Question

a). A spherical brass shell has an interior volume of 1.65 × 10-3 m³. Within this interior volume is a solid steel ball that has a volume of 5.00 × 10-4 m³. The space between the steel ball and the inner surface of the brass shell is filled completely with mercury. A small hole is drilled through the brass, and the temperature of the arrangement is increased by 13 °C. What is the volume of the mercury that spills out of the hole? THE ANSWER IS NOT 0.0000019604 m

b). A 85.0-gram lead bullet hits a steel plate, both initially at 23 °C. The bullet melts and splatters on impact. Assuming that the bullet receives 84% of its kinetic energy as heat energy, at what minimum speed must it be traveling to melt on impact? (The melting temperature of Lead is 327°C) THE ANSWER IS NOT 315.549m/s

Answer 1

a)

volume of the brass shell, V1=1.65*10^-3 m^3

volume of the steel ball, V2=5.00*10^-4 m^3

and

alpa_brass=19*10^-6 /K

alpa_steel=13*10^-6 /K

alpa_mercury=60.33*10^-6 /K

use,

volume expansion of brass, dV1=v1*(3*alpa_brass)*(dT)

dV1=1.65*10^-3*(3*19*10^-6)*(13)

dV1=1.222*10^-6 m^3

and

volume expansion of steel, dV2=v2*(3*alpa_steel)*(dT)

dV2=5*10^-4*(3*13*10^-6)*(13)

dV2=0.253*10^-6 m^3

volume expansion of mercury, dV3=v3*(3*alpa_mercury)*(dT)

dV3=(v1-v2)*(3*alpa_mercury)*(dT)

dv3=(1.65*10^-3-5*10^-4)*(3*60.33*10^-6)*(13)

dV3=2.706*10^-6 m^3

volume of mercury spilling out=dv3-(dv1-dv2)

=2.706*10^-6-(1.222*10^-6-0.253*10^-6)

=1.74*10^-6 m^3

b)

use,

m*C*dT+m*L=(84/100)*1/2*m*v^2

C*dT+L=0.84*1/2*v^2

128*(327-23)+2.45*10^4=0.84*1/2*v^2

====> v=388.56 m/sec

bullet speed, v=388.56 m/sec