A 0.0500-kg lead bullet of volume 5.00 × 10-6 m3 at 20.0 °C hits a block that is
ID: 1484269 • Letter: A
Question
A 0.0500-kg lead bullet of volume 5.00 × 10-6 m3 at 20.0 °C hits a block that is made of an ideal thermal insulator and comes to rest at its center. At that time, the temperature of the bullet is 327 °C. Use the following information for lead: coefficient of linear expansion: a = 2.0 × 10-5/C° specific heat capacity: c = 128 J/(kg • C°) latent heat of fusion: Lf = 23 200 J/kg melting point: T melt = 327 °C
A) What is the speed of the bullet
B) What is the volume of the bullet when it comes to rest
Explanation / Answer
Solution:From the question we have
A 0.0500-kg lead bullet of volume 5.00 × 10-6 m3 at 20.0 °C hits a block that is made of an ideal thermal insulator and comes to rest at its center
At that time, the temperature of the bullet is 327 °C. Use the following information for lead: coefficient of linear expansion: a = 2.0 × 10-5/C° specific heat capacity: c = 128 J/(kg • C°) latent heat of fusion: Lf = 23 200 J/kg melting point: T melt = 327 °C
Now,
Heat gained by bullet = Initial KE of the bullet = 0.5*0.05*v², where v is its initial velocity = 0.05*128*(327-20) + 0.05*23200 or
0.5v² = (128*307)+23200 or
v = ?[2*{(128*307)+23200}
V=152.31m/s
density of lead at 20 degree c = 0.05/(5.00*10??) = 10000
The required volume = (5.00*10??)*[1+{(3*2*10??)*320}] = 5.009*10?? m3
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.