A 3.00x10^-3 kg lead bullet enters a target with a speed of 2.00x10^2 m/s and comes to rest within the target. Calculate the rise in temperature of the bullet assuming that 80.0% of the heat produced is absorbed by the bullet. The Specific heat of lead is 1.30x10^2 J/kg K

Answer 1

KE=(1/2)mv^2
KE=(1/2)(.003)(200^2)
KE=60J*(1cal/4.184J)
KE=14.34cal

Q=(%)(KE)
Q=(.80)(14.34)
Q=11.47 cal

Q=mcT
T=Q/mc
T=(11.47)/(.003)(1.3x10^2)
T=29.41 degree celsius