Question

A ball at ${40}^{\circ }C$ is dropped from a height of $5\text{km}$. The ball is heated due to air resistance and it completely melts just before reaching the ground. The specific heat capacity of the ball is $125{\text{J kg}}^{-1}{{}^{\circ }C}^{-1}$ and melting point is ${340}^{\circ }\text{C}$. Then, the latent heat of the fusion of the ball (in ${\text{kJ kg}}^{-1}$) is

[Take $g=10{\text{m/s}}^2$]

Answer 1

Let the mass of the ball be $m\text{kg}$.
Gravitational potential energy $Q=mgh$
$\Rightarrow Q=m\times 5\times {10}^4\text{J}$
Energy required to heat the ball from ${40}^{\circ }\text{C}$ to ${340}^{\circ }\text{C}$ is given by
$Q_1=ms\Delta T$
$Q_1=m\times 125\times [340-40]=m\times 37500\text{J}$
Energy required to melt the ball completely
$Q_2=mL$, where $L$ = Latent heat of fusion of the ball

For the ball to melt completely, gravitational P.E must be converted to the required heat due to the presence of air resistance.
From the data given in the question,
$Q=Q_1+Q_2$
$\Rightarrow m\times 5\times {10}^4=m\times 37500+mL$
$\Rightarrow L=12500{\text{J kg}}^{-1}$
The latent heat of the fusion of ball is $12500{\text{J kg}}^{-1}=12.5\text{kJ/kg}$