A 3.00-g bullet (c = 0.0305 cal/g . C° = 128 J / kg . C°) moving at 180 m/s enters a bag of sand and stops. By what amount does the temperature of the bullet change if 80% its KE becomes thermal energy that is transferred to the bullet?

Answer 1

1/2 m v^2 * .8=mcDeltaTemp
DeltaTemp=.4*180^2/128

notice mass of the bullet divides out, so no change of units is needed for specific heat.