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)`.

Explanation

Here, `m=50g=(50)/(1000)kg=(1)/(20) kg,c=0.02`
`theta_(1)=30^(@)C, upsilon=840m//s, theta_(2)=0^(@)C,M=?`
`l=80cal//gram`.
In going up, KE of bullet is converted into P.E. and in falling down, P.E. is converted into same K.E.
`:.` Energy converted into heat`=(1)/(2)m upsilon^(2)`
`=(1)/(2)xx(1)/(2)(840)^(2)J`
`E=(1)/(40)xx(840xx840)/(4.2)cal`.
Heat spent by bullet to cool to `0^(@)C`
`=cm Deltatheta=0.02xx50xx(30-0)cal. `
`=30cal`.
If M is mass of ice melted, heat spent `=MxxL = Mxx80` cal.
`:. E=30+Mxx80`
`(840xx840)/(40xx4.2) = 30+Mxx80`
`4200-30=80M`
`M=(4170)/(80)=52.1 gram`