A-Level · Physics · AQA · Mark scheme decoded
AQA A-Level Physics: Internal Energy and Heat Transfer — mark scheme explained
The short answer
Understanding internal energy and heat transfer is crucial in the study of thermodynamics, which forms a significant part of A-Level Physics. Internal energy is the sum of the randomly distributed kinetic energies and potential energies of the particles within a system.
The question
A 2 kg block of ice at -10°C is heated until it completely melts. The specific heat capacity of ice is 2100 J/kg°C, and the specific latent heat of fusion for ice is 334,000 J/kg. Calculate the total energy required.
[Paraphrased for study — not reproduced from any exam paper.]
Mark scheme, decoded
How the examiner actually awards the marks on this topic.
Gradora's own decode of the marking approach — not the exam board's published mark scheme.
How marks are awarded
For calculation questions, marks are typically awarded for correct use of formulas, substitution of values, and final answers. For explanation questions, marks are given for clarity, accuracy, and completeness of the response.
What the command words demand
- Calculate
- Perform a numerical calculation using given data and appropriate formulas.
- Explain
- Provide a detailed account of why or how something happens, including relevant concepts and principles.
- Identify
- Recognize and name specific elements, errors, or factors in a given context.
- Describe
- Give a detailed account of the characteristics or features of a concept or process.
Model answer
A full-mark response to the question above, worked through step by step.
Timing: Allocate about 5-7 minutes per question to ensure you have enough time to read, understand, and answer each part accurately.
- Calculate the energy required to raise the temperature of the ice from -10°C to 0°C using Q = mcΔθ.0 marks
- Q 1 = 2 kg × 2100 J/kg°C × (0°C - (-10°C))1 mark
- Q 1 = 2 kg × 2100 J/kg°C × 10°C0 marks
- Q 1 = 42,000 J1 mark
- Calculate the energy required to melt the ice at 0°C using Q = ml.0 marks
- Q 2 = 2 kg × 334,000 J/kg1 mark
- Q 2 = 668,000 J1 mark
- Add the two energies to find the total energy required.0 marks
- Total Q = Q 1 + Q 20 marks
- Total Q = 42,000 J + 668,000 J0 marks
- Total Q = 710,000 J1 mark
Final answer: 710,000 J
Work through every step correctly and you earn all 5 marks.
Another worked example
A 5 kg block of copper is heated from 20°C to 80°C. The specific heat capacity of copper is 385 J/kg°C. Calculate the energy required.
- Use the formula Q = mcΔθ to calculate the energy required.1 mark
- Q = 5 kg × 385 J/kg°C × (80°C - 20°C)1 mark
- Q = 5 kg × 385 J/kg°C × 60°C0 marks
- Q = 115,500 J1 mark
Final answer: 115,500 J
Work through every step correctly and you earn all 3 marks.
Common mistakes
Confusing specific heat capacity with specific latent heat.
Why it happens: Students often mix up the definitions and formulas for these two concepts, leading to incorrect calculations.
Fix: Review the definitions: Specific heat capacity is used for temperature changes (Q = mcΔθ), while specific latent heat is used for phase changes (Q = ml).
Forgetting to convert units, especially when dealing with mass and time.
Why it happens: Students sometimes forget to ensure all units are consistent before performing calculations, leading to incorrect answers.
Fix: Always check that the units for mass (kg), temperature (°C), and time (s) are consistent. Convert if necessary.
Using the wrong formula for continuous flow systems.
Why it happens: Students may use the standard Q = mcΔθ formula instead of the continuous-flow relationship (power = mass flow rate × c × Δθ + rate of heat loss).
Fix: Remember to use the correct continuous-flow approach: P = (m/t)cΔθ + H, eliminating heat loss by comparing two different flow rates (P₁ − P₂ = ((m₁ − m₂)/t)cΔθ).
Neglecting to account for phase changes in energy calculations.
Why it happens: Students sometimes forget that during a phase change, all added energy goes into changing the state of the substance, not increasing its temperature.
Fix: Always check if the problem involves a phase change and use the appropriate formula (Q = ml) for that part of the calculation.
Failing to identify and correct systematic errors in experiments.
Why it happens: Students may not recognize or address systematic errors, leading to inaccurate experimental results.
Fix: Calibrate equipment, use more accurate instruments, and correct for known biases. Always check for consistency in measurements.
Incorrectly applying the first law of thermodynamics.
Why it happens: Students may misinterpret or misuse the formula ΔU = Q - W, leading to incorrect calculations of internal energy changes.
Fix: Ensure a clear understanding of the first law: The change in internal energy is equal to the heat added minus the work done by the system. Practice applying this concept in various scenarios.
Where the marks go
The question types you’ll meet on this topic and the marks each one carries — so you know what to expect and where to focus.
| Question type | What you’re asked to do | Marks |
|---|---|---|
| Calculate Thermal Energy | Find total energy to heat ice then melt it using specific heat capacity and latent heat. | 5 |
| Calculate Thermal Energy | Use Q = mcΔθ to find the energy needed to heat a copper block. | 3 |
| Calculate Temperature Change | Use energy, mass and specific heat capacity to find a temperature rise over time. | 4 |
| Calculate Thermal Equilibrium | Apply conservation of energy to find the final temperature when two substances reach thermal equilibrium. | 5 |
| Total across these question types | 17 | |
Question types and mark tariffs are Gradora’s guidance based on how this topic is typically examined — not the board’s official paper structure.