Two Runs, One Underestimated Specific Heat
A student determines the specific heat capacity of aluminium shot (accepted value 0.90 J g-1 °C-1) by the method of mixtures. She heats 200.0 g of the shot in boiling water, which at her school boils at 99.0 °C, then pours the shot into a polystyrene cup holding 100.0 g of water at 21.4 °C, stirs, and records the highest temperature reached. Run A gives a final temperature of 42.1 °C; the cup had not been pre-warmed and the transfer took about 8 s in a cold metal scoop. Run B repeats the procedure with the cup pre-warmed to about 40 °C, a lid, and a plastic scoop, and gives 43.6 °C. She uses c(water) = 4.18 J g-1 °C-1 and the relation heat lost by the metal = heat gained by the water, then averages the two runs and reports 0.80 J g-1 °C-1. Her teacher says that the second run, properly corrected, is within 1 % of the accepted value.
1. Calculate the value of c she obtains from each run using her simple relation, then explain why both lie below 0.90 J g-1 °C-1 and why Run B is closer to it.
Read the worked answer
Run A: q(water) = mcΔT = 100.0 × 4.18 × (42.1 − 21.4) = 418 × 20.7 = 8.65 × 10^3 J. The metal cools from 99.0 °C to 42.1 °C, a change of 56.9 °C, so c = 8.65 × 10^3 / (200.0 × 56.9) = 0.760 J g-1 °C-1.
Run B: q = 418 × 22.2 = 9.28 × 10^3 J and ΔT(metal) = 55.4 °C, so c = 9.28 × 10^3 / (200.0 × 55.4) = 0.838 J g-1 °C-1.
Both are low because the relation assumes every joule the metal loses is absorbed by the water. In reality some heat warms the cup, lid, thermometer and scoop, and some escapes to the room, especially during the 8 s transfer in a metal scoop that conducts well. The water therefore gains less heat than the metal releases, the numerator is too small, and c is systematically (not randomly) underestimated. Run B shrinks those losses - pre-warmed cup, lid, plastic scoop - so less heat is diverted and the estimate moves towards 0.90. This is a biased error, so averaging the two runs cannot fix it; it just averages two biased numbers.
2. Her teacher notes that the cup, lid and thermometer together have a heat capacity equivalent to 8.0 g of water. Recalculate Run B on that basis and state which assumption of the simple relation is still violated and whether it matters at this precision.
Read the worked answer
Treat the calorimeter as an extra 8.0 g of water, so the effective mass being warmed is 108.0 g:
q = 108.0 × 4.18 × 22.2 = 1.002 × 10^4 J c = 1.002 × 10^4 / (200.0 × 55.4) = 0.905 J g-1 °C-1
That is within about 0.5 % of 0.90, matching the teacher's claim. The correction matters because 8 g out of 108 g is roughly 7 % of the heat sink - ignoring it forces c low by a similar fraction.
Still violated: the mixture never stops exchanging heat with the room, and the shot is assumed to arrive at exactly 99.0 °C and to reach thermal equilibrium instantly. Does it matter? A thermometer read to ±0.1 °C on a 22.2 °C rise is already ±0.45 %, so a residual leak would have to exceed roughly 1 % of q to be visible. Reducing the leak below the reading resolution is exactly what the lid and pre-warmed cup achieved; this is why the corrected Run B is trustworthy and Run A is not.
3. She repeats the method unchanged with 200.0 g of a metal whose accepted c is 0.13 J g-1 °C-1. Predict the final temperature (ignoring losses) and explain with numbers why the percentage error in c would be far larger than for aluminium.
Read the worked answer
Heat balance: 200.0 × 0.13 × (99.0 − T) = 100.0 × 4.18 × (T − 21.4) 26.0(99.0 − T) = 418(T − 21.4) 2574 − 26T = 418T − 8945 11519 = 444T, so T = 25.9 °C - a water rise of only 4.5 °C.
Three consequences:
- A ±0.1 °C reading error is 2.2 % of a 4.5 °C rise, against 0.45 % of Run B's 22.2 °C rise.
- The 8 g water-equivalent calorimeter absorbs 8.0 × 4.18 × 4.5 ≈ 150 J while the water absorbs only about 1880 J. Ignoring the cup is now an ~8 % error, against ~0.5 % before.
- Any heat lost during transfer is a larger fraction of the much smaller heat content of this metal.
So the same apparatus that gives roughly 1 % precision for aluminium would give about 10 % for this metal. The fix is to change the ratio of heat capacities: use more metal or less water so the temperature rise stays around 20 °C, rather than trusting the fixed masses that suited aluminium.