Question

A lead bullet at 27°C just melts when stopped by an obstacle. Assuming that 25 % of heat is absorbed by the obstacle, find the velocity of the bullet if its initial temperature is 27^{\circ}C. [For lead, melting point = 327^{\circ}C, specific heat = 0.03 cal/g-^{\circ}C, latent heat of fusion = 6 cal/g and J = 4.2 J/cal.]

Answer 1

409.87803 m/s

Step-by-step explanation:

v = Velocity of bullet

L = Latent heat of fusion = 6 cal/g

c = Specific heat of lead = 0.03 cal/g°C

`\Delta T` = Change in temperature = (327-27)

m = Mass of bullet

`1\ J=4.2\ J/cal`

The heat will be given by the kinetic energy of the bullet

`Q=(1)/(2)mv^2`

According to the question

`Q=0.75(1)/(2)mv^2`

This heat will balance the heat going into the obstacle

`Q=mc\Delta T+mL\\\Rightarrow 0.75(1)/(2)mv^2=m(c\Delta T+L)\\\Rightarrow v^2=(2)/(0.75)* (0.03* (327-27)+6)\\\Rightarrow v^2=40\ kcal\\\Rightarrow v^2=40* 4.2* 10^3\\\Rightarrow v^2=168000\ m^2s^2\\\Rightarrow v=√(168000)\\\Rightarrow v=409.87803\ m/s`

The speed of the bullet is 409.87803 m/s