Question

A lead bullet penetrates into a solid object and melts. Assuming that 50% of its kinetic energy was used to heat it, calculate the initial speed of the bullet. The initial temperature of the bullet is `27^@C` and its melting point is `327^@C`. Latent heat of fusion of lead `= 2.5 xx 10^4 J//kg` and specific heat capacity of lead `=125 J//kg.K`.

Answer 1

Let the mass of the bullet = m.
Heat required to take the bullet from `27^(@)`C to `327^(@)`C
Q = ms(`T_(2)-T_(1))` = `mxx125 xx(327-27))`
`=mxx(125)(300)=m xx(3.75 xx 10^(4)Jkgi^(-1)`
Heat required to melt the bullet = `mL=mxx(2.5 xx10^(4)Jkg^(-1)`
If the initial speed be v, the kinetic energy is `1/2mv^(2)` and hence, the heat developed is `1/4(1/2mv^(2))` = `1/8mv^(2)`.
Thus, `1/8mv^(2)`=Q=`(m(3.75+2.5)xx10^(4)Jkg^(-1)`
or `v=707.11ms^(-1)`