A `50` g lead bullet (specific heat `0.02`) is initially at `30^(@)C`. It is fired vertically upward with a speed of `840 ms^(-1)`. On returning to the starting level it strikes a cake of ice at `0^(@)C`. How much ice is melted ? Assume that all energy is spent in melting only. Latent heat of ice `=336 j g^(-1)`.

Answer

`[52.875 g]`

Explanation

When bullet falls back to its original level its downward velocity is same as given upward velocity at that location, i.e. ,`840 ms^(-1)` . KE of bullet on returening to the initial level
`W=(1)/(2)mv^(2) =(1)/(2)xx((50)/(1000))xx(840)^(2)=17640 j`
This KE will be converted into heat energy on striking the cake of ice. So heat produced `Q_(1)=W=17640j`
Heat lost by bullet in cooling from `30^(@)C` to `0^(@)C` is
`Q_(2)=mc Delta T=50xx0.02xx(30-0)=30 cal =30xx4.2 j=126 j`
Total heat given by bullet to icec is
`Q=Q_(1)+Q_(2)=17640+126=17766 j`
Let m be the mass of ice melted. Then
`17766=mL` or `m=(17766j)/(Ljg^(-1))=(17766)/(336)g =52.875 g`