Search any question & find its solution
Question:
Answered & Verified by Expert
A lead ball at $30^{\circ} \mathrm{C}$ 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. If the specific heat of lead $=126 \mathrm{Jkg}^{-10} \mathrm{C}^{-1}$ and melting point of lead $=130 \quad{ }^{\circ} \mathrm{C}$ and suppose that any mechanical energy lost is used to heat the ball, then the latent heat of fusion of lead is
Options:
Solution:
2282 Upvotes
Verified Answer
The correct answer is:
$2.4 \times 10^4 \mathrm{Jkg}^{-1}$
The gravitational potential energy of the ball $=m g h$
$\begin{aligned} & =m \times 10 \times 6.2 \times 10^3 \\ & =m \times 62 \times 10^4 \mathrm{~J}\end{aligned}$
Now energy required to take ball from $30^{\circ} \mathrm{C}$ to $330^{\circ} \mathrm{C}$ is
$m \times 126 \times 300=m \times 37800$
Energy required to melt the ball $=m L$
where, $L=$ latent heat
$\begin{aligned} \Rightarrow m \times 6.2 \times 10^4 \times L & =m \times 37800+m L \\ L & =2.4 \times 10^4 \mathrm{Jkg}^{-1}\end{aligned}$
$\begin{aligned} & =m \times 10 \times 6.2 \times 10^3 \\ & =m \times 62 \times 10^4 \mathrm{~J}\end{aligned}$
Now energy required to take ball from $30^{\circ} \mathrm{C}$ to $330^{\circ} \mathrm{C}$ is
$m \times 126 \times 300=m \times 37800$
Energy required to melt the ball $=m L$
where, $L=$ latent heat
$\begin{aligned} \Rightarrow m \times 6.2 \times 10^4 \times L & =m \times 37800+m L \\ L & =2.4 \times 10^4 \mathrm{Jkg}^{-1}\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.