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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.
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.
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