Calorimetry and an Overconfident Identification
A student heats a 52.3 g unknown metal sample in boiling water at 98.5 °C, then transfers it quickly to a coffee-cup calorimeter containing 100.0 g of water at 22.4 °C. The mixture reaches 26.1 °C. The thermometer reads to ±0.2 °C and the balance to ±0.01 g. The student ignores the calorimeter's heat capacity, which a separate test shows to be 12 J/°C, and also assumes no heat is lost during transfer. Using c_water = 4.18 J/(g·°C), the student calculates a specific heat and compares it with tabulated values: copper 0.385, zinc 0.388, iron 0.450, and aluminum 0.900 J/(g·°C). The student's value seems closest to copper and reports the metal as copper. A partner objects that the identification is not justified. The teacher asks the group to quantify the result and decide what can and cannot be concluded.
1. Calculate the specific heat without and with the calorimeter heat-capacity correction. Compare the result with the tabulated values.
Read the worked answer
Water gained Q_w = (100.0)(4.18)(26.1 − 22.4) = 1546.6 J. The metal cooled by 98.5 − 26.1 = 72.4 °C. Ignoring the cup, c = 1546.6/[(52.3)(72.4)] = 0.408 J/(g·°C). Including the cup's heat capacity, the water-plus-cup gain is (418 + 12)(3.7) = 1591 J, so c = 1591/3786.5 = 0.420 J/(g·°C). The correction raises the result by about 3%. The value is about 9% above copper (0.385) and 7% below iron (0.450), and essentially at zinc (0.388) within the crude comparison. It is nowhere near aluminum (0.900). A single corrected number near 0.42 does not uniquely identify copper; it is closer to copper or zinc than to iron, but the gap to copper is not negligible.
2. Estimate the uncertainty in the corrected specific heat from the thermometer and balance readings. Does the result distinguish copper, zinc, iron, and aluminum?
Read the worked answer
Temperature uncertainty dominates. For the water, ΔT = 3.7 ± sqrt(0.2² + 0.2²) = 3.7 ± 0.28 °C, a 7.6% relative uncertainty; the metal's ΔT = 72.4 ± 0.28 °C is only 0.4%. Mass uncertainties are under 0.1%. Thus c = 0.420 ± 0.032 J/(g·°C), roughly 0.388 to 0.452, before including heat loss during transfer or imperfect mixing. That interval includes zinc (0.388) and iron (0.450), and copper (0.385) is just outside the lower edge; aluminum is excluded. Because systematic heat loss would shift the result and is not included, the identification as copper is not defensible. The honest conclusion is a specific heat near 0.42 J/(g·°C), with a range that does not uniquely identify the metal.
3. Propose improvements to the experiment and explain how each reduces a specific error.
Read the worked answer
Use a larger sample, such as 150 to 200 g, so the water temperature rise is larger and the ±0.2 °C reading error is a smaller fraction. Use a well-insulated calorimeter with a lid and a measured heat capacity; pre-equilibrate the cup; transfer the metal quickly with tongs; stir gently and record a temperature-time curve, extrapolating to the instant of mixing to correct for heat loss. Repeat at least three trials and report mean and standard deviation. To identify the metal, compare the corrected value with a calibration curve made from known samples under the same apparatus, or measure another property such as density rather than relying on one thermal measurement. This design reduces both random reading error and systematic heat exchange, which is the main weakness of the single trial.