A 5.30-g lead bullet traveling at 390 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 temperature of the bullet?
KE=mv²/2
½KE=mcΔT
ΔT= KE/2mc= mv²/4mc=
= v²/4c=390²/4•130=292.5℃
To calculate the increase in temperature of the bullet, we need to use the equation:
Q = mcΔT,
where Q is the heat absorbed by the bullet, m is the mass of the bullet, c is the specific heat capacity of lead, and ΔT is the change in temperature.
First, let's find the initial kinetic energy of the bullet:
KE = (1/2)mv²,
where m is the mass of the bullet and v is its velocity.
Given:
Mass of the bullet (m) = 5.30 g (convert to kg: 0.0053 kg)
Velocity of the bullet (v) = 390 m/s
Calculating the initial kinetic energy:
KE = (1/2)(0.0053 kg)(390 m/s)²
= 0.5 * 0.0053 kg * (390 m/s)²
= 5.12585 J (rounded to 5.13 J)
Since half of the kinetic energy is transformed into internal energy, the heat absorbed by the bullet can be written as:
Q = (1/2) KE
Q = (1/2) * 5.13 J
= 2.565 J (rounded to 2.57 J)
Next, we need to calculate the specific heat capacity of lead (c). The specific heat capacity of lead is approximately 128 J/kg·K.
Now we can rearrange the formula:
Q = mcΔT
To find ΔT, we rearrange the equation as follows:
ΔT = Q / mc
Substituting the values:
ΔT = (2.57 J) / (0.0053 kg * 128 J/kg·K)
= 2.57 J / 0.6784 J/K
≈ 3.79 K
Therefore, the increase in the temperature of the bullet is approximately 3.79 degrees Celsius.