300 g of water, the initial temperature of which was 40 * C, cooled down to 20 * C. how much heat was released during this?

Task data: m (mass of water) = 300 g (0.3 kg); t0 (temperature before cooling) = 40 ºС; t (final temperature) = 20 ºС.
Reference values: Cw (specific heat of water) = 4200 J / (kg * K).
The heat released during the cooling of 3300 g of water is determined by the formula: Q = Cw * m * (t – t0).
Calculation: Q = 4200 * 0.3 * (40 – 20) = 25 200 J (25.2 kJ).
Answer: When 300 grams of water were cooled, 25.2 kJ of heat was released.