# When struck by a hammer, a steel part weighing 0.5 kg heats up by 10 ° C. What is the mechanical work performed

**When struck by a hammer, a steel part weighing 0.5 kg heats up by 10 ° C. What is the mechanical work performed by a hammer if 20% of this work went to increase the internal energy of the part? The specific heat capacity of steel is 500 J / (kg ° C).**

To calculate the mechanical work done when a hammer hits a taken part, we apply the formula: η = Q / A = Cc * m * Δt / A, from where we express: A = Cc * m * Δt / η.

Constant and variable values: Cc – heat capacity of steel (Cc = 500 J / (kg * K)); m is the mass of the steel part (m = 0.5 kg); Δt – temperature rise (Δt = 10 ºС); η is the efficiency of the process under consideration (η = 20% = 0.2).

Calculation: A = 500 * 0.5 * 10 / 0.2 = 12.5 kJ (12.5 * 10 ^ 3 J).

Answer: The hammer had to perform mechanical work of 12.5 kJ.