Where did the missing 300 joules go? Diagnosing a calorimetry result
A student measures the specific heat capacity of a metal block by the method of mixtures. The block, mass 100.0 g, is held in steam at 98.0 degrees C, then carried across the bench with tongs and dropped into 80.0 g of water at 24.0 degrees C held in a copper calorimeter of mass 50.0 g (c_Cu = 0.385 J g^-1 K^-1). The mixture settles at 30.5 degrees C; the thermometer reads to 0.1 degree. The student computes c = 0.32 J g^-1 K^-1 while ignoring the calorimeter, finds that the handbook value for this metal is 0.385, and concludes that either the handbook or the thermometer must be wrong. A demonstrator suggests that the two most likely systematic errors are the neglected heat capacity of the calorimeter and heat lost by the block during the transfer from the steam bath. Take c_water = 4.18 J g^-1 K^-1.
1. Recompute c exactly as the student did, then with the calorimeter included. Compare the size of the two systematics and decide which of them can and which cannot explain the discrepancy.
Read the worked answer
Temperature changes: water and calorimeter +6.5 K, block -67.5 K.
Heat gained by water = 80.0 x 4.18 x 6.5 = 2174 J. Heat gained by calorimeter = 50.0 x 0.385 x 6.5 = 125 J.
- Student (calorimeter ignored): c = 2174/(100.0 x 67.5) = 0.322 J g^-1 K^-1.
- Calorimeter included: c = (2174 + 125)/(100.0 x 67.5) = 0.341 J g^-1 K^-1.
- Handbook: 0.385 J g^-1 K^-1.
Deficit: 16 percent as the student computed, 11.5 percent after correction.
The calorimeter term is 125/2299 = 5.4 percent of the total heat absorbed. It must certainly be included - it is a systematic bias in the same direction - but it can account for only about 5 of the 16 percentage points. Roughly 11 percent of the heat is still unaccounted for, and that must come from heat lost by the block (or from the block being cooler than 98 degrees C on arrival). The handbook is not the problem.
2. Estimate how much heat was lost and what that implies physically. Then propose two design changes that would reduce this error.
Read the worked answer
Expected heat transfer if c = 0.385:
Q = 100.0 x 0.385 x 67.5 = 2599 J, against 2299 J actually absorbed.
Lost = 300 J, i.e. 11.5 percent. That is equivalent to the block arriving at
98.0 - 300/(100.0 x 0.385) = 98.0 - 7.8 = 90.2 degrees C.
A bare block of surface area roughly 0.015 m^2, sitting 60-70 K above room temperature, loses of order 10 W by convection plus radiation; 300 J corresponds to about 30 s of handling, which is entirely plausible for carrying a block with tongs, finding the calorimeter and getting the lid on. So the diagnosis is quantitatively realistic.
Design changes:
- Transfer the block in a closed, lagged holder (or a steam-jacketed tube that opens directly into the calorimeter) so it is never exposed to room air; keep the lid on except for the instant of transfer.
- Apply a cooling correction: record temperature versus time before and after mixing and extrapolate the fore-aft lines to the mixing instant by Newton's law of cooling, so the heat leak during mixing is corrected rather than ignored.
Secondary: use a larger mass of water so the temperature rise is smaller and the block spends less time above ambient? That lengthens the run, so the better lever is shortening the transfer, not changing the water mass.
3. On a repeat the same student obtains a final temperature 0.6 degree ABOVE the value predicted from the handbook value. Give two plausible causes consistent with the sign of the error, and state one check that would distinguish them.
Read the worked answer
A final temperature that is too high means more heat than expected reached the water (or the water was effectively smaller).
- The block was wet. A film of water from the steam bath, say 2 g, enters at 98 degrees C and contributes 2 x 4.18 x 67.5 = 564 J, raising the mixture by roughly 564/(80 x 4.18 + 19) = 1.6 K. Even 1 g gives about 0.8 K, the right order. The extra water also slightly changes the water mass.
- The block was hotter than 98 degrees C - it rested on the bottom of the vessel in contact with the hotplate, or the steam was superheated, so the assumed starting temperature is too low. A 2 K error in the block temperature changes the delivered heat by about 100 x 0.385 x 2 = 77 J, only about 0.2 K of final temperature, so this is a smaller effect unless the error is large.
- Thermometer calibration offset at 30 degrees C; this would shift the final reading without any heat anomaly.
Discriminating check: weigh the block (and the calorimeter contents) before and after the transfer. A wet block shows a reproducible mass gain of 0.5-2 g, whereas a temperature/calibration fault leaves the masses unchanged. A second check is to heat the block in an oven at a measured 100 degrees C and repeat with the same protocol.