Find out the minimum height from which a piece of ice at `0^(@)C` should be dropped so that it melts completely due to its impact with the ground. Assume that half of the energy loss due to the fall is responsible for the fusion of ice. (Latent heat of fusion of ice = `80 cal cdot g^(-1), g = 980 cm cdot s^(-2) , J = 4.2 J cdot cal^(-1))`

Explanation

Let the piece of ice of mass m be allowed to fall from height h.
Now, energy loss due to the fall = loss in potential energy = mgh = work done, W.
Half of it, i.e. `W/2` amount of energy is converted into heat and it is responsible for the fusion of ice.
`:.` Heat produced, `H = (W//2)/(J) = (mgh)/(2)`
As m g of ice melts into water, required latent heat = 80 m cal
So, `(mgh)/(2) = 80 m`
or, `h = (80 xx 2J)/(g) = (80 xx 2 xx 4.2 xx 10^(7))/(980)`
`[ J = 4.2 J cdot cal^(-1) = 4.2 xx 10^(7) erg cdot cal^(-1)`]
`= 6.857 xx 10^(6)cm = 68.57 km`.