Calorimetry, Sign Conventions, and a Suspicious ΔH
A student dissolves 4.00 g of NH4NO3 (molar mass 80.04 g/mol) in 96.0 g of water in a coffee-cup calorimeter (assume the solution's specific heat is 4.18 J/g·°C and the calorimeter heat capacity is negligible). The initial temperature is 22.5 °C; after stirring, the final temperature is 17.3 °C. The student reports ΔH_solution = +21.7 kJ/mol and claims this means 'the reaction released heat because the temperature dropped, so it's exothermic.' A second student says the sign is wrong and the process is endothermic. The instructor asks for a rigorous reconciliation.
1. Calculate q for the process and ΔH per mole, then resolve the sign dispute.
Read the worked answer
Total mass: 4.00 g + 96.0 g = 100.0 g. Temperature change of solution: ΔT = 17.3 − 22.5 = −5.2 °C.
Heat absorbed by the solution: q_solution = m·c·ΔT = 100.0 × 4.18 × (−5.2) = −2174 J ≈ −2.17 kJ. The negative sign means the solution lost thermal energy.
Heat of the dissolving process (system): q_system = −q_solution = +2.17 kJ. The system absorbed heat from the water, so the water cooled.
Per mole: 4.00 g / 80.04 g/mol = 0.0500 mol. ΔH = +2.17 kJ / 0.0500 mol = +43.4 kJ/mol.
Resolution: The process is endothermic (ΔH > 0). The student's sign convention is backwards - a temperature drop of the surroundings means the system absorbed heat. The magnitude is also off: the student likely used the wrong mass or forgot the factor. The correct value is about +43 kJ/mol, not +21.7.
2. The student's reported magnitude (+21.7 kJ/mol) is roughly half the correct value. Propose a plausible arithmetic error that produces exactly this factor.
Read the worked answer
A factor-of-two error typically arises from using the solute mass only (4.00 g) instead of the total solution mass (100.0 g), or from halving ΔT. Check: using m = 4.00 g gives q = 4.00 × 4.18 × 5.2 = 87 J, far too small - not this. Using only the water mass 96.0 g gives q ≈ 2087 J, close to correct, not half.
The clean explanation: the student computed q = 2.17 kJ but then divided by 0.100 mol (perhaps mistaking 8.00 g or doubling the moles) instead of 0.0500 mol, halving ΔH. Or the student used ΔT = 2.6 °C (half the true 5.2 °C) by reading a thermometer with 1 °C graduations and misreading the span. Either way, the correct ΔH ≈ +43 kJ/mol, and the sign must be positive (endothermic). The instructor should have the student recheck the mole count and the full temperature span.
3. If the calorimeter had a non-negligible heat capacity, how would the true ΔH compare to the calculated value, and why?
Read the worked answer
With a real calorimeter heat capacity C_cal, the heat exchanged includes heating/cooling the cup itself: q_total = (m·c + C_cal)·ΔT. Since ΔT is negative here, including C_cal makes |q_total| larger, so the magnitude of ΔH increases (more endothermic). The simple calculation ignoring C_cal underestimates |ΔH|.
Judgment point: the problem explicitly says to neglect the calorimeter, but a careful student notes the direction of the error. Also, assuming the solution's specific heat equals pure water's (4.18 J/g·°C) introduces a small systematic error, and heat loss to the room would make the measured ΔT smaller in magnitude, again underestimating |ΔH|. Real calorimetry requires calibration to bound these errors.