A 5 g lead bullet traveling at 300 m/s is stopped by a large tree. If half the kinetic energy of the bullet is transformed into internal energy and remains with the bullet while the other half is transmitted to the tree, what is the increase in the temperature of the bullet? Assume the specific heat of lead is 128 J/kg°C.

the conservation of energy allows to find the result for the change in temperature of the bullet is:

            ΔT = 195ºC

Given parameters

To find

The conservation of energy is an important law of conservation of physics, stable that energy cannot be created, not destroyed only transformed.

Let's look for the mechanical energy of the bullet that is totally kinetic.

          Em = K = ½ m v²

          Em = ½ 5 10-3 (300) ²

          Em = 2.25 10² J

They indicate that half is transformed into internal energy.

          U = Em / 2

          U = 2.25 10² / 2

          U = 1.25 10² J

The energy is

           U = Q =  

          ΔT=    

Let's calculate.

          ΔT =  

          ΔT = 1.95 10²

          ΔT = 195ºc

In conclusion using the conservation of energy we can find the result for the change in temperature of the bullet is:

            ΔT = 195ºC

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∆temperature = 175781.25°

Explanation:

Kinetic energy = 1/2 * mv²

Kinectic energy= 1/2 * 5 * 300²

Kinectic energy = 2.5 * 90000

Kinectic energy= 225000 joules.

If half is transmitted to tree..

If 225000/2 = 112500 joules

Mc∆temperature = energy

Mc∆temperature = 112500

∆temperature = 112500/mc

∆temperature= (112500)/((5/1000)*128)

∆temperature= 112500/0.64

∆temperature = 175781.25°