Question

A 4.20-g lead bullet traveling at 360 m/s is stopped by a large tree. If half the kinetic energy of the bullet is transformed into internal energy and remains with the bullet while the other half is transmitted to the tree, what is the increase in temperature of the bullet?

Answer 1

Given that half the kinetic energy is transformed into internal energy, So

(1/2)*KE = Q

KE = (1/2)*m*V^2

Q = internal energy of bullet due to temperature increase = m*C*dT

C = specific heat of bullet of lead = 128 J/kg-C

So,

m*C*dT = (1/2)*[(1/2)*m*V^2]

dT = V^2/(4*C)

V = Speed of bullet = 360 m/sec

So,

dT = 360^2/(4*128)

dT = 253 C = Increase in temperature

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