Question

A lead ball at $30{}^{\circ }C$ 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 = $126Jk{g}^{-1}{}^{\circ }{C}^{-1}$ and melting point of lead = $330{}^{\circ }C$. Assume that any mechanical energy lost is used to heat the ball. Use $g$ = $10m{s}^{-2}$.

• A. $2.4\times {10}^{4}cal/kg$
• B. $2.4\times {10}^{5}cal/kg$
• C. $2.4\times {10}^{4}J/kg$
• D. $4.4\times {10}^{5}J/kg$

Answer 1

The correct option is C

$2.4\times {10}^{4}J/kg$

The initial potential energy of the ball,

Potential Energy = $mgh$

= $m\times \left(10m{s}^{-2}\right)\times (6.2\times {10}^{3}m)$

= $m\times (6.2\times {10}^4{m}^2{s}^{-2})$
= $m\times (6.2\times {10}^{4}Jk{g}^{-1})$ -----(i)

Energy required to take the ball from $30{}^{\circ }C$ to $330{}^{\circ }C$
= 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),

$m\left(6.2\times {10}^4\right)=m(L\times 37800)$

$\frac{\left(}{6.2\times {10}^4}=L$

$L$ = $2.4\times {10}^{4}Jk{g}^{-1}$