A bullet of mass 5gram , travelling with a speed of 210m s ^ - 1...

As per given condition Half of kinetic energy is transferred into heat energy and raises the temperature of bullet

$\dfrac{1}{2} \times \dfrac{1}{2}m{v}^{2} = ms\DeltaT$

$\Deltat = \dfrac{{V}^{2}}{4s}$

$= \dfrac{210 \times 210}{4 \times 4 . 2 \times 0 . 3 \times 1000} = 87 . 5 ° C$