Calorimetry Quandary: Heat Loss in a Coffee-Cup Calorimeter
A student dissolves 4.00 g of ammonium nitrate (molar mass 80.04 g/mol) in 50.0 g of water in a coffee-cup calorimeter. The initial temperature of the water is 22.4 °C and the final temperature after dissolution is 14.8 °C. The student calculates the enthalpy of solution and gets a value about 20% smaller in magnitude than the accepted value of +25.7 kJ/mol. The student's thermometer reads to ±0.2 °C, and the balance reads to ±0.01 g. The calorimeter is open to the air, and the room is at 22.0 °C. The student asks why their result is off and whether the discrepancy is a random error or a systematic error.
1. Calculate the student's experimental enthalpy of solution per mole of NH4NO3, stating your assumptions about the specific heat and the heat capacity of the calorimeter.
Read the worked answer
Assume the solution has the specific heat of water (4.18 J/g·°C) and ignore the calorimeter's own heat capacity.
Total mass ≈ 50.0 g water + 4.00 g solute = 54.0 g.
q_solution = m·c·ΔT = (54.0 g)(4.18 J/g·°C)(14.8 − 22.4 °C) = (54.0)(4.18)(−7.6) ≈ −1715 J.
Because the dissolution is endothermic, q_reaction = −q_solution = +1715 J absorbed from surroundings.
Moles = 4.00 g / 80.04 g/mol = 0.0500 mol.
ΔH = 1715 J / 0.0500 mol ≈ +34,300 J/mol ≈ +34.3 kJ/mol.
Note that ignoring the calorimeter heat capacity and using the water specific heat for the whole solution both introduce approximations, but the sign and rough magnitude are correct.
2. The calculated magnitude (+34 kJ/mol) is actually larger than, not smaller than, the accepted +25.7 kJ/mol. Explain a plausible source of systematic error that would inflate the magnitude, and how the open calorimeter and room temperature contribute.
Read the worked answer
The room is at 22.0 °C while the solution cools to 14.8 °C. Because the solution becomes colder than the room, heat flows in from the surroundings through the open calorimeter. That heat input warms the solution, making the measured temperature drop (7.6 °C) smaller than the true drop would be if the system were perfectly insulated. A smaller ΔT would make the calculated |ΔH| smaller, not larger - so heat leak alone cannot explain an inflated magnitude.
A more likely inflating factor: the accepted value assumes dilute solution and the specific heat of the actual solution, which differs from pure water. Also, ignoring the calorimeter's heat capacity means the measured temperature change reflects only the solution, but part of the reaction's heat also warms the cup/thermometer. Correcting for a finite calorimeter heat capacity would require a larger q for the same ΔT, so neglecting it typically under-corrects. The key reasoning point: heat leak from a warmer room reduces the observed ΔT (biasing magnitude low), whereas the student's result is high, so the dominant error must be in the assumed heat capacity/specific heat or in the mass accounting, not the open lid.
3. Classify the ±0.2 °C thermometer uncertainty and the ±0.01 g balance uncertainty as random or systematic, and estimate the propagated uncertainty in ΔT and in the molar enthalpy.
Read the worked answer
Readability uncertainties on a thermometer and balance are best treated as random (they vary trial to trial), though a miscalibrated thermometer would be systematic.
ΔT = T_final − T_initial, so the uncertainty in ΔT is the quadrature sum of two readings: σ_ΔT = √(0.2² + 0.2²) ≈ 0.28 °C.
Relative uncertainty in ΔT ≈ 0.28 / 7.6 ≈ 3.7%.
Mass uncertainty: 0.01 g on 4.00 g ≈ 0.25%; on 50.0 g ≈ 0.02%. Moles uncertainty ≈ 0.25%.
Combined relative uncertainty in ΔH ≈ √(3.7² + 0.25²) ≈ 3.7%, so ΔH ≈ 34.3 ± 1.3 kJ/mol.
This propagated uncertainty (±1.3 kJ/mol) is far too small to explain the gap from 25.7 kJ/mol. Therefore the discrepancy is dominated by systematic model error (heat capacity/specific heat assumptions, heat exchange with the room), not by instrument precision - an important judgment the exam rewards.