A 25-g lead bullet is shot with a speed of 240 m/s into a wooden wall. Assuming that 70% of the kinetic energy is absorbed by the bullet as heat (and 30% by the wall), what is the final temperature of the bullet? (Assume the bullet is initially at room temperature of 20°C.
Look at the first problem below, in "Related Questions"
To find the final temperature of the bullet, we need to use the principle of conservation of energy. We will calculate the heat absorbed by the bullet using the equation:
Q = mcΔT
Where:
Q = Heat absorbed (in joules)
m = Mass of the bullet (in kg)
c = Specific heat capacity of lead (in J/kg·°C)
ΔT = Change in temperature (final temperature - initial temperature) (in °C)
First, let's find the heat absorbed by the bullet.
Given:
Mass of the bullet (m) = 25 g = 0.025 kg
Initial temperature (T1) = 20 °C
Specific heat capacity of lead (c) = 130 J/kg·°C
ΔT = T2 - T1
Now, the kinetic energy of the bullet is given by:
KE = (1/2)mv^2
Where:
KE = Kinetic energy
m = Mass of the bullet
v = Velocity of the bullet
Given:
Mass of the bullet (m) = 0.025 kg
Velocity of the bullet (v) = 240 m/s
KE = (1/2) * 0.025 kg * (240 m/s)^2
KE = 0.72 kJ (kilojoules)
Since we know that the bullet absorbs 70% of the kinetic energy as heat:
Heat absorbed by the bullet = 0.7 * 0.72 kJ
Heat absorbed by the bullet = 0.504 kJ (kilojoules)
Now, we can calculate the final temperature of the bullet using the equation:
Q = mcΔT
0.504 kJ = 0.025 kg * 130 J/kg·°C * ΔT
Solving for ΔT:
ΔT = 0.504 kJ / (0.025 kg * 130 J/kg·°C)
ΔT ≈ 0.155 °C
Therefore, the final temperature of the bullet is approximately 0.155 °C.
Note: In this calculation, we assumed that no heat is lost to the surroundings, which is an idealized scenario. In real-life situations, there may be some heat loss to the environment, affecting the final temperature.