How much heat is required to heat a 200 kg steel part by 50K? Specific heat capacity of steel 460 J (kgK).

univerkov | September 21, 2021 | education

Initial data: m (weight of the steel part taken) = 200 kg; Δt (required increase in part temperature) = 50 K.

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

The required amount of heat to increase the temperature of the steel part is determined by the formula: Q = Сс * m * Δt.

Calculation: Q = 460 * 200 * 50 = 4,600,000 J (4.6 MJ).

Answer: 4.6 MJ of heat is required to heat the steel part taken.

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