What amount of heat is transferred by a steel bar with a volume of 0.02 m3 to surrounding bodies when cooled from 520 to 20 degrees Celsius. (The density of steel is 7800 kg \ m3, and its specific heat capacity is 500 J \ (kg C Celsius).
Given: V (volume of the cooled steel bar) = 0.02 m3; t1 (initial temperature of the bar) = 520 ºС; t2 (final temperature) = 20 ºС.
Constants: by condition ρс (steel density) = 7800 kg / m3; Сс (specific heat capacity of steel) = 500 J / (kg * ºС).
The heat transferred by the steel bar to the surrounding bodies is calculated by the formula: Q = Cc * m * (t1 – t2) = Cc * ρс * V * (t1 – t2).
Calculation: Q = 500 * 7800 * 0.02 * (520 – 20) = 39 * 10 ^ 6 J (39 MJ).
Answer: The steel bar will transfer 39 MJ of heat to the surrounding bodies.
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