Specific Heat Capacity and an Over-Confident Conclusion
A student measures the specific heat capacity of water using an immersion heater. The water has mass 0.500 kg, measured on a balance reading to ±1 g. The heater is connected to a 12.0 V supply, with a current of 4.00 A; the voltmeter and ammeter each have a reading uncertainty of one unit in the last digit (0.1 V and 0.05 A). The heater is switched on for 300 s, timed with a stopwatch to ±0.1 s. The water temperature rises from 21.0 °C to 27.4 °C; the thermometer has an estimated uncertainty of ±0.2 °C. The student calculates c = VIt/(mΔT) and obtains 4500 J kg⁻¹ K⁻¹. The accepted value is 4180 J kg⁻¹ K⁻¹. The student concludes that the value is too high because energy is lost to the surroundings, and that the experiment has 'proved' the accepted value is wrong.
1. Show the calculation of c and determine its absolute uncertainty. State whether the accepted value lies within this uncertainty.
Read the worked answer
Energy supplied: E = VIt = 12.0 × 4.00 × 300 = 14 400 J.
Temperature rise: ΔT = 27.4 − 21.0 = 6.4 K.
c = E/(mΔT) = 14 400/(0.500 × 6.4) = 4500 J kg⁻¹ K⁻¹.
Percentage uncertainties: V: 0.1/12.0 = 0.83%; I: 0.05/4.00 = 1.25%; t: 0.1/300 = 0.03%; m: 0.001/0.500 = 0.20%; ΔT: each reading ±0.2 °C, so the difference has ±0.28 °C, giving 0.28/6.4 = 4.4%.
Combining in quadrature: √(0.83² + 1.25² + 0.03² + 0.20² + 4.4²) ≈ 4.7%.
Absolute uncertainty: 0.047 × 4500 ≈ 210 J kg⁻¹ K⁻¹, so c = (4500 ± 210) J kg⁻¹ K⁻¹.
The accepted 4180 differs by 320, about 1.5 times the uncertainty. It is close to the edge of the range, so the data do not establish a real discrepancy.
2. Evaluate the student's conclusion that heat loss explains the discrepancy. What further evidence would distinguish heat loss from random measurement error?
Read the worked answer
Heat loss would indeed make the calculated c too high: if some supplied energy leaves the water, ΔT is smaller than it should be for the measured electrical energy, so E/(mΔT) is larger. The size is plausible: the energy needed to warm the water by 6.4 K is 0.500 × 4180 × 6.4 = 13 376 J, while 14 400 J was supplied; the difference is 1024 J, about 7.1%. A 7.1% loss would explain the 4500 value almost exactly.
However, the random uncertainty is also about 4.7%, and the discrepancy is only about 1.5σ. The single run cannot distinguish a consistent 7% heat loss from a random fluctuation. To distinguish them: repeat the experiment several times and see whether the result is consistently high (systematic heat loss) or scattered around 4180 (random error). Alternatively, measure the cooling rate after switching off the heater and extrapolate, or insulate the vessel and see whether c falls toward the accepted value. The claim that the experiment has 'proved' the accepted value wrong is not justified.
3. Suggest two improvements to the experiment, and explain for each whether it reduces random uncertainty, systematic error, or both.
Read the worked answer
Insulate the vessel and add a lid. This mainly reduces systematic heat loss, so it should lower the calculated c if heat loss was the cause. It also reduces variation between runs, so it can reduce random uncertainty slightly.
Use a larger temperature rise (for example, heat to 35–40 °C) or a thermometer with smaller uncertainty. A larger ΔT reduces the fractional uncertainty in ΔT, which is the dominant random term here. The trade-off is that a larger ΔT may increase heat loss, so insulation is needed too.
Other acceptable improvements: stir the water continuously to make the temperature uniform (reduces systematic error from a non-uniform temperature), measure V and I continuously rather than assuming constant values, or use a joule meter to measure energy directly.