## ERQ · 14 marks · Topics: E.4 Fission + A.3 Work, energy and power
**Stem.** A pressurised water reactor (PWR) at a coastal power station uses uranium-235 as fuel. The reactor core operates at a steady thermal power output of 3200 MW. A representative fission reaction in the core is:
$$^{235}_{92}\text{U} + ^{1}_{0}\text{n} \rightarrow ^{141}_{56}\text{Ba} + ^{92}_{36}\text{Kr} + 3\,^{1}_{0}\text{n}$$
Each fission event releases 200 MeV of energy on average. The thermal energy is transferred to pressurised water in a primary loop, which drives turbines coupled to electrical generators. Coolant water is pumped from the sea at a rate of 5.0 × 10⁴ kg s⁻¹, entering at 12 °C and leaving the condenser at 22 °C. The specific heat capacity of seawater is 4000 J kg⁻¹ K⁻¹. The electrical power delivered to the grid is measured at 1100 MW.
### Part (a) Outline [2 marks] · AO1 · Topic: E.4
Outline why energy is released in the fission of a uranium-235 nucleus.
### Part (b)(i) Calculate [3 marks] · AO2/AO3 · Topic: E.4
Calculate the rate at which uranium-235 nuclei must undergo fission to sustain the 3200 MW thermal power output.
### Part (b)(ii) Show that [2 marks] · AO2/AO3 · Topic: E.4
Show that the mass of uranium-235 consumed by fission in a 24-hour period is approximately 3.4 kg.
### Part (c) Determine [3 marks] · AO3 · Topic: E.4 + A.3
Using the seawater coolant data, determine the overall efficiency of the power station in converting fission thermal energy into electrical energy delivered to the grid.
### Part (d) Suggest [4 marks] · AO3 · ASSUMPTIONS DISCRIMINATOR
The theoretical maximum (Carnot) efficiency for the turbine cycle operating between 285 °C (primary loop) and 22 °C (condenser) is approximately 47%. The efficiency calculated in (c) is significantly lower. Suggest **two** distinct physical reasons for this discrepancy, explaining the energy pathway in each case.
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## Mark Scheme
### Part (a) [2 marks]
- Mark 1: Identifies that the binding energy per nucleon of the fission products (Ba-141, Kr-92) is greater than that of U-235 / products lie nearer the peak of the binding-energy-per-nucleon curve [ECF: no]
- Mark 2: States that the resulting mass defect Δm is released as energy via E = Δmc² (or: total rest mass of products < rest mass of reactants) [ECF: no]
### Part (b)(i) [3 marks]
- Mark 1: Converts 200 MeV per fission to joules: 200 × 10⁶ × 1.60 × 10⁻¹⁹ = 3.20 × 10⁻¹¹ J [ECF: no]
- Mark 2: Uses rate = P / E_per_fission = 3.20 × 10⁹ / 3.20 × 10⁻¹¹ [ECF: yes]
- Mark 3: Answer = 1.0 × 10²⁰ fissions s⁻¹ (2 s.f., correct units) [ECF: yes]
### Part (b)(ii) [2 marks]
- Mark 1: Number of fissions in 24 h = (rate from b(i)) × 86400 = 8.64 × 10²⁴ AND mass per nucleus = 235 × 1.66 × 10⁻²⁷ = 3.90 × 10⁻²⁵ kg [ECF: yes from b(i)]
- Mark 2: Total mass = 8.64 × 10²⁴ × 3.90 × 10⁻²⁵ ≈ 3.37 kg ≈ 3.4 kg (must show working to at least 3 s.f. to justify "show that") [ECF: yes]
### Part (c) [3 marks]
- Mark 1: Calculates heat rejected to sea: P_rejected = ṁcΔT = (5.0 × 10⁴)(4000)(10) = 2.0 × 10⁹ W = 2000 MW [ECF: no]
- Mark 2: Recognises useful work output is the 1100 MW electrical (not thermal); efficiency η = P_electrical / P_thermal = 1100 / 3200 [ECF: yes]
- Mark 3: η = 0.34 or 34% (2 s.f.) [ECF: yes]
### Part (d) [4 marks]
Award 2 marks per distinct reason (1 for identification, 1 for energy-pathway explanation). Maximum 4. Accept any two of:
- **Carnot is an upper bound for ideal reversible cycles:** real turbine/steam cycle is irreversible — friction in turbine blades, turbulence, throttling losses cause entropy generation, so work output < ideal W = QΔT/T_hot [1 ID + 1 explanation]
- **Resistive (I²R) losses in generator windings and step-up transformers:** mechanical work delivered by turbine shaft is partially dissipated as heat in copper coils rather than transferred to the grid [1 ID + 1 explanation]
- **Auxiliary power demand (parasitic load):** coolant pumps, control systems, and cooling-tower fans consume mechanical/electrical work drawn from the cycle, reducing net electrical output below gross generator output [1 ID + 1 explanation]
- **Heat losses to surroundings from pipework / pressure vessel:** thermal energy escapes by conduction and radiation before reaching the turbine, so Q_into_turbine < 3200 MW [1 ID + 1 explanation]
- **Incomplete heat transfer in steam generator / condenser ΔT pinch:** finite temperature differences across heat exchangers mean the working fluid never reaches 285 °C nor cools to 22 °C, narrowing the effective T_hot − T_cold gap [1 ID + 1 explanation]
[ECF: yes — accept reasons consistent with candidate's value in (c)]
### Marker notes
- Alternative method for (b)(i): candidates who first compute mass-energy per fission in kg then divide power are accepted provided arithmetic is consistent.
- For (c), candidates who incorrectly equate "thermal power" with the 2000 MW rejected (rather than 3200 MW source) lose Mark 2 but may earn Mark 1 and an ECF Mark 3.
- For (d), do **not** award marks for vague answers such as "energy is lost as heat" without identifying *where* or *how*. The explanation mark requires a named pathway (friction, I²R, parasitic load, conduction, pinch).
- Reject "Carnot is just a theoretical limit" as a standalone reason — candidate must identify a specific irreversibility.