Question

A lead ball at ${30}^{\circ }C$ 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 of lead = $126J/k{g}^{\circ }C$ and melting point of lead = ${330}^{\circ }C$. Assume that any mechanical energy lost is used to heat the ball. Use \(g = 10 m/s^{2}\).

• A. $24.7\times {10}^{4}J/kg$
• B. $210.4\times {10}^{4}J/kg$
• C. $2.4\times {10}^{4}J/kg$
• D. $12.4\times {10}^{4}J/kg$

Answer 1

The correct option is C $2.4\times {10}^{4}J/kg$

The initial gravitational potential energy of the ball = mgh
$=m\times \left(10m/s^2\right)\times (6.2\times {10}^{3}m)$
$=m\times (6.2\times {10}^4{m}^2/s^2)=m\times (6.2\times {10}^{4}J/kg)$
All this energy is used to heat the ball as it reaches the ground with a small velocity. Energy required to take the ball from ${30}^{\circ }C$ to ${330}^{\circ }C$ is
$m\times \left(126J/k{g}^{\circ }C\right)\times \left({300}^{\circ }C\right)$
$=m\times 37800J/kg$
and energy required to melt the ball at
${330}^{\circ }C=mL$
where l = latent heat of fusion of lead.
Thus,
$m\times (6.2\times {10}^{4}J/kg)=m\times 37800J/kg+mL$
$orL=2.4\times {10}^{4}J/kg$.