Calorimetry: A Specific Heat That Matches Iron Too Well
Students determine the specific heat of an unknown metal. A 52.3 g sample is heated in boiling water, its temperature measured as 98.5 °C with a digital probe, then transferred with tongs into a coffee-cup calorimeter holding 150.0 g of water at 22.0 °C. The aluminum inner cup has a mass of 45.0 g and a specific heat of 0.900 J/(g·°C). The mixture is stirred and reaches 24.6 °C. Using only the water, one group computes c = 0.422 J/(g·°C); after the instructor points out the cup, they recompute and get 0.449 J/(g·°C), which is within 0.2% of the handbook value for iron (0.449), and they report the metal as iron. The sample sat in air for about three seconds during transfer, and no correction was made for heat lost to the room. The instructor asks whether the identification is justified.
1. Reproduce both calculations and state, with reasoning, the direction in which the ignored cup and the transfer delay each push the result.
Read the worked answer
Water only. Heat gained by water = 150.0 × 4.184 × (24.6 − 22.0) = 150.0 × 4.184 × 2.6 = 1632 J. Heat lost by metal = m c ΔT = 52.3 × c × (98.5 − 24.6) = 3865 c. Setting equal: c = 1632/3865 = 0.422 J/(g·°C).
Including the cup. The cup gains 45.0 × 0.900 × 2.6 = 105 J, so total heat absorbed is 1737 J and c = 1737/3865 = 0.449 J/(g·°C).
Direction of each error.
- Ignoring the cup: heat that actually went into the aluminum is credited to the metal, so the metal appears to have released more energy than it did, and its c is computed too low. Correcting it raises c by about 6%.
- Three seconds in air: the sample is cooler than 98.5 °C when it enters the water, but the calculation still uses ΔT = 73.9 °C, overstating the temperature drop and therefore understating c. This error also pushes low.
Both neglected effects push c downward, so the reported 0.449 is a lower bound on the true value, not an unbiased estimate. Treating it as exact is unjustified.
2. The group's value matches iron to 0.2%. Explain why that agreement is not strong evidence for the identification.
Read the worked answer
The match is only meaningful if the measurement is unbiased and its uncertainty is smaller than the gaps between candidate materials. Neither condition holds here.
Both known systematic errors - the unaccounted cup and the transfer cooling - bias c downward. That means the true specific heat of this sample is somewhat greater than 0.449, by an amount at least as large as the transfer loss. So the sample is not iron; a material whose handbook c is 0.449 is ruled out precisely because the corrected value must exceed the one that matched. Candidates with c in the 0.46–0.50 range (for example, some steels, or a copper-zinc alloy) are consistent with the data.
A 0.2% match against a handbook table also claims far more precision than the apparatus supports: a 0.1 °C error in the final temperature alone shifts c by about 0.4%, and a 0.5 °C error in the initial metal temperature shifts it by about 0.7%. Real agreement should be quoted as something like c = 0.45 ± 0.02 J/(g·°C), which does not distinguish iron from several alloys. Identification by specific heat needs corroboration - density, appearance, magnetic behavior, or a spark test - not a single thermal measurement.
3. Describe how the procedure could be improved so that the systematic bias is measured rather than assumed.
Read the worked answer
Eliminate or quantify the cup. Run a blank with the same cup and same water volume but no metal, adding a known amount of heat electrically (a small immersion heater of measured power and time), and determine the calorimeter's effective heat capacity, C_cal = Q/ΔT, once. Use that measured C_cal instead of the handbook aluminum value, since the cup, lid, and stirrer all participate.
Minimize and measure transfer loss. Transfer quickly with pre-warmed tongs and a lid that opens minimally; better, heat the sample in a wire basket that goes straight into the water. To estimate the residual, repeat with transfer times of about 1 s, 3 s, and 6 s and extrapolate c back to zero delay - a linear extrapolation against transfer time gives the loss-free value with an uncertainty.
Control the rest. Use a sealed, insulated cup; stir gently and consistently; record temperature continuously and use the extrapolated mixing temperature from the pre- and post-mixing slopes rather than a single reading; calibrate the thermometer against a known transition (ice point).
Report properly. Quote c with a combined uncertainty and state the method used for the calorimeter constant, so a reader can judge whether a match to a handbook value is meaningful.