During cold treatment, a steel part weighing 0.54 kg at a temperature of 20 degrees was placed in a refrigerator

During cold treatment, a steel part weighing 0.54 kg at a temperature of 20 degrees was placed in a refrigerator, the temperature of which was 80 degrees. How much heat was generated when the part was cooled?

Task data: m (mass of the steel part processed by cold) = 0.54 kg; t0 (initial temperature of the part) = 0 ºС; t (temperature of the refrigerator, final temperature of the part) = -80 ºС.

Reference values: C (specific heat capacity of steel) = 462 J / (kg * K).

The amount of heat released when the part is cooled in the refrigerator is determined by the formula: Q = C * m * (t0 – t).

Calculation: Q = 462 * 0.54 * (0 – (-80)) = 24 948 J.

Answer: 24 948 J of heat was released.