Specific Heat Capacity of Aluminium: Random Error or Systematic?
In a specific-heat-capacity experiment, a 1.000 kg aluminium block has a 12.0 V, 2.00 A electrical heater and a thermometer embedded in it, and is wrapped in insulation. In run A the heater is switched on for 480 s and the temperature rises from 18.0 °C to 29.9 °C. In run B, starting at 30.5 °C, the supply is adjusted to 12.0 V, 4.00 A for 240 s and the temperature rises to 41.8 °C. The room temperature is 22.0 °C and the accepted value for aluminium is 900 J kg⁻¹ K⁻¹. The voltmeter, ammeter and stopwatch are each accurate to about 0.5%, the balance to ±0.005 kg, and each thermometer reading to ±0.1 °C. A student claims the two results agree 'within experimental error' and that any discrepancy with the accepted value is due to random error in the temperature readings.
1. Calculate the value of c from each run and decide whether the student's claim about random error is defensible.
Read the worked answer
Run A: Q = VIt = 12.0 × 2.00 × 480 = 1.152 × 10⁴ J; Δθ = 11.9 K c = 11 520/(1.000 × 11.9) = 968 J kg⁻¹ K⁻¹ (+7.6% on 900)
Run B: Q = 12.0 × 4.00 × 240 = 1.152 × 10⁴ J; Δθ = 11.3 K c = 11 520/11.3 = 1019 J kg⁻¹ K⁻¹ (+13.3%)
Uncertainty budget: Δθ is a difference of two readings each ±0.1 °C, so ±0.14 K - 1.2% in A. Voltage, current and time contribute about 1% in total; mass 0.1%. Combined, roughly 1.5%, i.e. ±15 J kg⁻¹ K⁻¹.
The excess is 68 and 119 J kg⁻¹ K⁻¹, about 5 and 8 standard deviations. The difference between A and B (51 J kg⁻¹ K⁻¹) is also more than twice the combined uncertainty. The claim fails: the scatter is far too large and, crucially, it is one-sided - both values lie above 900. Random error scatters about the true value; a consistent excess points to a systematic effect. The sign is diagnostic: heat escaping from the block lowers Δθ and therefore inflates the calculated c.
2. Explain why runs A and B give different values even though the same energy was supplied.
Read the worked answer
The energy is identical (11.52 kJ) but the two runs sit at different temperatures relative to the 22.0 °C room.
- Run A: 18.0 → 29.9 °C, mean 23.95 °C, mean excess ≈ 2 K
- Run B: 30.5 → 41.8 °C, mean 36.2 °C, mean excess ≈ 14 K
The rate of heat loss is roughly proportional to (T − T_room), so run B loses heat about seven times faster than run A for the same duration pattern, and a larger fraction of the 11.52 kJ never reaches the block. A smaller Δθ (11.3 K instead of the loss-free 12.8 K) gives a larger apparent c.
Run A is closer not because it is more careful but because its working range nearly straddles room temperature, so heat gained at the start roughly cancels heat lost at the end. That is a design principle worth naming: choose the temperature range so that it is centred on room temperature.
The correct analysis is to plot apparent c against mean excess temperature and extrapolate to zero excess; the intercept, not either individual run, is the best estimate.
3. Identify the most likely systematic correction that would close the remaining gap, and state one design improvement.
Read the worked answer
Even with heat loss corrected, the block is not the only thing being heated. The embedded heater, the thermometer and its pocket have heat capacity C_i, so the correct statement is:
Q = (mc + C_i)Δθ
Run A gives a total heat capacity of 11 520/11.9 = 968 J K⁻¹. If losses were negligible, the unaccounted capacity would be 968 − 900 = 68 J K⁻¹. A 0.04 kg copper heater contributes 0.04 × 385 ≈ 15 J K⁻¹, so this explains only part of the gap; the rest is residual loss. Cooling-correction or extrapolation methods would remove the remainder.
Design improvement (choose one and justify):
- Pre-cool the block to about 14 °C so that the range 14 → 26 °C is centred on the 22 °C room, making net heat exchange approximately zero.
- Use a lower power (e.g. 12 V, 1.00 A, 960 s) to reduce temperature gradients between the heater and the outer surface.
- After switching off, wait for the lagging to equilibrate before taking the final reading (stirring is not possible; the thermometer pocket must be deep enough).
- Include the heater and thermometer in the effective mass, or calibrate the block once with a material of known c.
Option 1 is best: it converts a systematic error into a near-cancellation without needing extra apparatus.