Calorimetry with a Wet Metal Sample
In a calorimetry lab, a student heats a 0.152 kg metal sample in boiling water at 99.2°C and then quickly transfers it to a coffee-cup calorimeter. The cup has an aluminum inner liner of mass 0.045 kg and contains 0.120 kg of water at 21.5°C. The mixture reaches a maximum of 24.8°C. The student uses 4186 J/(kg·°C) for water and 900 J/(kg·°C) for aluminum, and calculates the metal’s specific heat. The sample is listed as lead, whose accepted specific heat is 128 J/(kg·°C), but the student’s value is noticeably higher. The student suspects heat loss to the room is the problem. A classmate notices several droplets of hot water clinging to the sample when it leaves the boiling bath. The thermometer has 0.2°C markings, and the balance reads to 0.1 g. The cup lid is loose, and the room is 22°C.
1. Calculate the student’s experimental specific heat and compare it with lead. Does heat loss to the room explain the discrepancy? Explain the sign of the effect.
Read the worked answer
Heat gained by the water:
(Q_w=(0.120)(4186)(24.8-21.5)=1.66\times10^3\ \mathrm{J}).
Heat gained by the aluminum cup:
(Q_{\mathrm{Al}}=(0.045)(900)(3.3)=1.34\times10^2\ \mathrm{J}).
Total heat from the metal is about (1.79\times10^3\ \mathrm{J}). The metal cools by (99.2-24.8=74.4\ ^\circ\mathrm{C}), so
(c_{\mathrm{exp}}=\frac{1790}{(0.152)(74.4)}\approx158\ \mathrm{J/(kg,^\circ C)}).
This is about 24% above 128 J/(kg·°C). Heat loss to the room would make the final temperature lower, which would reduce the calculated heat gained by the water and cup and therefore reduce (c_{\mathrm{exp}}). It cannot explain a value that is too high; it would bias the result in the opposite direction.
2. Estimate how much adhering water would account for the difference from lead, and explain how ignoring it biases the result.
Read the worked answer
For lead, the expected heat released by the metal is
(Q_{\mathrm{Pb}}=(0.152)(128)(74.4)\approx1.45\times10^3\ \mathrm{J}).
The excess heat in the student’s calculation is about (1790-1450=340\ \mathrm{J}). If that heat came from hot water cooling through 74.4°C, its mass would be
(m=\frac{340}{(4186)(74.4)}\approx1.1\times10^{-3}\ \mathrm{kg}),
or about 1.1 g. That is plausible for droplets clinging to the sample. Ignoring the droplets adds an unaccounted hot-water heat source, so the numerator is too large and the computed specific heat is too high. The sample should be blotted dry before transfer, and the lid should be seated to reduce heat loss.