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.