## ERQ · 10 marks · Topics: A.3 Work, energy and power + B.5 Current and circuits · Archetype: theory_application
**Integration:** primary=A.3 Work, energy and power, secondary=B.5 Current and circuits (strength: supporting)
**Stem.** A portable electric heater is connected to a household mains supply of 230 V. The heater contains a single resistive element and is labelled "2.00 kW at 230 V". When switched on, the element warms a small room, and the manufacturer claims that the heater transfers thermal energy to the surrounding air at a constant rate equal to its rated power. A student uses the heater for 15 minutes and treats the heater as a closed system that converts all electrical energy directly into thermal energy delivered to the room air. The resistance of the element at its normal working temperature is to be taken as constant in parts (a)–(c).
### Part (a) Define [2 marks] · AO1 · Topic: A.3 Work, energy and power
Define *power* and state its SI unit.
### Part (b) Calculate [3 marks] · AO2 · Topic: A.3 Work, energy and power
Using the manufacturer's claim, calculate the thermal energy, in MJ, delivered to the room air during the 15-minute period.
### Part (c) Show that [3 marks] · AO2 · Topic: A.3 Work, energy and power + B.5 Current and circuits
The heater is connected directly to the 230 V mains supply. Show that the resistance of the heating element at its working temperature is about 26 Ω.
### Part (d) Discuss [2 marks] · AO3 · ASSUMPTIONS DISCRIMINATOR
The calculation in part (b) assumes the heater delivers thermal energy to the room air at a constant rate of 2.00 kW from the instant it is switched on. Discuss one reason why this assumption is not realistic, and comment on whether your answer in (b) is an overestimate or an underestimate of the actual thermal energy delivered to the air.
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## Mark Scheme
### Part (a) [2 marks] — Define
- M1: Power is the rate of doing work / rate of energy transfer (per unit time) [no ECF]
- M2: SI unit is the watt (W) OR J s⁻¹ [no ECF]
### Part (b) [3 marks] — Calculate
- M1: Recognises E = Pt with correct conversion t = 15 × 60 = 900 s [no ECF]
- M2: Substitution: E = 2000 × 900 [ECF from (a) not applicable]
- M3: E = 1.80 × 10⁶ J = 1.80 MJ (accept 1.8 MJ) [no ECF]
### Part (c) [3 marks] — Show that
- M1: Identifies P = V²/R (or equivalent route via P = VI and V = IR) [no ECF]
- M2: Rearranges to R = V²/P and substitutes: R = (230)² / 2000 [no ECF]
- M3: R = 26.45 Ω ≈ 26 Ω (student-derived value must lie in 26.4–26.5 Ω before rounding) [no ECF]
### Part (d) [2 marks] — Discuss (model critique: limitation + implication for (b))
*Pathway A — transient warm-up of element:*
- M1a: Identifies that when switched on, the element is cold and must itself absorb thermal energy before reaching working temperature, so not all electrical energy is delivered to the air at the rated rate from t = 0 / the assumption of constant 2.00 kW transfer to the air from switch-on is invalid
- M2a: Therefore the answer in (b) is an **overestimate** of the thermal energy actually delivered to the air during the 15 minutes (some energy is stored as internal energy of the element)
*Pathway B — temperature dependence of resistance:*
- M1b: Identifies that the resistance of the element is lower when cold than when hot, so the initial electrical power dissipated P = V²/R exceeds 2.00 kW (or, the rated 2.00 kW only applies at working temperature, not throughout)
- M2b: Therefore the answer in (b) is an **underestimate** of the thermal energy delivered, because the heater dissipates more than 2.00 kW during the warm-up phase
*Pathway C — heat losses through casing / conduction to surroundings other than room air:*
- M1c: Identifies that some thermal energy is transferred to the casing, walls, or lost by conduction rather than being delivered to the room air, so the heater is not a perfectly closed system delivering all energy to the air
- M2c: Therefore the answer in (b) is an **overestimate** of the thermal energy delivered specifically to the room air
Award M1 for any one valid limitation of the constant-2.00-kW-to-air model AND M2 for the matching directional comment on (b)'s validity. M2 is only awarded if it is consistent with the M1 mechanism chosen.
### Marker notes
- Alternative method accepted for (c): P = VI gives I = 2000/230 = 8.70 A, then R = V/I = 230/8.70 = 26.4 Ω.
- Show-that target in (c): R = 26.45 Ω given to 4 sig figs; student-derived 26.4–26.5 Ω acceptable before rounding to 26 Ω.
- (d) accept any one of the three pathways above; do not award M2 if the direction of the error contradicts the mechanism in M1.
- Do not credit vague statements such as "the heater is not 100% efficient" without identifying where the energy goes or why the rate is not constant.