Aluminum bullet embedded in ice: why isn't velocity used?

[Boozehound]

A 2.65g aluminum bullet at 31.4°C is fired at a speed of 209 m/s into a large block of ice at 0°C, in which it embeds itself. What quantity of ice melts?

Q=(9.00E2)(.00265)(31.4)=74.889J
74.889J=m(33.5E4)=2.23E-4kg

so i get a final answer of 2.23E-4kg but its wrong. i think I am either missing a step because I am not using the velocity of the bullet, or I am plugging wrong numbers in. any help is appreciated.

 

[danago]

I think you need to consider the change in kinetic energy, which i assume is all converted to heat energy, thus, increasing the temperature of the bullet.

 

[Boozehound]

what formula would i use for that? i looked in my book and i couldn't come up with a formula...

 

[danago]

Well kinetic energy is given by $$E_k = \frac{{mv^2 }}{2}$$.

You can work out the initial kinetic energy by simply using m=0.00265, v=209, and then find a final kinetic energy, using the same mass, but v=0, since it has stopped in the ice. Then find the change in energy from those.

 

[Boozehound]

alright i got it thank you very much!