A steel part weighing 20 kg heated up from 10 to 60 degrees. How much has the internal energy

A steel part weighing 20 kg heated up from 10 to 60 degrees. How much has the internal energy of the part increased if the specific heat capacity of steel is 460 J / kg degrees?

Task data: m (weight of the steel part) = 20 kg; t0 (part temperature before heating) = 10 ºС; t (final temperature) = 60 ºС.

Reference values: according to condition C (specific heat capacity of steel) = 460 J / (kg * K).

The increase in the internal energy of the steel part is determined by the formula: Q = C * m * (t – t0).

Calculation: Q = 460 * 20 * (60 – 10) = 460 000 J (460 kJ).

Answer: The internal energy of the steel part has increased by 460 kJ.