How much heat is required to heat a 200 g steel part from 35 to 1235 ‘C?

Initial data: m (weight of the steel part taken) = 200 g (0.2 kg); t0 (initial temperature) = 35 ºС; t (final temperature of the part) = 1235 ºС.

Reference data: Сс (specific heat capacity of steel) = 500 J / (kg * K).

The required amount of heat for heating the steel part taken is determined by the formula: Q = Cc * m * (t – t0).

Let’s make the calculation: Q = 500 * 0.2 * (1235 – 35) = 120,000 J or 120 kJ.

Answer: To heat the steel part taken, 120 kJ of heat is needed.