A lead bullet has initial temperature 27C,melting point 327C,latent heat at melting 2.1x10^4J/kg and specific heat capacity of 126J/(kgC). Assuming that all the kinetic energy convert to heat calculate the speed of a bullet that hit a steel slab.
kinetic energy of bullet (`E_k` ) will be equal to the sum of heat required to get the bullet to melting point(`Q_m` ) and latent heat required for melting (`Q_l ` )
`E_k = Q_m+Q_l`
Assume that bullet has a mass of m.
`E_k = 1/2xxmxxv^2`
`Q_m = mxx126xx(327-27) = mxx126xx300`
`Q_l = mxx2xx10^4 `
`1/2xxmxxv^2 = mxx126xx300+mxx2xx10^4 `
`v^2 = 2(126xx300+20000) `
`v^2 = 115600 `
`v = 340m/s`
So the bullet need to strike the steel at a velocity of 340m/s.