A piece of ice at `0^(@)C` is dropped to the ground from some height. The piece of ice melts completely due to its impact with the ground. Find the height from which the piece was dropped considering that 60% of its energy is converted into heat. `(J = 4.2 J cdot cal^(-1))`

Explanation

The potential energy of the piece of ice of mass m at height h = mgh , kinetic energy = 0
So, total mechanical energy = mgh
This energy is conserved till it touches the ground. Due to the impact with the ground, 60% of this energy.
i.e, `mgh xx 60/100 or 0.6 mgh` is converted into heat energy.
`:.` Heat produced = `(0.6 mgh)/(J)`
Again, heat required to melt m g of ice = mL
(L = latent heat of fusion of ice)
`:. (0.6 mgh)/(J) - mL`
or, `h = (jL)/(0.6 g) [L = 80 cal cdot g^(-1) = 80 xx 1000 cal cdot kg^(-1)]`
`= (4.2 xx (80 xx 1000))/(0.6 xx 9.8) = 5.71 xx 10^(4) m = 57.1 m`