    Question

# A lead ball at 30 ∘C is dropped from a height of 6.2 km. 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 = 126 J kg−1 ∘C−1 and melting point of lead = 330 ∘C. Assume that any mechanical energy lost is used to heat the ball. Use g = 10 ms−2.

A

2.4 × 104 cal/kg

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B

2.4 × 105 cal/kg

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C

2.4 × 104 J/kg

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D

4.4 × 105 J/kg

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Solution

## The correct option is B 2.4 × 104 J/kg The initial potential energy of the ball, Potential Energy = mgh = m × (10 m s−2) × (6.2 × 103 m) = m × (6.2 × 104 m2 s−2) = m × (6.2 × 104 J kg−1) -----(i) Energy required to take the ball from 30 ∘C to 330 ∘C = mL + mc(∆ T) = m ( L + c(∆ T)) = m ( L + (126*300 )) = m ( L + 37800) ---- (ii) Since potential energy of the ball is converted to heat energy to melt the ball, equating (i) and (ii), m(6.2 × 104) =m(L × 37800) (6.2 × 104)37800 = L L = 2.4 × 104 J kg−1  Suggest Corrections  0      Related Videos   What is Heat and Temperature
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