From what height did the lead ball fall without initial velocity if its temperature increased by 12

From what height did the lead ball fall without initial velocity if its temperature increased by 12 ° C upon impact on the surface?

Data: Δt (change in temperature of a given lead ball after impact) = 12 ºС; V0 (initial ball velocity) = 0 m / s.

Constants: Cc (specific heat capacity of lead) = 140 J / (kg * K); g (acceleration due to gravity) ≈ 10 m / s2.

To determine the initial height of a given lead ball, we apply the formula: Сс * m * Δt = m * g * h, whence h = Сс * Δt / g.

Calculation: h = 140 * 12/10 = 168 m.

Answer: The target ball was 168 m above the surface.



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