Question

An ice block at $0^{\circ }\text{C}$ and of mass $m$ is dropped from height ${}^{\prime }{h}^{\prime }$ such that the loss in gravitational potential energy of the block is exactly equal to the heat required to completely melt the ice. Taking latent heat of fusion of ice, $L=80\text{cal/gm}$, acceleration due to gravity $g=10{\text{m/s}}^2$ and mechanical equivalent of heat $J=4.2\text{J/cal}$. find the value of ${}^{\prime }{h}^{\prime }$.

• A. $18.20\text{km}$
• B. $9.10\text{km}$
• C. $33.60\text{km}$
• D. $45.50\text{km}$

Answer 1

The correct option is C $33.60\text{km}$

Given:
Loss in gravitational potential energy $\left(W\right)=mgh$
Heat required to melt the ice $\left(H\right)=mL$
where $L$ is latent heat of vapourisation
We know that, $W=JH$
From the data given in the question, we can write,
$mgh=mL\times J$
$\Rightarrow h=\frac{JL}{g}$
Given,
Latent heat of fusion of ice $L=80\text{cal/gm}=80\times {10}^3\text{cal/Kg}$,
Acceleration due to gravity $g=10{\text{m/s}}^2$
and Mechanical equivalent of heat $J=4.2\text{J/cal}$
$\therefore h=\frac{4.2\times 80\times {10}^3}{10}$
$=33.6\times {10}^3\text{m}=33.60\text{km}$
Thus, option (c) is the correct answer.