Given data
mass of bullet m= 0.042 kg
velocity of bullet v= 456 m/s
now here energy conserved so the kinetic energy of the
bullet is (k.e) is converted in to heat energy results the change
in temperature of the bullet
so k.e= 1/2*m*v^2Â Â Â =>
1/2*0.042*456^2 =4366.656 J
                                                              Â
= 4366.656*0.239 cal =1043.630 cal
now from the relation  dQ=
m*C*dt     C is the specific heat of
alluminium 0.22 Kcal/kg C= 220 cal/kg C
now change in temperature is  dt=
dQ/m*CÂ Â Â => dt= 1043.630/(0.042*220) = 112.946
celcius
