine/Show that; part (d) is AO3 with Suggest. The progression from foundational knowledge through calculation to critical evaluation is sound and meets IB standards.
cross_theme pass

Part (c) genuinely integrates both A.3 (work, energy, power) and B.5 (current, circuits). The solution requires calculating heater resistance from rated power (A.3 concept), then finding series circuit current using Ohm's law (B.5), and finally computing power dissipation in the cable (A.3 applied to B.5 context). Both physics concepts are essential to the calculation; neither is filler.
discriminator pass

Part (d) explicitly asks students to 'suggest three further reasons why the experimentally determined efficiency is below 100 %', which requires them to critique the assumptions and limitations of the idealized model (perfect insulation, heater as pure ohmic resistor, no parasitic losses, constant resistance). This is genuine model-critique work beyond calculation.
rule_of_one pass

Each part's mark count aligns well with its award criteria. Parts (a)–(c) have 2–3 marks matching 2–3 distinct operations (definition, formula/substitution/answer). Part (d) awards max 3 marks for up to 3 distinct physical reasons, which is appropriate for a 'Suggest' AO3 question. No divergence exceeds 1.
topic pass

Part (c) genuinely integrates both topics: it requires circuit analysis (B.5—series resistance, current calculation, Ohm's law) to determine the current flowing through the cable, and then applies energy/power concepts (A.3—P = I²R) to quantify power dissipation. Neither topic is merely nominal; both are essential to solving the problem. The scenario naturally bridges them via the two-core cable's resistance within the electrical circuit feeding the heater.

Planner output (stage 1)

Archetype: real-world device
Scenario: A portable electric heater operates on a 230 V AC mains supply and is advertised to deliver 2000 W of thermal power. The heater contains a heating element (resistor) and an internal circuit with a switch and protective fuse. A student investigates the energy conversion efficiency by measuring the temperature rise of water in a chamber, calculating work done by the heating element, and evaluating power losses in the circuit due to resistance in connecting wires and internal components.
Cross-pollination part: c
Discriminator part: d

Part	Marks	AO	Command
a	2	AO1	Define
b(i)	2	AO2	Calculate
b(ii)	2	AO2	Determine
c	3	AO3	Show that
d	3	AO3	Suggest

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