Specific Heat Capacity: When Errors Cancel
A student places 200 g of water at 18.5 °C in a lagged copper calorimeter of mass 60 g and heats it with an immersion heater marked '50 W, 12 V'. The laboratory power supply actually delivers 11.0 V. After 180 s the water temperature is 29.0 °C. The thermometer reads to the nearest 0.5 °C. The student ignores the calorimeter's heat capacity and assumes the heater transfers all its energy to the water, then compares the result with 4200 J/(kg K). Copper has a specific heat capacity of 390 J/(kg K). The calorimeter is lagged, but the lid is not a tight fit and the water is not stirred during the run.
1. Calculate the specific heat capacity the student obtains, and compare it with the accepted value. Then state what the answer would have been if the supply voltage had been ignored.
Read the worked answer
Power actually delivered = 50 × (11.0/12.0)² = 42.0 W.
Energy supplied = 42.0 × 180 = 7560 J.
Temperature rise ΔT = 29.0 − 18.5 = 10.5 K.
c = E/(mΔT) = 7560/(0.200 × 10.5) = 3600 J/(kg K).
That is 14% below 4200 J/(kg K).
If the rated 50 W had been used: E = 9000 J and c = 9000/2.10 = 4290 J/(kg K), apparently only 2% high. The 'better' answer comes from two errors cancelling: the power was overestimated and the heat losses were ignored. Agreement with the accepted value is therefore not evidence that the method was sound.
2. Estimate whether heat taken by the calorimeter and heat lost to the room can plausibly account for the missing 14%.
Read the worked answer
Calorimeter heat capacity = 0.060 × 390 = 23.4 J/K.
Energy it absorbs = 23.4 × 10.5 ≈ 250 J, about 3% of 7560 J.
Heat lost to the room is harder to pin down. With a poor lid and no stirring, the water surface may be several kelvin above room temperature for most of the run, so a few watts of loss is realistic - say 3 W for 180 s = 540 J, about 7%. The two together, roughly 10%, are the right order to explain most of the 14%.
But the measurement itself is coarse: each reading is ±0.25 °C, so ΔT = 10.5 ± 0.5 K, a 5% uncertainty. A defensible statement is c = 3600 ± 250 J/(kg K), consistent with 4200 once the known systematic effects are included.
3. Recommend improvements, and identify which one would most reduce the error.
Read the worked answer
Useful changes:
- Measure current and voltage with meters during the run and compute P = VI, rather than trusting the rating or assuming the supply is exactly 12 V.
- Stir continuously and record temperature every 30 s, then correct for cooling by extrapolating the warming and cooling trends to the mid-point of the run.
- Use a larger temperature rise, such as 15 K, so that the ±0.5 °C thermometer resolution becomes a smaller fraction.
- Include the calorimeter's heat capacity in the calculation, and improve the lid and insulation.
The largest single gain is measuring V and I directly. The power estimate was the biggest identified systematic error, and it also produced the false agreement when the supply was assumed to be 12 V. Improving the lid and stirring addresses the next largest term.