Ice into Warm Water: When Nothing Melts Completely
A student drops 200 g of ice at −10 °C into 500 g of water at 30 °C contained in a calorimeter whose water equivalent is 50 g, and stirs until nothing further changes. Take c(ice) = 2100 J kg⁻¹ K⁻¹, c(water) = 4186 J kg⁻¹ K⁻¹, L(fusion) = 3.34 × 10⁵ J kg⁻¹, and assume no heat exchange with the room. She expects the final temperature to lie somewhere between 0 °C and 30 °C, and intends to use the result as a demonstration of conservation of energy. The laboratory assistant warns her that the answer is 'boring and interesting at the same time' - the thermometer may not land where she expects, and a small change in the starting conditions will flip the outcome completely. A second bench repeats the run with 250 g of ice at −10 °C but the same water and calorimeter.
1. Determine the final state of the first mixture: final temperature, and the masses of ice and water present.
Read the worked answer
Heat needed to bring the ice to 0 °C and melt it all:
- warming: 0.200 × 2100 × 10 = 4200 J
- melting: 0.200 × 3.34 × 10⁵ = 66 800 J
- total = 71 000 J
Heat available from cooling water + calorimeter from 30 °C to 0 °C: Q = (0.500 + 0.050) × 4186 × 30 = 0.550 × 4186 × 30 = 69 069 J
Since 69 069 J < 71 000 J, the available heat is not enough to melt all the ice, so the final temperature cannot be above 0 °C. It is exactly 0 °C, with ice and water coexisting.
Unmelted ice: deficit = 71 000 − 69 069 = 1931 J; m = 1931/(3.34 × 10⁵) = 5.8 × 10⁻³ kg ≈ 5.8 g.
Final state: 5.8 g ice + (500 + 194.2) = 694 g water, all at 0 °C.
The lesson is that the latent term (66 800 J) dwarfs every sensible-heat term; a mixture can be 'almost' all melted and still be pinned at 0 °C.
2. Work out the final state for the second bench, which uses 250 g of ice at −10 °C with the same 500 g of water and the same calorimeter.
Read the worked answer
Heat needed for all the ice:
- warming: 0.250 × 2100 × 10 = 5250 J
- melting: 0.250 × 3.34 × 10⁵ = 83 500 J
- total = 88 750 J
Heat available from cooling 0.550 kg equivalent from 30 °C to 0 °C = 69 069 J (unchanged).
Deficit = 88 750 − 69 069 = 19 681 J, so the final state is again 0 °C, but with much more ice left: m = 19 681/(3.34 × 10⁵) = 0.0589 kg ≈ 59 g of ice.
Water present = 500 + (250 − 59) = 691 g.
Judgement. Adding 50 g more ice removes 19.7 kJ more heat but the water had only 69 kJ to give, so the extra ice mostly survives. Notice how insensitive the final temperature is (0 °C in both runs) while the ice remaining is very sensitive to the initial mass. A student who reports a temperature alone has thrown away the informative part of the result; the ice residue is the real measurement.
3. Find the minimum initial water temperature (same 500 g, same calorimeter) that would just melt all the ice in the first run, and comment on what this implies for the experiment.
Read the worked answer
Setting up the balance. For all the ice to melt and the mixture to finish at exactly 0 °C, the water + calorimeter must supply 71 000 J in cooling from T to 0 °C:
0.550 × 4186 × T = 71 000 → T = 71 000/2302 = 30.84 °C.
So about 30.8 °C is the threshold; at the recorded 30 °C the mixture misses it by a mere 0.8 °C.
Why this matters. A sub-degree change in starting temperature (or a slightly warm calorimeter, or 5 g of water left clinging to the ice) decides between 'a little ice floats at 0 °C' and 'everything melts and the temperature rises a few tenths of a degree'. The experiment is therefore extremely sensitive to the quantities students usually wave away:
- the water equivalent of the calorimeter (50 g here contributes 10 % of the heat capacity);
- heat leaking in from the room during stirring, which supplies of order the 2 kJ deficit in a couple of minutes;
- the ice being wet when transferred.
A defensible report must state not just 0 °C but the ice residue with its uncertainty, and must estimate the heat-leak error rather than assume it away.