Specific Heat Capacity of Aluminium: Correcting a Discrepancy
A student determines the specific heat capacity of aluminium using a 24 W electrical heater embedded in a 0.500 kg aluminium block. The block is wrapped in lagging. The heater is switched on for 300 s. The temperature rises from 18.0 °C to 30.0 °C. The student calculates c = Pt/(mΔT) and obtains 1200 J/(kg °C). The accepted value is about 900 J/(kg °C). The thermometer reads 0.5 °C too high throughout. The power supply is actually delivering 22 W, not 24 W, because the heater has a resistance of 12 Ω and the supply is 16.2 V. The block loses heat to the surroundings, and the room is at 18 °C. The temperature is read every 30 s; the temperature-time graph is a straight line after the first minute but curves near the start.
1. Recalculate the specific heat capacity using the actual power and the corrected temperature rise, and decide whether the zero error explains the discrepancy.
Read the worked answer
Using the actual power:
c = Pt/(mΔT) = 22 × 300/(0.500 × 12.0) = 6600/6.00 = 1100 J/(kg °C).
The thermometer reads 0.5 °C too high throughout. Both the initial and final readings are shifted by the same amount, so the temperature rise is still 12.0 °C. The zero error therefore cancels and does not explain the discrepancy. The calculated value is still about 22% above the accepted 900 J/(kg °C).
2. Explain how heat loss and the shape of the temperature-time graph account for the remaining discrepancy.
Read the worked answer
Heat lost to the surroundings means that not all of the 6600 J supplied raises the block's temperature. If the block were perfectly lagged, its temperature rise would be larger. In the calculation the measured ΔT is too small, so c is overestimated.
At the start the block is at room temperature, so the rate of heat loss is small; this is why the graph curves while the heater and thermometer settle. As the block warms, the temperature difference from the room increases and the heat loss becomes larger. The later straight-line section does not mean there is no heat loss; it means the block is warming at a steady rate while losing heat at a steady rate. A cooling correction can be made by recording the cooling curve after switching off and extrapolating it, or by using the rate of temperature rise at the instant the block is at room temperature.
3. Suggest two improvements and evaluate whether simply heating for longer would improve the experiment.
Read the worked answer
Two useful improvements are:
- Improve the lagging and fit a lid, so the measured temperature rise is closer to the true rise.
- Use a data logger or a thermometer with finer resolution and take readings more frequently, reducing reading uncertainty.
Heating for longer would give a larger temperature rise and reduce the percentage effect of the thermometer resolution, but it would also increase the heat loss because the block would be further above room temperature. It is not automatically better. The best judgement is to combine good insulation with a cooling correction, rather than simply increasing the heating time.