Answer:
409.87803 m/s
Explanation:
v = Velocity of bullet
L = Latent heat of fusion = 6 cal/g
c = Specific heat of lead = 0.03 cal/g°C
[tex]\Delta T[/tex] = Change in temperature = (327-27)
m = Mass of bullet
[tex]1\ J=4.2\ J/cal[/tex]
The heat will be given by the kinetic energy of the bullet
[tex]Q=\dfrac{1}{2}mv^2[/tex]
According to the question
[tex]Q=0.75\dfrac{1}{2}mv^2[/tex]
This heat will balance the heat going into the obstacle
[tex]Q=mc\Delta T+mL\\\Rightarrow 0.75\dfrac{1}{2}mv^2=m(c\Delta T+L)\\\Rightarrow v^2=\dfrac{2}{0.75}\times (0.03\times (327-27)+6)\\\Rightarrow v^2=40\ kcal\\\Rightarrow v^2=40\times 4.2\times 10^3\\\Rightarrow v^2=168000\ m^2s^2\\\Rightarrow v=\sqrt{168000}\\\Rightarrow v=409.87803\ m/s[/tex]
The speed of the bullet is 409.87803 m/s
