The bullet will be brought down in temperature from 28 C to 0 C. That will add heat to the ice equal to
M*Clead*28 = 0.868 calories
where Clead is the specific heat of lead,0.031 cal/(g C)
In addition to this, the kinetic energy of the bullet, will be converted to heat, equal to
m V^2/2 = 0.001 kg*230^2/2 = 26.45 J = 6.33 calories
Total heat added to ice = 6.33 + 0.87 = 7.20 calories
Divide that by the heat of fusion of ice, 80 cal/g, to find out how much ice melts. Not very much, apparently.
A 1.00 g lead bullet at 28.0°C is fired at a speed of 230 m/s into a large block of ice at 0°C, in which it becomes embedded. What quantity of ice melts?
Can someone walk me through this step by step?
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