Calorimetry Result That Is 6% Low: Four Effects, One Number
In a calorimetry lab, a 50.0 g metal sample is boiled for several minutes, lifted out with tongs and dropped into a nested aluminum cup of mass 50.0 g holding 100.0 g of water at 21.0 °C. A thermometer already in the water reads a maximum of 27.5 °C after stirring. The local boiling point is 99.0 °C, and the group uses 4186 J/(kg·K) for water and 900 J/(kg·K) for aluminum. Their calculation gives c ≈ 843 J/(kg·K), about 6% below the accepted value for aluminum, and they write that the sample is aluminum with a 6% experimental error. The transfer takes 3–4 s through 22 °C air; the cup has no lid; roughly 0.05 g of water evaporates during the run; the thermometer bulb and the stirring rod are not included in the heat-capacity accounting; and the sample carries a visible film of water (about 0.5 g) into the cup. The instructor asks which of these could plausibly account for the shortfall and how the group could tell them apart.
1. Reproduce the group's calculation and check their arithmetic.
Read the worked answer
Heat capacity of the receiving system (water plus cup):
C = m_w c_w + m_cup c_Al = (0.100)(4186) + (0.050)(900) = 418.6 + 45.0 = 463.6 J/K
Temperature rise = 27.5 − 21.0 = 6.5 K, so
Q_gained = 463.6 × 6.5 = 3013 J
For the sample, ΔT = 99.0 − 27.5 = 71.5 K, so
c = Q/(m ΔT) = 3013/(0.0500 × 71.5) = 3013/3.575 = 843 J/(kg·K)
The arithmetic is correct. The result is 6.3% below 900 J/(kg·K). It is also far from any other common metal - iron 450, copper 385, zinc 388, lead 128 - so the identification of the sample as aluminum is not in doubt; only the interpretation of the 6% is. A useful sensitivity to carry forward: each joule of heat that went somewhere unaccounted shifts the answer by Δc = 1/(0.0500 × 71.5) ≈ 0.28 J/(kg·K).
2. For each of the four unaccounted effects - sample cooling during transfer, water evaporation, the thermometer and stirrer's heat capacity, and the wet sample - state the sign of the bias and estimate its size in J/(kg·K). Then say what the group can legitimately conclude.
Read the worked answer
Use Δc ≈ ΔQ/3.575, i.e. about 0.28 J/(kg·K) per joule.
Cooling during transfer (negative bias). If the sample enters at 95 °C instead of 99 °C, the real ΔT is 67.5 K while the calculation uses 71.5 K, so c_true = c_meas × 71.5/67.5 = 1.059 c_meas. The reported value is low by about 5.6%, roughly 50 J/(kg·K).
Evaporation (negative bias). 0.05 g vaporising removes 0.05 × 2260 ≈ 113 J from the cup, so Q_gained is understated: Δc ≈ −32 J/(kg·K).
Thermometer and stirrer (negative bias). A 10 g glass thermometer (c ≈ 750) plus a 5 g steel rod (c ≈ 450) give about 9.8 J/K; warmed through 6.5 K that is 64 J, so Δc ≈ −18 J/(kg·K).
Wet sample (positive bias). About 0.5 g of water at 99 °C gives up 0.0005 × 4186 × 71.5 ≈ 150 J that is credited to the metal: Δc ≈ +42 J/(kg·K).
Summing: −50 − 32 − 18 + 42 ≈ −58 J/(kg·K), which is almost exactly the observed −57. That agreement is a warning, not a triumph, because the four terms were estimated after the fact and all scale with quantities the group never recorded. The legitimate conclusion is that 843 J/(kg·K) is a lower bound contaminated by several percent of unquantified systematics; the data are consistent with aluminum but do not test the accepted value at the 6% level, and the write-up should say so rather than claiming a 6% experimental error.
3. Design one experiment that would settle the calorimetry without relying on any of these estimates.
Read the worked answer
Calibrate the apparatus electrically, in situ, under the same conditions as the metal runs. Place a known resistor (for example 10 Ω, 5 W) in the cup with the same 100.0 g of water, pass a measured current for a measured time - 1.00 A at 10.0 V for 300 s delivers 3000 J - and record the temperature rise with the same stirring, the same lid state and the same thermometer. Then
C_eff = IVt/ΔT
is the effective heat capacity of the entire loaded system. Because the thermometer, stirrer, cup, lid, evaporation and heat leak are all present in both the calibration and the metal run, they are absorbed into C_eff instead of being guessed at. The metal's specific heat then follows as c = C_eff ΔT_water/(m ΔT_metal) with no assumed heat capacities at all, and the wet-sample film is the only remaining correction, which is easy to control by blotting the sample on a dry cloth before transfer.
To handle transfer loss independently, record temperature for two minutes before mixing and two minutes after, fit straight lines to the fore-drift and after-drift, and extrapolate both to the mixing instant - the Newton-cooling correction - which yields the true ΔT rather than the observed peak. Together these two changes replace every estimate in the previous answer with a measurement.