How much heat is required to heat a lead part weighing 200 g at 70 C?

univerkov | May 11, 2021 | education

Task data: m (mass of a given lead part) = 200 g (in SI m = 0.2 kg); Δt (required change in part temperature) = 70 ºС.

Constants: Сс (specific heat capacity of lead) = 140 J / (kg * K).

The required heat for heating a given lead part is determined by the formula: Q = Cc * m * Δt.

Let’s calculate: Q = 140 * 0.2 * 70 = 1960 J.

Answer: It takes 1960 Joules of heat to heat a given lead part.

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