A 0.047 kg aluminum bullet traveling at 459 m/s strikes an armor plate and comes to a stop. If all its energy is converted to heat that is absorbed by the bullet, what is the bullet's temperature change in degrees Celsius?
KE= (1/2)mv^2
= (1/2)(0.47)(456)^2
= ( 1/2)(0.47)(2007936) =
(1/2)(9772992) = 4886.496J.
Assuming that all of this energy is converted into the
temperature change of the bullet:
U=4886=m*c*dT=0.047kg*900J/(kg*oC)*dT
Solving for the change in temperature dT:
dT=U/(m*c)=4886/(0.047*900)=116oC
KE= (1/2)mv^2
= (1/2)(0.47)(456)^2
= ( 1/2)(0.47)(2007936)
= (1/2)(9772992) = 4886 J
Assuming that all of this energy is converted into the
temperature change of the bullet:
U=4886=m*c*dT=0.047kg*900J/(kg*oC)*dT
Solving for the change in temperature dT:
dT=U/(m*c)=4886/(0.047*900)=116oC
KE= (1/2)mv^2
= (1/2)(0.47)(459)^2
= ( 1/2)(0.47)(210681) =
(1/2)(99020.07) = 4951.035J
Assuming that all of this energy is converted into the
temperature change of the bullet:
U=4951=m*c*dT=0.047kg*900J/(kg*oC)*dT
Solving for the change in temperature dT:
dT=U/(m*c)=4951/(0.047*900)=117oC
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