Ice in Warm Water: Deciding the Final State Before Calculating
A student puts 0.300 kg of water at 30.0 °C into a polystyrene cup and adds 0.050 kg of ice taken straight from a freezer at −5.0 °C. She stirs gently and records the lowest temperature reached with a thermometer that reads to 0.5 °C. The laboratory is at 29 °C and the air is humid, so the outside of the cup quickly becomes wet with condensation. She then repeats the experiment using 0.150 kg of ice at −5.0 °C with the same 0.300 kg of water at 30.0 °C, and is surprised that the thermometer settles at a very different value. Take c_water = 4200 J kg⁻¹ °C⁻¹, c_ice = 2100 J kg⁻¹ °C⁻¹ and the specific latent heat of fusion of ice = 3.34 × 10⁵ J kg⁻¹. Treat the cup as having negligible heat capacity and ignore the spoon in your first calculation.
1. For the first experiment, determine whether all the ice melts and calculate the final temperature of the mixture.
Read the worked answer
Never assume the final state - test it by comparing energies. Use 0 °C as the reference point.
Energy the water can give up cooling to 0 °C: (Q_{water} = mc\Delta\theta = 0.300\times4200\times30.0 = 37,800\text{ J})
Energy needed to warm the ice to 0 °C: (Q_{warm} = 0.050\times2100\times5.0 = 525\text{ J})
Energy needed to melt all the ice: (Q_{melt} = mL_f = 0.050\times3.34\times10^5 = 16,700\text{ J})
Total required to produce 0.050 kg of water at 0 °C: (525 + 16,700 = 17,225\text{ J}).
Since 37 800 J > 17 225 J, all the ice melts and there is 37 800 − 17 225 = 20 575 J left over to warm the whole mixture.
Mass of mixture = 0.300 + 0.050 = 0.350 kg.
(\Delta\theta = \frac{Q}{mc} = \frac{20,575}{0.350\times4200} = 14.0\text{ °C})
Final temperature ≈ 14.0 °C, quoted to two significant figures to match the 0.5 °C thermometer.
2. In the repeat with 0.150 kg of ice at −5.0 °C, predict the final state and temperature, and explain why the same method cannot be used again.
Read the worked answer
Test the state again - this is the crucial step.
Available energy from the water cooling to 0 °C: (37,800\text{ J}) (unchanged).
Required to warm the ice to 0 °C: (0.150\times2100\times5.0 = 1575\text{ J}).
Remaining available for melting: (37,800 - 1575 = 36,225\text{ J}).
Required to melt all 0.150 kg: (0.150\times3.34\times10^5 = 50,100\text{ J}).
Since 36 225 J < 50 100 J, the water cannot melt all the ice. The mixture therefore ends at exactly 0 °C with ice and water together in equilibrium.
Mass melted: (m = Q/L_f = 36,225/3.34\times10^5 = 0.108\text{ kg}), so about 0.042 kg of ice remains.
The trap is that the leftover-energy formula (\Delta\theta = Q/mc) would give a negative or impossible temperature if applied blindly here, because the assumption "all ice melted, mixture is all water" is false. Always compare the available energy with the energy needed to melt all the ice first; only then choose the equation.
3. The final temperature in the first experiment is 14.0 °C while the room is at 29 °C. Explain how heat exchange with the surroundings and the metal spoon affect the measured value, and say which effect is likely larger.
Read the worked answer
Direction first, magnitude second. Because the mixture ends colder than the room (14 °C vs 29 °C), the temperature gradient across the cup wall points inwards. Heat flows from the warm room into the cold cup, and the condensation of water vapour on the outside releases latent heat. Both add energy to the mixture that the ideal calculation did not allow for, so the measured final temperature is higher than the true value - the calculated value 14.0 °C is a lower bound.
The spoon. A metal spoon has significant heat capacity and starts at room temperature, so it gives up heat to the mixture, again raising the final temperature. However, a 20 g steel spoon ((m c \Delta\theta \approx 0.020\times450\times15 \approx 135\text{ J})) transfers well under 1% of the ~20 000 J involved, whereas the leakage through the walls acts over the whole several minutes of the experiment. So the insulation failure dominates.
Consequences for the design: use a lid, dry the outside of the cup, work quickly, and plot a cooling-correction graph (temperature against time, extrapolated back to the moment of mixing) so that the reading is corrected rather than assumed.