A 0.010 kg (10 gram) round ball of solid lead at 280 K initially moves at 400 m/

Question

A 0.010 kg (10 gram) round ball of solid lead at 280 K initially moves at 400 m/s passing

horizontally through atmospheric air. Aerodynamic drag reduces the speed to 300 m/s before the ball

suddenly impacts an immovable and impenetrable solid wall. The frictional work (by the ball on the air)

causes the air to increase in temperature. Assume that heat transfer from the air to the ball is one-third the

value of the frictional work. On impact, the mass of the lead does not change, but the impact causes the ball

to flatten, partially melt and stick to the wall. In the instant this happens there is very little time for significant

heat transfer to occur. Find a) the diameter of the ball, b) the ball temperature just before impact, and c) the

fraction of the lead that likely melted upon impact.

Note: The molar mass of lead is 207 kg/kmol, the density of solid lead is 11300 kg/m3

, the specific heat (change in specific internal energy per degree) is 0.13 kJ/kg-K and at the melting point of lead of 600.6 K. at std. atm. pressure it requires 23 kJ/kg of energy input to melt solid lead at the melting temperature into liquid.

Points only awarded to best explained answer PLEASE SHOW WORK.

Explanation / Answer

a) let's hope that they want the diameter before impact, which is easy.
mass divided by density equals volume.
volume of a sphere is (4/3)Pi r^3
0.010kg/11300kg/m^3 = volume = 8.85 x 10 ^-7 m^3
r = [(3/4Pi)v]^1/3 = 0.005955858m

diameter = 0.012m

b) The loss in kinetic energy is due to wind resistance.
Let's find the change in energy
KE start = KE finish + frictional work
frictional work = KE start - KE finish = 1/2*0.010kg(400^2 - 300^2)m^2/s^2 = 350J
The air transfers 350/3 J to the lead
It takes 0.13 kJ to heat 1kg of lead 1 deg (Kelvin and C degrees are the same size)
Energy = m cp delta T
let's watch our units
delta T = energy/(m cp) = .350/3kJ/(0.010kg*0.13 kJ/kg-K) = 89.7K
The bullet started at 280K so,

immediately before impact the ball temperature is 369.7K

c) When the bullet hits, the remaining KE is converted to thermal energy. This thermal energy heats the bullet to its melting point and then actually melts some of the lead.
KE = (1/2)m v^2 = = (1/2)*0.010kg*(300m/s)^2 = Thermal energy = 450J = .450 kJ
to heat the lead to its melting point requires:
m cp delta T = 0.010kg*0.13kJ/kgK*(600.6-369.7)K = 0.300 kJ
This leaves us with .450 kJ - 0.300 kJ = 0.150 kJ to melt some of the lead. How much lead can we melt?
Energy = heat of fusion times mass
mass = energy/heat of fusion = 0.150 kJ/23kJ/kg = 0.0065 kg
so, 0.0065 of the 0.010kg of the ball could melt. This is about 65/100 or 13/20 in fractional form or 65%

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