A bullet travelling horizontally with a speed of 250m/s hits a very big stationary block of ice such that the?

Question

resulting movement of the block of ice due to impact of the bullet is negligible. If the bulllet is made up of copper initially at 25(superscript)oC and all its kinetic energy is imparted on melting parts of the block of ice initially at 0.00(superscript)oC, find the amount of water that will result due to the bullet. (specific heat capacity of copper is 390 J/kg-K, heat of fusion of water is 3.34 x 10^3 J/kg).

Answer 1

Ok, here you need to do this :

There will be some amount of water, that will result due to the bullet, then :

You know that : Heat = m*3.34*10^3 J/kg

In other words, "m" is the mass of water that will result, so we need to multiply it by the heat of fusion of water.

Then, the energy that the bullet will trasmit to the ice is the kinetic energy plus the heat, the bullet is gonna lose from 25º to 0º.

Considering mass of the bullet as "m2"

That will be : m2*(250)^2 / 2 + m2*390*25

so : m*3.34*10^3 = m2*(250)^2 / 2 + m2*390*25

That's the equation, an as the first question said, you need the mass of the bullet.

Answer 2

you need the mass of the bullet