Coffee-Cup Calorimetry and an Unreliable Specific Heat
Students calibrate a coffee-cup calorimeter by mixing 50.0 g of water at 60.0°C with 50.0 g of water at 20.0°C; the final temperature is 38.6°C. They then heat a 25.0 g metal slug in boiling water (99.0°C) and transfer it to 75.0 g of water at 21.0°C in the same calorimeter. The final temperature is 26.4°C. Using c_water = 4.184 J/(g·°C), the class calculates the metal's specific heat as 0.61 J/(g·°C). The accepted value for the metal they think they have, aluminum, is 0.90 J/(g·°C). A student says the result is close enough because 0.61 and 0.90 differ by only 0.29 J/(g·°C). Another says the metal must not be aluminum. The class did not use the calibration result in the metal calculation, and the calorimeter had no lid during the metal trial.
1. Evaluate the calorimeter calibration. What does the 38.6°C result indicate, and how would you estimate the calorimeter's heat capacity?
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Ideal adiabatic mixing of equal masses at 60.0°C and 20.0°C would reach 40.0°C. The observed 38.6°C is lower, so heat was lost to the cup and surroundings, or the cup started cooler than 20.0°C. Estimate the heat lost by the hot water: q_hot = 50.0 × 4.184 × (60.0 − 38.6) = 4.48 kJ. Heat gained by the cold water is q_cold = 50.0 × 4.184 × (38.6 − 20.0) = 3.89 kJ. The difference, about 0.59 kJ, is absorbed by the calorimeter. Its heat capacity is approximately 0.59 kJ / (38.6 − 20.0)°C = 31.5 J/°C. This is an operational calibration that includes the cup and thermometer. It shows the apparatus is not perfectly adiabatic and should be included in the metal calculation or corrected by extrapolating a time-temperature curve.
2. Recalculate the metal's specific heat from the data. The class reported 0.61 J/(g·°C); identify the likely calculation error.
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Heat gained by the water is q_water = 75.0 × 4.184 × (26.4 − 21.0) = 1.69 kJ. Including the calorimeter, q_cal = 31.5 × 5.4 = 0.17 kJ, so the metal released about 1.86 kJ. The metal's temperature change is 99.0 − 26.4 = 72.6°C. Thus c = 1.86 kJ / (25.0 g × 72.6°C) = 1.02 J/(g·°C); ignoring the calorimeter gives 0.93 J/(g·°C). The reported 0.61 is not consistent with these data. A likely error is using 50.0 g of water, the calibration mass, instead of 75.0 g: 50.0 × 4.184 × 5.4 = 1.13 kJ, and 1.13 kJ / (25.0 × 72.6) = 0.62 J/(g·°C). The class appears to have carried over the wrong water mass.
3. Is the metal aluminum? What improvement would you make?
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The corrected value, about 0.93 J/(g·°C) ignoring the calorimeter or 1.02 including it, is much closer to aluminum's 0.90 than the reported 0.61, but it is not exact. A single trial with heat loss cannot confirm identity. Improve the method by using a lid and insulation, pre-equilibrating the cup, transferring the metal quickly, stirring gently, and recording a time-temperature curve. Extrapolate the curve back to the mixing instant to correct for heat lost during measurement, and apply the calorimeter heat capacity from the calibration. Repeat at least three trials and report a mean and range. Confirming density or comparing with a known aluminum sample would provide independent evidence.