The Cooling Curve Experiment
Students heat a 200 g aluminium block to 90 °C and place it in 150 g of water at 20 °C in an insulated cup. They record the mixture reaching 38 °C. The specific heat capacity of water is 4200 J/(kg·K); for aluminium the accepted value is 900 J/(kg·K). One student calculates the aluminium's specific heat from the data and gets about 640 J/(kg·K). Another says the insulation must be perfect because the cup is insulated. A third suggests the block lost heat to the air during transfer. The group must decide what the result means and how to improve the method.
1. Carry out the heat-balance calculation and explain why the result is lower than the accepted value.
Read the worked answer
Heat gained by water = 0.150 × 4200 × (38 − 20) = 0.150 × 4200 × 18 = 11 340 J. If all of this came from the aluminium: 11 340 = 0.200 × c × (90 − 38) = 0.200 × c × 52 = 10.4c, so c ≈ 1090 J/(kg·K). That is higher than the accepted 900, not lower. If the student got 640, they likely used the wrong temperature change or mass. Taking the calculation at face value, the measured c exceeds the accepted value, meaning less heat was transferred to the water than the block actually released. This is consistent with heat lost to the surroundings during transfer and to the cup. The judgement is that the method systematically underestimates the heat reaching the water, inflating the calculated c.
2. Critique the claim that the insulated cup makes heat loss negligible, and propose two concrete improvements.
Read the worked answer
An insulated cup reduces loss but does not eliminate it: the cup has thermal mass, the lid may be imperfect, and the block is exposed to air during transfer from the oven to the cup. The measured temperature rise also lags because the mixture takes time to equilibrate, so the recorded final temperature may be low. Improvements: (1) transfer the block quickly and use tongs to minimise air exposure, and pre-warm or account for the cup's heat capacity; (2) stir and record temperature over time, extrapolating the cooling curve back to the moment of mixing to correct for ongoing loss. A further refinement is to repeat with several starting temperatures and plot heat gained against temperature change to average out losses.