The temperature of a piece of lead is `27^(@)C`. Find out the minimum velocity of its impact with a wall so that it melts completely. Suppose 58% of the heat generated is dissipated. Given, `J = 4.2 J cdot cal^(-1)` , melting point, specific heat and latent heat of fusion of lead are `327^(@)C, 0.03 cal cdot g^(-1) cdot^(@)C^(-1)` and 5 cal cdot g^(-1)`, respectively.

Explanation

`H xx (100-58)/(100) = mst + ml`
or, `H = (mst + ml) xx 100/42`
Again, `W = 1/2 mv^(2)`
Now, `W = JH`
or, `1/2 mv^(2) = J(mst + ml) xx 100/42`
or, `v^(2) = 2J(st + l) xx 100/42`
`= 2 xx (4.2 xx 10^(7)) xx [0.03 xx (327-27) + 5] xx 100/42`
`=28 xx 10^(8)`
So, `v = 5.29 xx 10^(4) cm cdot s^(-1) = 529 m cdot s^(-1)`.