What amount of heat was released when water weighing 5 kg freezes, if its initial temperature is 12 ° C?

Initial data: m (mass of water) = 5 kg; t0 (initial water temperature) = 12 ºС.

Reference values: tк (water crystallization temperature) = 0 ºС; C (specific heat of water) = 4200 J / (kg * K); λ (specific heat of crystallization of water) = 3.4 * 10 ^ 5 J / kg.

Heat quantity: Q = Q1 (water cooling) + Q2 (transition to ice) = C * m * (t0 – tк) + λ * m.

Let’s calculate: Q = 4200 * 5 * (12 – 0) + 3.4 * 10 ^ 5 * 5 = 1,952,000 J = 1.952 MJ.

Answer: When 5 kg of water freeze, 1.952 MJ of heat was released.