The steel striker of the pneumatic hammer weighing 1.2 kg heated up by 20 degrees during operation

The steel striker of the pneumatic hammer weighing 1.2 kg heated up by 20 degrees during operation for 1.5 minutes. Assuming that 40% of the hammer’s total energy went into heating the striker, determine the work performed and the power developed at the same time.

Task data: m (weight of the steel striker) = 1.2 kg; t (the duration of the pneumatic hammer operation (t = 1.5 min = 90 s); Δt (change in the temperature of the steel striker) = 20 ºС; η (energy losses for heating) = 40% (0.4).

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

1) The work performed: η = Q / A, whence we express: A = Q / η = Cc * m * Δt / η = 500 * 1.2 * 20 / 0.4 = 30 * 10 ^ 3 J = 30 kJ.

2) Power of the pneumatic hammer during operation: N = A / t = 30 * 10 ^ 3/90 = 333.3 W.