Asked by Anand
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?
Can someone walk me through this step by step?
Answers
Answered by
drwls
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.
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.
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