When 0.5 kg of water is converted at 10 C into ice at 0 C, an amount of heat is released.

Sud (H2O) = 4183 J / (kg * K);
Qud. pl. (H2O) = 330 kJ / kg;
The total heat will be the sum of the heat released during the cooling of water and the heat of freezing of water:
Q = Csp (H2O) * m (H2O) * (T1 – Tzam.) + Qsp. pl. (H2O) * m (H2O) = 4183 * 0.5 * (10 – 0) + 330,000 * 0.5 = 185915 J.
Answer: Q = 185915 J.