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A lead ball initially at 30 deg C is dropped from a height of 6.2 km. Air resistance heats the ball, and it just completely melts on reaching the ground. Taking all the lost mechanical energy as heat in the ball, find the latent heat of fusion of lead. (Specific heat of lead = 126 J/(kg*deg C), melting point of lead = 330 deg C, g = 10 m/s².)

1. 6 * 10³ J/kg
2. 1.2 * 10⁴ J/kg
3. 1.8 * 10⁴ J/kg
4. 2.4 * 10⁴ J/kg

Correct answer: 2.4 * 10⁴ J/kg

Solution

Energy per kg = g*h = 62000 J; the heating part c*Deltatheta = 126*300 = 37800 J, so L = 62000 - 37800 = 24200 ~ 2.4 * 10⁴ J/kg.

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