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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