Method of mixtures: computing c and diagnosing the direction of systematic errors
An undergraduate determines the specific heat of a metal by the method of mixtures. She heats 100 g of metal shot in boiling water at 98.0 degrees C, transfers it quickly into a copper calorimeter of mass 60 g containing 150 g of water at 25.0 degrees C, stirs, and records a maximum temperature of 30.2 degrees C. She uses c_water = 4.18 J g^-1 K^-1 and c_copper = 0.385 J g^-1 K^-1, neglects heat exchange with the surroundings and the heat capacity of the thermometer, and obtains a value close to 0.50 J g^-1 K^-1. The demonstrator notes that the room was at 22 degrees C, that the transfer took about four seconds, and that the shot carried a little water with it. He asks her to state the sign of each systematic error separately rather than quoting a blanket percentage uncertainty. All temperatures are quoted to 0.1 K.
1. Reproduce the calculation of the specific heat from the data, and check the arithmetic.
Read the worked answer
Heat gained by the water and calorimeter as they warm from 25.0 to 30.2 degrees C (Delta T = 5.2 K):
Q_gain = (150 x 4.18 + 60 x 0.385) x 5.2 = (627.0 + 23.1) x 5.2 = 650.1 x 5.2 = 3380.5 J.
Heat lost by the metal cooling from 98.0 to 30.2 degrees C (Delta T = 67.8 K):
Q_loss = 100 x c x 67.8 = 6780 c (with c in J g^-1 K^-1).
Equating: c = 3380.5/6780 = 0.499 J g^-1 K^-1, i.e. 0.50 to two figures, consistent with her quoted value.
A sanity check on sensitivity: the water term dominates (627 of 650 J/K, about 96 per cent), so the copper heat capacity and its uncertainty barely matter; the limiting quantities are the two temperature differences and the mass of water.
2. State the direction in which each of the following biases c: (a) heat lost by the hot shot during the four-second transfer, (b) heat exchange between calorimeter and room air during the measurement, (c) water carried over with the shot. Treat each separately.
Read the worked answer
(a) If the shot cools on the way, it enters the calorimeter at a temperature below 98.0 degrees C, so the true Delta T of the metal is smaller than 67.8 K. Using too large a Delta T in the denominator makes c come out too small (negative bias).
(b) The calorimeter sits above room temperature throughout (25 to 30.2 degrees C, room 22 degrees C), so it loses heat to the air; the observed Delta T of the water is smaller than it would be in an ideal adiabatic run. In the numerator this makes c too small (negative bias).
(c) Water carried with the shot adds to the water mass that absorbs heat and also evaporates, but if it is not accounted for, the extra heat sink is attributed to the metal; the effective water mass is underestimated, so c comes out too large (positive bias).
The point of the question is that the errors do not all push the same way, so the net bias cannot be stated without knowing the sizes involved.
3. The demonstrator says the result 'may be closer to the truth than the raw number suggests'. Which competing errors could partially cancel, and what experimental changes would let her quantify rather than guess?
Read the worked answer
Errors (a) and (b) both push c down; error (c) pushes it up. If a little water is carried over on the shot, the upward bias can partly offset the downward bias from transfer loss and room loss, so the raw 0.50 may be closer to the true value than a one-sided analysis implies. Equally, ignoring the thermometer's heat capacity is a further small positive bias on c (extra heat sink not credited to the water).
To replace guesswork with measurement: (i) repeat the run with several starting water temperatures on either side of room temperature (e.g. 18, 22 and 26 degrees C) and plot the corrected value against the mean working temperature - Newton's-law extrapolation to zero heat exchange; (ii) pre-warm the transfer vessel or use a short, shielded transfer and time it, so the cooling correction can be estimated; (iii) blot the shot dry and reweigh, or include the carried water mass; (iv) calibrate the calorimeter with a known electric heater to determine its effective water equivalent, including the thermometer. Reproducibility alone is not accuracy: precise repeats with the same systematic error will agree with each other and still be wrong.