After cooking, 200 g of butter remained in the pan at a rate of 120 degrees.

After cooking, 200 g of butter remained in the pan at a rate of 120 degrees. how much heat will be released when the oil cools if the room temperature is 20 degrees?

Initial data: m (weight of sunflower oil) = 200 g; t0 (initial oil temperature) = 120 ºС; t (room temperature) = 20 ºС.

Reference values: C (specific heat capacity of sunflower oil) = 1700 J / (kg * K).

SI system: m = 200 g = 0.2 kg.

The amount of heat released when the oil is cooled can be calculated using the formula: Q = C * m * (t0 – t).

Let’s calculate: Q = 1700 * 0.2 * (120 – 20) = 34,000 J = 34 kJ.

Answer: When the sunflower oil cools down, 34 kJ of heat will be released.