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A lead ball at 30 c is dropped from a height of 6.2km.The ball is heated due to the air resistance and it completely melts just before reaching the ground.The molten substance falls slowly on the ground.Calculate the latent heat of fusion of lead.Specific heat capacity of lead=126J/kg-c and melting point of lead =330 c assume that any mechanical energy lost is used to heatt the ball.Use g=10m/s^2
Using Energy Conservation : mgh = ms?T + mL 10x6.2x10³ = 126(300) + L (62 - 37.8)x10³ = L L = 24.2x10³ J/kg
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