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AQA A-Level Physics includes several required practicals related to thermal physics. This lesson covers the practical skills, techniques, and analysis needed for the three key experiments: measuring specific heat capacity, verifying Boyle's law, and estimating absolute zero. Strong practical skills are tested in Paper 3 and also appear in context questions on Papers 1 and 2.
Spec mapping (AQA 7408 Required Practical 8 — Investigation of Boyle's law and Charles's law / determination of specific heat capacity): This lesson covers the practical skills assessed under RP8, including determination of specific heat capacity by an electrical method (using a heater of known power, measuring temperature rise vs energy supplied), verification of Boyle's law (pressure-volume relationship at constant temperature), and graphical extrapolation to absolute zero (using Charles's law or the pressure law). Candidates must be able to identify variables, evaluate uncertainty, propose improvements, and interpret graphs. (Refer to the official AQA specification document for exact wording.)
Synoptic links: RP8 integrates the gas laws (§3.6.2.2), specific heat capacity (§3.6.2.2), electrical energy E = IVt (§3.5), and uncertainty propagation (§3.1.3). The graphical extrapolation to absolute zero connects directly to the kelvin temperature scale (§3.6.2) and reinforces the thermodynamic interpretation of T = 0 K as the state of minimum mean molecular kinetic energy (§3.6.2.3). Practical evaluation skills assessed here recur in every AQA required practical across all topics, making RP8 a high-value preparation point for Paper 3.
To determine the specific heat capacity of a solid material (typically aluminium or copper) using an electrical heater.
Energy supplied: E = VIt (or read from joulemeter)
Temperature change: Δθ = θ₂ − θ₁
Specific heat capacity: c = E / (mΔθ) = VIt / (mΔθ)
Question: An aluminium block of mass 1.00 kg is heated for 300 s with a heater running at V = 12.0 V and I = 4.00 A. The temperature rises from 21.0 °C to 37.2 °C. Calculate the specific heat capacity.
Solution:
E = VIt = 12.0 × 4.00 × 300 = 14 400 J
Δθ = 37.2 − 21.0 = 16.2 K
c = E / (mΔθ) = 14 400 / (1.00 × 16.2) = 889 J kg⁻¹ K⁻¹
The accepted value for aluminium is 900 J kg⁻¹ K⁻¹, so this result is within 1.2% — indicating a well-conducted experiment with good insulation.
| Source of Error | Type | Effect on c | Improvement |
|---|---|---|---|
| Energy lost to surroundings | Systematic | Measured c is too high (temperature rise is smaller than expected, so c = E/(mΔθ) gives a larger value) | Insulate the block thoroughly |
| Poor thermal contact | Systematic | Temperature reading lags behind true block temperature | Use oil in thermometer and heater holes |
| Non-uniform temperature distribution | Random | Temperature reading may not represent the whole block | Allow time for heat to spread; use a data logger to monitor continuously |
| Reading thermometer before equilibrium | Random | Final temperature may be inaccurate | Continue monitoring after switching off; record the maximum temperature |
| Assuming all electrical energy goes into heating | Systematic | Overestimates energy used for heating | Difficult to eliminate completely; good insulation minimises the error |
If you plot temperature against time, you can use the initial gradient of the line (before significant heat losses occur) to calculate c:
Gradient = Δθ/Δt = P/(mc) (where P = VI is the heater power)
Therefore: c = P / (m × gradient)
This method can give a more accurate result than using the total time and total temperature change, because it uses data from early in the experiment when heat losses are smallest.
Exam Tip: In exam questions about this practical, always discuss heat losses as the main source of systematic error. State that the measured value of c will be higher than the true value because some energy was lost to the surroundings, making the temperature rise smaller than expected.
To verify that pressure is inversely proportional to volume for a fixed mass of gas at constant temperature.
Method 1: Plot p against 1/V
If Boyle's law holds, p ∝ 1/V, so a plot of p against 1/V should give a straight line through the origin.
Calculate 1/V for each data point and plot the graph.
A straight line through the origin confirms Boyle's law.
Method 2: Plot pV against p
If pV = constant, a plot of pV against p should give a horizontal straight line.
Method 3: Plot log(p) against log(V)
If p ∝ V^n, then log(p) = −n log(V) + constant.
For Boyle's law, n = −1, so the gradient should be −1.
A student records the following data:
| p (×10⁵ Pa) | V (cm³) | 1/V (cm⁻³) | pV (×10⁵ Pa cm³) |
|---|---|---|---|
| 1.00 | 50.0 | 0.0200 | 50.0 |
| 1.25 | 40.2 | 0.0249 | 50.3 |
| 1.50 | 33.5 | 0.0299 | 50.3 |
| 2.00 | 25.0 | 0.0400 | 50.0 |
| 2.50 | 20.1 | 0.0498 | 50.3 |
| 3.00 | 16.8 | 0.0595 | 50.4 |
The pV values are approximately constant (50.0–50.4 × 10⁵ Pa cm³), confirming Boyle's law within experimental uncertainty.
| Source of Error | Type | Improvement |
|---|---|---|
| Gas not at constant temperature after compression | Systematic | Wait for temperature to return to room temperature between readings |
| Friction in piston/syringe | Systematic | Take readings increasing and decreasing pressure; use average |
| Air leak from system | Systematic | Check all seals; use silicon grease |
| Parallax error reading scale | Random | Read scale at eye level; use digital sensor |
| Gas column includes dead volume | Systematic | Account for volume of tubing and fittings |
Exam Tip: The most common pitfall in Boyle's law experiments is not allowing the gas to cool between compressions. Fast compression heats the gas, causing the volume to be larger than it should be at constant temperature. Always state that you waited for thermal equilibrium after each change.
To estimate the value of absolute zero by extrapolating the relationship between the volume (or pressure) of a fixed mass of gas and its temperature.
Plot L (or V) against θ (in °C). The data should form a straight line (Charles's law: V ∝ T).
Extrapolate the line backwards (towards lower temperatures) until it crosses the horizontal axis (L = 0 or V = 0). The x-intercept gives an estimate of absolute zero in °C.
If the experiment is well-conducted, the intercept should be close to −273 °C.
A student records the following data:
| Temperature (°C) | Length of air column (mm) |
|---|---|
| 20 | 52 |
| 30 | 54 |
| 40 | 56 |
| 50 | 58 |
| 60 | 60 |
| 70 | 62 |
| 80 | 64 |
The gradient of the line = (64 − 52) / (80 − 20) = 12/60 = 0.20 mm °C⁻¹.
The equation of the line is: L = 0.20θ + c
At θ = 20, L = 52: 52 = 0.20 × 20 + c → c = 52 − 4 = 48
So: L = 0.20θ + 48
Setting L = 0: 0 = 0.20θ + 48 → θ = −48/0.20 = −240 °C
This estimate of absolute zero (−240 °C) is lower than the true value (−273 °C). The discrepancy could be due to experimental errors such as uneven heating, heat losses, or the air not being completely dry.
A better result would typically give −250 to −280 °C.
| Source of Error | Type | Improvement |
|---|---|---|
| Non-uniform temperature in water bath | Random | Stir the water bath continuously |
| Rapid cooling (readings taken before equilibrium) | Systematic | Allow plenty of time between readings; cool slowly |
| Water vapour in the air column | Systematic | Use concentrated sulfuric acid to keep the air dry |
| Air column length measured inaccurately | Random | Use a ruler with mm divisions; avoid parallax |
| Limited temperature range | Affects extrapolation | Use the widest possible range (e.g. 10 °C to 90 °C) |
| Capillary tube not vertical | Systematic | Clamp carefully; check with spirit level |
Exam Tip: The key weakness of the absolute zero experiment is the long extrapolation from the measured data range (~10–90 °C) back to −273 °C. Small errors in the gradient have a large effect on the extrapolated intercept. Always discuss this when evaluating the experiment.
Percentage uncertainty in a single measurement: % uncertainty = (absolute uncertainty / measured value) × 100%
Combining uncertainties:
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