A hammer weighing 2 tons falls on a steel bar weighing 1 kg from a height of 3 m.

A hammer weighing 2 tons falls on a steel bar weighing 1 kg from a height of 3 m. How many degrees will the bar heat up on impact, if 50% of the hammer’s total energy goes into heating?

Given: m1 (mass of the taken hammer) = 2 t (in SI m1 = 2000 kg); m2 (weight of steel bar) = 1 kg; h (drop height) = 3 m; η (the percentage of energy of the hammer that went to heat the blank) = 50% (0.5).

Constants: g (acceleration due to gravity) ≈ 10 m / s2; Cc (specific heat capacity of steel) = 500 J / (kg * K)

We express the change in temperature of the steel ingot from the equality: η = Q / Ep = Cc * m2 * Δt / (m1 * g * h), whence Δt = m1 * g * h * η / (Cc * m2).

Calculation: Δt = 2000 * 10 * 3 * 0.5 / (500 * 1) = 60 ºС.