A 3.00x10^-3kg lead bullet travels at a speed of 2.40x10^2m/s and hits a wooden post. if half the heat energy generated remains with the bullet, what is the increase in temperature of the embedded bullet?(ci=1.28x10^2 j/kg degrees C
Compute the kinetic energy of the bullet. Take half of that value as heat added to the bullet, Q.
Q = (1/2) M V^2
From that Q and the specific heat C and mass, compute the temperature rise, delta T.
delta T = Q /(M C) = (1/2) V^2/C
225
To calculate the increase in temperature of the embedded bullet, we can use the equation:
Q = mcΔT
Where:
Q is the heat energy transferred
m is the mass of the bullet
c is the specific heat capacity of the material (lead)
ΔT is the change in temperature
First, let's calculate the initial kinetic energy of the bullet using the formula:
KE = 1/2 * mv^2
Where:
m is the mass of the bullet
v is the velocity of the bullet
Plugging in the given values:
m = 3.00x10^-3 kg
v = 2.40x10^2 m/s
KE = 1/2 * (3.00x10^-3 kg) * (2.40x10^2 m/s)^2
KE = 1/2 * (3.00x10^-3 kg) * (57600 m^2/s^2)
KE = 1.728 J
Since half of the heat energy generated remains with the bullet, the heat energy transferred can be calculated as:
Q = 0.5 * 1.728 J
Q = 0.864 J
Now we can rearrange the equation Q = mcΔT to solve for ΔT:
ΔT = Q / (mc)
Plugging in the known values:
Q = 0.864 J
m = 3.00x10^-3 kg
c = 1.28x10^2 J/kg degrees C
ΔT = (0.864 J) / ((3.00x10^-3 kg) * (1.28x10^2 J/kg degrees C))
ΔT = 0.864 / (3.00x10^-3 * 1.28x10^2) degrees C
ΔT = 0.864 / (3.84x10^-1) degrees C
ΔT = 2.25 degrees C
Therefore, the increase in temperature of the embedded bullet is 2.25 degrees Celsius.
To find the increase in temperature of the embedded bullet, we can use the equation for heat energy:
Q = mcΔT
Where:
Q is the heat energy transferred
m is the mass of the bullet
c is the specific heat capacity of lead (given as 1.28x10^2 J/kg degrees C)
ΔT is the change in temperature
First, let's calculate the heat energy transferred (Q). We know that half the heat energy remains with the bullet, so we can modify the equation as follows:
Q = (1/2)mcΔT
Next, we need to find the mass of the bullet (m) and the change in temperature (ΔT).
Given:
Mass of the bullet (m) = 3.00x10^-3 kg
Speed of the bullet = 2.40x10^2 m/s
The kinetic energy (KE) of the bullet can be calculated using the equation:
KE = (1/2)mv^2
In this case, the kinetic energy is converted into heat energy. So, we can equate it to the heat energy transferred (Q):
KE = Q
(1/2)mv^2 = (1/2)mcΔT
Now, we can solve for the change in temperature (ΔT):
v^2 = cΔT
ΔT = v^2 / c
Plugging in the given values:
ΔT = (2.40x10^2 m/s)^2 / (1.28x10^2 J/kg degrees C)
ΔT ≈ 4.5 degrees Celsius
Therefore, the increase in temperature of the embedded bullet is approximately 4.5 degrees Celsius.