The method of mixtures. Ice is added to warm water in a calorimeter. The water and the calorimeter lose heat, and the ice gains heat in three stages: it warms to 0 °C (if it started below 0 °C), it melts at 0 °C, and the melted water warms from 0 °C to the final temperature. If no heat is exchanged with the surroundings, the heat lost equals the heat gained:

(m_wc_w + m_cc_c)(T_w − T_f) = m_ic_i(0 − T_i) + m_iL_f + m_ic_wT_f

L_f = [ (m_wc_w + m_cc_c)(T_w − T_f) ] / m_i − c_i(0 − T_i) − c_wT_f

The temperatures are in °C and the differences are the same as in kelvin. Use c_w = 4186 J kg⁻¹ K⁻¹ for water and c_i = 2100 J kg⁻¹ K⁻¹ for ice. The specific heat capacities of the calorimeters are 900 (aluminium), 385 (copper), 840 (glass) and 1300 (polystyrene) J kg⁻¹ K⁻¹. The accepted value of the latent heat of fusion of water is 334 kJ kg⁻¹ (3.34 × 10⁵ J kg⁻¹).

Measuring the masses. The mass of the water is the mass of the calorimeter with the water minus the mass of the empty calorimeter, so you do not need the density of water. The mass of the ice is found the same way: weigh the calorimeter again after the ice has been added, and subtract. (Mass is conserved when the ice melts, so the balance reading does not change as it melts.)

Choosing the amounts. All the ice must melt, or the final temperature is not the temperature at which the whole mixture is in equilibrium. The water and the calorimeter must be able to supply enough heat: m_i(c_i|T_i| + L_f) must be less than (m_wc_w + m_cc_c)T_w. If there is too much ice or the water is too cold, the temperature falls to 0 °C and ice is left floating. A final temperature only a few degrees above 0 °C makes the result very sensitive to the thermometer error, because T_w − T_f is large but T_f is small; a final temperature near the middle of the range is a good compromise.

Sources of error to investigate. Try the experiment again with one change at a time and see what happens to the result. (1) Heat exchanged with the room. If the calorimeter is colder than the room (22 °C) it gains heat, which helps to melt the ice, so the water gives up too little heat and L_f comes out too low. If it is warmer than the room it loses heat, the water seems to give up too much and L_f comes out too high. With the water starting at 50 °C and finishing near 25 °C it is warmer than the room all the time, so an open calorimeter gives a result that is too high. A lid and jacket reduce the effect, and starting with the water as far above room temperature as it finishes below it makes the loss and the gain cancel. The effect is larger if you do not stir, because the experiment takes longer. (2) Wet ice. Ice taken from a bath of ice and water at 0 °C carries water with it. That water is weighed as ice but needs no heat to melt, so L_f comes out too low. Drying the ice first fixes this. (3) Leaving out the calorimeter, the warming of the ice, or the warming of the melted ice. The results panel shows how big each of these omissions is. (4) Reading the lowest temperature too early or too late. (5) Thermometer and balance uncertainty. The results panel gives the uncertainty of L_f from the uncertainties of the readings (±0.1 g and ±0.1 °C), combined as independent errors.

How the simulation works. The water, melted ice and calorimeter are at one temperature, because they are stirred (or nearly). The ice takes heat from the water at a rate proportional to the temperature difference and to its surface area, which is why the ice melts quickly at first and then slows. Without stirring the heat transfer is five times slower. The calorimeter exchanges heat with the 22 °C room at a rate proportional to the difference in temperature (0.05 W K⁻¹ with a lid and jacket, 0.35 W K⁻¹ when open). Evaporation, the thermometer and the stirrer are ignored. Energy is conserved at every step of the calculation.