It has been said that sometimes lead bullets melt upon impact. Assume that a bullet with an initial temperature of 301 K receives 67.9 % of the work done on it by a wall on impact as an increase in internal energy. (The melting point of lead is 601 K, the heat of fusion for lead is L = 23.2 kJ/kg, and the specific heat of lead is c = 0.129 kJ/kg K.)
a) What is the minimum speed with which a 13.3-g lead bullet would have to hit a surface (assuming the bullet stops completely and all the kinetic energy is absorbed by it) in order to begin melting?
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b) What is the minimum impact speed required for the bullet to melt completely?
To find the minimum speed with which the lead bullet would have to hit the surface in order to begin melting, we need to consider the energy balance equation for the bullet upon impact.
a) The energy balance equation is given by:
Initial energy = Work done + Increase in internal energy
The initial energy of the bullet is the kinetic energy, which can be calculated using the formula:
Initial energy = (1/2)mv^2
where m is the mass of the bullet and v is its velocity.
The work done on the bullet by the wall on impact is given as 67.9% of the total work done, so we can calculate it as:
Work done = (67.9%/100) * Total work done
The increase in internal energy is given as a percentage of the total work done, so:
Increase in internal energy = (67.9%/100) * Total work done
The total work done is equal to the kinetic energy of the bullet. Therefore, we can rewrite the energy balance equation as:
(1/2)mv^2 = (67.9%/100) * (1/2)mv^2 + Increase in internal energy
Now, let's solve for the minimum speed (v) with which the bullet would have to hit the surface in order to begin melting.
The increase in internal energy can be calculated using the specific heat of lead and the change in temperature:
Increase in internal energy = mcδT
where c is the specific heat of lead, m is the mass of the bullet, and δT is the change in temperature.
Since the bullet reaches its melting point of 601 K, the change in temperature is:
δT = 601 K - 301 K = 300 K
Substituting the values and rearranging the equation, we get:
(1/2)mv^2 = (67.9%/100) * (1/2)mv^2 + mcδT
Simplifying further:
(1/2)mv^2 - (67.9%/100) * (1/2)mv^2 = mcδT
Rearranging the equation and canceling out terms:
0.5 * ((100 - 67.9%)/100) * mv^2 = mcδT
0.5 * (32.1%/100) * mv^2 = mcδT
0.1605 * mv^2 = mcδT
Dividing both sides by mc:
0.1605v^2 = δT
Now, substitute the values for δT and solve for v:
0.1605v^2 = 300 K
v^2 = (300 K) / 0.1605
v^2 = 1863.92
v = √1863.92
v ≈ 43.14 m/s
Therefore, the minimum speed with which the 13.3-g lead bullet would have to hit the surface to begin melting is approximately 43.14 m/s.
b) To find the minimum impact speed required for the bullet to melt completely, we need to consider the heat of fusion.
The heat of fusion (L) is the amount of energy required to melt a unit mass of a substance.
For the bullet to melt completely, the total heat energy must be equal to or greater than the heat of fusion (L) multiplied by the mass of the bullet.
The total heat energy is equal to the work done on the bullet by the wall, which is equal to the kinetic energy (1/2)mv^2.
So, the equation becomes:
(1/2)mv^2 ≥ mL
Substituting the values and rearranging the equation, we have:
(1/2)(0.0133 kg)(v^2) ≥ (0.0133 kg)(23.2 kJ/kg)
Dividing both sides by (0.0133 kg), we get:
(1/2)v^2 ≥ (23.2 kJ/kg)
v^2 ≥ (23.2 kJ/kg) / (1/2)
v^2 ≥ 46.4 kJ/kg
v^2 ≥ 46.4 * 10^3 J/kg
Taking the square root of both sides, we find:
v ≥ √(46.4 * 10^3) m/s
v ≥ √46.4 * √(10^3) m/s
v ≥ 6.815 * 10 m/s
v ≥ 681.5 m/s
Therefore, the minimum impact speed required for the bullet to melt completely is approximately 681.5 m/s.
a) To calculate the minimum speed with which the lead bullet would have to hit a surface in order to begin melting, we need to determine the amount of thermal energy required to reach the melting point.
Given:
Initial temperature of the bullet (T1) = 301 K
Melting point of lead (Tm) = 601 K
Specific heat of lead (c) = 0.129 kJ/kg K
Heat of fusion for lead (L) = 23.2 kJ/kg
Mass of the bullet (m) = 13.3 g = 0.0133 kg
First, we need to determine the change in internal energy (ΔU) required to raise the temperature of the bullet from T1 to Tm:
ΔU = mcΔT
ΔU = (0.0133 kg)(0.129 kJ/kg K)(601 K - 301 K)
ΔU = 1.314 kJ
Next, we need to determine the work done on the bullet by the wall on impact (W):
W = ΔU / 0.679
W = 1.314 kJ / (0.679)
W = 1.936 kJ
The work done on the bullet is equal to the change in kinetic energy (KE) since the bullet stops completely:
W = ΔKE
W = (1/2)mv^2
1.936 kJ = (1/2)(0.0133 kg)v^2
Now, we can solve for the minimum speed (v) required for the bullet to begin melting:
v^2 = (2 * 1.936 kJ) / (0.0133 kg)
v^2 = 292.744 kJ / kg
v = √(292.744 kJ / kg)
v ≈ 17.1 m/s
Therefore, the minimum speed with which a 13.3-g lead bullet would have to hit a surface in order to begin melting is approximately 17.1 m/s.
b) To calculate the minimum impact speed required for the bullet to melt completely, we need to determine the total thermal energy required for complete melting.
The total thermal energy required for complete melting can be calculated as follows:
Total thermal energy = ΔU + mL
Where ΔU is the change in internal energy and mL is the heat of fusion.
Using the values calculated from part a:
ΔU = 1.314 kJ
m = 0.0133 kg
L = 23.2 kJ/kg
Total thermal energy = 1.314 kJ + (0.0133 kg)(23.2 kJ/kg)
Total thermal energy ≈ 1.314 kJ + 0.30776 kJ
Total thermal energy ≈ 1.62276 kJ
Now, we can calculate the minimum impact speed required for the bullet to melt completely by equating the total thermal energy to the change in kinetic energy:
Total thermal energy = ΔKE
1.62276 kJ = (1/2)(0.0133 kg)v^2
Solving for v:
v^2 = (2 * 1.62276 kJ) / (0.0133 kg)
v^2 = 243.008 kJ / kg
v = √(243.008 kJ / kg)
v ≈ 15.6 m/s
Therefore, the minimum impact speed required for the bullet to melt completely is approximately 15.6 m/s.