Calculating Melted Ice from a Lead Bullet

[AdnamaLeigh]

Given: latent heat of fusion of water is 333000 J/kg.
A 3.37 g lead bullet at 27 degrees C is fired at a speed of 272 m/s into a large block of ice at 0 degrees C, in which it embeds itself. What quantity of ice melts? Assume the specific heat of lead is 128 J/kg x C. Answer in units of g.

KE = 1/2mv^2
KE = 1/2(.00337)(272)^2 = 124.663 J

KE = Q = mL
124.663 = m(333000) m = .374 g

I don't think that this is correct because I didn't take into account change in temperature of the bullet.

(bullet) mc∆T = mL (ice)

I would set it up like that but I don't have the change in temperature for the bullet. I tried using this formula to figure out the temperature but I got a weird answer:

124.663 = mc∆T = .00337(128)(27 - 0) T= -262 C

That obviously can't be correct. I really need some help.

 

[Fermat]

I think you can assume that ...

1) All the mechanical energy (1/2mv²) is transformed into heat energy used to melt the ice.
2) Everything, the ice, the melted ice/water and the bullet are all at the same temperature, which must be 0 deg C.
3) The mechanical energy plus the heat loss from the bullet are both used to melt the ice.

So, the change in temp of the bullet is 27 deg.

Mind you, I haven't done any thermo in a long time, so I can't guarantee that I'm correct!.