Calorimetry with a Cold Sample: Finding the Specific Heat
A student places a 0.150 kg metal sample, heated in boiling water (assumed 100 degrees C), into a calorimeter containing 0.200 kg of water at 20.0 degrees C. The calorimeter cup has a heat capacity equivalent to 30 g of water. The final equilibrium temperature is 26.5 degrees C. The accepted specific heat of the metal is 0.385 kJ/(kg K) (copper). The student computes c and gets a value about 25% too high. The room is 22 degrees C, and the transfer takes several seconds.
1. Compute the specific heat from the data (including the calorimeter) and compare with the accepted value. Identify at least two systematic errors that would push the result high.
Read the worked answer
Heat gained by water + cup = heat lost by metal. Water: m c dT = 0.200 x 4.18 x (26.5 - 20.0) = 0.200 x 4.18 x 6.5 = 5.43 kJ. Cup: equivalent 0.030 kg water: 0.030 x 4.18 x 6.5 = 0.815 kJ. Total gained = 6.25 kJ. Metal: 0.150 x c x (100 - 26.5) = 0.150 x c x 73.5 = 11.03 c. So c = 6.25/11.03 = 0.567 kJ/(kg K), about 47% above 0.385. (Even ignoring the cup, c = 5.43/11.03 = 0.492, still ~28% high.)
Systematic errors that inflate c: (1) the metal cooled during transfer, so its true starting temperature was below 100 C, making the actual dT smaller and the true heat loss smaller than assumed - the calculation then overestimates c; (2) heat gained from the warm room or from the hot sample's surroundings; (3) water evaporation/condensation; (4) the sample was wet with boiling water, adding hot water mass. Each makes the water's temperature rise larger than the metal alone caused, inflating the computed c.
2. The student wants to 'correct' the answer by lowering the assumed initial metal temperature until c matches 0.385. Is that legitimate? What procedural improvements would you recommend?
Read the worked answer
No. Back-solving the initial temperature to force agreement with the accepted value is fitting the assumption to the answer - circular reasoning and a form of data manipulation. The initial temperature is an independent measured quantity (or a known boiling point), not a free parameter.
Legitimate improvements: transfer the sample quickly with tongs and a lid to minimize cooling; pre-warm or insulate the calorimeter; measure the metal's actual temperature with a thermometer just before transfer rather than assuming 100 C; dry the sample; stir and record a proper temperature-time curve and extrapolate to the moment of mixing to correct for heat exchange with the room (Newton's-law correction). Report c with an uncertainty budget and compare to 0.385 within that uncertainty. A 47% discrepancy is too large to dismiss; the experiment should be repeated with these controls before drawing conclusions.