How many degrees does an iron part weighing 10 kg heat up from a blow of a 500 kg hammer

How many degrees does an iron part weighing 10 kg heat up from a blow of a 500 kg hammer falling from a height of 2.3 m if the part absorbs 80% of the released heat?

Data: m1 (weight of the iron part) = 10 kg; m2 (hammer weight) = 500 kg; h (height from which the hammer fell) = 2.3 m; η (percentage of absorbed heat from impact) = 80% (0.8).

Reference values: Czh (specific heat of iron) = 460 J / (kg * K); g (acceleration due to gravity) ≈ 10 m / s2.

The change in the temperature of the iron part after impact is expressed from the formula: η = Q / ΔEp = Czh * m1 * Δt / (m2 * g * h), whence Δt = m2 * g * h * η / (Czh * m1).

Calculation: Δt = 500 * 10 * 2.3 * 0.8 / (460 * 10) = 2 ºС.