Solving the Ice Block Bullet Problem: Calculating Melting Quantity"

[nemzy]

i am stuck on this problem:

a 3.9 g lead bullet at 31.8 C is fired at a speed of 279 m/s into a large block of ice at 0 C, in which it becomes embedded. what is the quantity of ice that melts??

well all i know is that i can find the specific heat and kinetic energy of the bullet, but where do i go from there?

 

[BLaH!]

So when the ice and the bullet come to thermal equilibrium their temperatures will be identical, let's call this temperature $$T^*$$. Also, from conservation of energy we know that the energy lost by the bullet will be gained by the ice in the form of heat. Ok, so let's write the total energy change of the bullet.

$$\Delta E_{bullet} = -\frac{1}{2}mv^2 - Q_{transfer} = -\frac{1}{2}mv^2 - m c_{Pb} \ \Delta T$$​

Here, $$c_{Pb}$$ is the specific heat of lead measured in units of Joules/kg C and $$\Delta T = T^* - T_{bullet}^{i} = T^* - 31.8^0 C$$

Now here is where a trick comes in. Since the problem states that it is a "large block" of ice, we can assume that the temperature of the block of ice doesn't change. That is, the final temperature of both the bullet and block of ice is 0 C. Thus, $$T^* = 0 C$$.

From conservation of energy, $$\Delta E_{ice} = -\Delta E_{bullet}$$. Now the problem is basically finished. You know how much energy was transferred from the bullet to the ice. To determine how much ice was melted (that is, went from solid water at 0 C to liquid water at 0 C), you must use water's latent heat of fusion.

 

[nemzy]

why is kinetic energy negative, and why do u subtract the Qtransfer? shouldn't you add it to find the total energy change?