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.
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Ответ:
the conservation of energy allows to find the result for the change in temperature of the bullet is:
ΔT = 195ºC
Given parameters
Mass of the bullet m = 5 g = 5 10-3 kg Bullet velocity v = 300 m / s Half of the energy is transformed into internal energy in the bullet. Specific heat lead ce = 128 J / kg ºCTo find
Bullet temperature change.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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Ответ:
∆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°
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