Calorimetry Result That Is Too Low
In a Grade 10 lab, students determine the specific heat of an unknown metal. Each group heats a 50.0 g sample in boiling water, then transfers it to a 100.0 g aluminum calorimeter cup (c = 0.90 J/(g·°C)) holding 150.0 g of water at 22.0 °C. A digital thermometer has 0.1 °C resolution. Group 1 dries the sample quickly and records a final temperature of 24.0 °C, calculating c ≈ 0.38 J/(g·°C), close to copper. Group 2, using the same metal, does not dry the sample, transfers it slowly, and records 23.2 °C, calculating c ≈ 0.19 J/(g·°C). Their thermometer had been resting in the hot water bath and was not allowed to return to room temperature. The outside of Group 2's calorimeter cup is wet, and the room has a slight breeze. The teacher has accepted values, an analytical balance, and a second thermometer.
1. For Group 2, evaluate each suspected error (wet sample, slow transfer, hot thermometer, wet cup, breeze) by stating whether it would raise or lower the calculated specific heat. Identify which are consistent with the low result.
Read the worked answer
Use the energy-balance expression c_metal = (m_w c_w + m_c c_c)(T_f − T_i) / [m_m (T_hot − T_f)]. A lower T_f decreases the numerator and increases the denominator, so any effect that lowers the final temperature lowers the calculated c.
- Wet sample: water carried in evaporates and/or adds thermal mass; the sample cools and T_f is lower → c too low. Consistent.
- Slow transfer: the metal cools before entering the calorimeter, so the true initial metal temperature is below 100 °C while the calculation uses 100 °C; the denominator is too large and T_f is lower → c too low. Consistent.
- Thermometer left in hot bath: a hot thermometer adds heat to the calorimeter, raising T_f → c too high. This does not explain Group 2's low value; it would bias the other way.
- Wet cup and breeze: evaporative cooling lowers T_f → c too low. Consistent, but likely smaller than the sample effects.
Group 2's low result is consistent with a wet, slowly transferred sample and a wet cup, not with the hot thermometer.
2. Explain why the aluminum cup must be included and whether neglecting it would make the result too high or too low. Estimate its contribution for Group 1.
Read the worked answer
The cup is part of the system that gains heat. For Group 1, the water's heat capacity is 150.0 g × 4.18 J/(g·°C) = 627 J/°C; the aluminum cup is 100.0 g × 0.90 J/(g·°C) = 90 J/°C, about 13% of the total. Neglecting the cup removes 90 J/°C from the numerator, so the calculated c is about 13% too low-not too high. For Group 1, including the cup gives c ≈ 0.38 J/(g·°C); omitting it gives roughly 0.33 J/(g·°C).
The cup is not negligible at this precision. Students should either include it explicitly or use a styrofoam cup with a measured, small heat capacity and justify the approximation.
3. Propose a revised procedure and one quantitative check that would catch the error before students trust the unknown metal's specific heat.
Read the worked answer
Revised procedure: heat the sample in boiling water and measure its temperature just before transfer; dry it quickly with a paper towel; transfer with tongs in under 2 s; use a room-temperature thermometer and stir gently; cover the calorimeter; wipe the outside dry; shield the setup from drafts; record the final temperature after the reading stabilizes.
Quantitative check: run a known material (e.g., an aluminum cylinder with accepted c ≈ 0.90 J/(g·°C)) through the same procedure. If the method returns 0.90 within, say, 10%, the procedure is credible; if it returns 0.6, the class knows there is a systematic heat loss before trusting the unknown. Also compute the energy balance both ways and compare trial-to-trial spread. A single trial is not evidence; a method check is.