Calorimetry and the Missing Joules
Students determine the specific heat of an unknown metal by calorimetry. They heat a 50.0 g sample in boiling water at 98.0 °C, then transfer it quickly to a coffee-cup calorimeter containing 150.0 g of water at 22.0 °C. The final temperature is 26.5 °C. Using c_metal = (m_w c_w ΔT_w)/(m_m ΔT_m), they obtain 0.79 J/(g·°C). The accepted value for aluminum is 0.90 J/(g·°C), so the group concludes the sample is not aluminum. Their teacher points out that the calorimeter cup and thermometer absorbed some heat, that a few drops of hot water clung to the sample during transfer, that the transfer took about 4 s, and that the thermometer has a time constant of several seconds. The group must decide whether their data are consistent with aluminum within a reasonable uncertainty, estimate the size of the heat loss, and propose an improved procedure.
1. Calculate the heat gained by the water and the metal's specific heat from the raw data. Then estimate the heat absorbed by the calorimeter cup if its mass is 10.0 g and its specific heat is 1.4 J/(g·°C).
Read the worked answer
Water: m_w = 150.0 g, c_w = 4.18 J/(g·°C), ΔT_w = 26.5 − 22.0 = 4.5 °C. Q_w = 150.0 × 4.18 × 4.5 = 2821.5 J.
Metal: m_m = 50.0 g, initial T = 98.0 °C, final T = 26.5 °C, so ΔT_m = 71.5 °C. If all the metal's heat went into the water, c_m = Q_w/(m_m ΔT_m) = 2821.5/(50.0 × 71.5) = 0.789 J/(g·°C).
Cup: Q_cup = m_cup c_cup ΔT_cup = 10.0 × 1.4 × 4.5 = 63 J. This is about 2.2% of Q_w. Including it in the energy balance gives c_m = (2821.5 + 63)/(50.0 × 71.5) = 0.807 J/(g·°C). So the cup correction raises the result only from 0.789 to 0.807, not enough by itself to reach 0.90.
2. The group concludes the sample is not aluminum. Evaluate that conclusion. What is a reasonable uncertainty in the corrected specific heat, and is 0.90 J/(g·°C) within it?
Read the worked answer
The conclusion is too strong. The corrected value is about 0.81 J/(g·°C), roughly 10% below aluminum's 0.90 J/(g·°C). But several identified effects all bias the result low:
- Heat absorbed by the cup and thermometer lowers the final water temperature, so Q_w is underestimated and c_m is underestimated.
- The sample cools during the 4 s transfer, so its true initial temperature is below 98.0 °C; using 98.0 °C overestimates ΔT_m and underestimates c_m.
- The thermometer lag means the recorded final temperature may be below the true equilibrium value, again lowering Q_w and c_m.
- A few drops of hot water carried with the sample would bias the other way, but that effect is probably smaller.
A reasonable uncertainty from temperature readings (±0.5 °C), mass (±0.1 g), and heat-loss corrections is at least ±0.05 to ±0.08 J/(g·°C). The interval 0.81 ± 0.08 reaches about 0.89, which is essentially at 0.90. So the data are consistent with aluminum within the likely systematic uncertainty, or at least not strong evidence against it. The group should say 'our measured value is lower than the accepted value for aluminum, probably because of heat loss; the sample is consistent with aluminum within experimental uncertainty' rather than 'the sample is not aluminum.'
3. Propose an improved calorimetry procedure that would reduce the dominant systematic error. Include at least two specific changes and explain how you would analyze the new data.
Read the worked answer
Reduce heat loss and correct for what remains. Specific changes:
- Use a calibrated, insulated calorimeter with a lid and a known heat capacity; determine the cup's effective heat capacity by an electrical calibration or by mixing known amounts of hot and cold water. Include it explicitly in the energy balance.
- Transfer the metal quickly and directly into the water, use tongs and a guided chute, and keep the transfer time to about 1 s. Measure the metal's temperature with a thermocouple attached to the sample or read it immediately before transfer.
- Use a fast-response thermometer or a computer-interfaced sensor, stir continuously, and record the temperature until it reaches a stable maximum; take the final temperature from the plateau, not a single reading.
- Run a blank trial with no metal to estimate the heat leak, or use a cooling correction based on Newton's law of cooling.
Analysis: set Q_metal = Q_water + Q_calorimeter + Q_loss and solve for c_m. Propagate uncertainties from masses, temperatures, and the calibrated heat capacity. Compare the corrected c_m with 0.90 J/(g·°C) using an uncertainty interval, and report the result as consistent or inconsistent rather than as a bare conclusion.