A 3.00x10^-3 kg lead bullet enters a target with a speed of 2.00x10^2 m/s and comes to rest within the target. Calculate the rise in temperature of the bullet assuming that 80.0% of the heat produced is absorbed by the bullet. The Specific heat of lead is 1.30x10^2 J/kg K
KE=(1/2)mv^2
KE=(1/2)(.003)(200^2)
KE=60J*(1cal/4.184J)
KE=14.34cal
Q=(%)(KE)
Q=(.80)(14.34)
Q=11.47 cal
Q=mcT
T=Q/mc
T=(11.47)/(.003)(1.3x10^2)
T=29.41 degree celsius
To calculate the rise in temperature of the bullet, we can use the equation:
Q = mcΔT
where:
Q = heat absorbed by the bullet (in J)
m = mass of the bullet (in kg)
c = specific heat capacity of lead (in J/kg K)
ΔT = change in temperature (in K)
First, let's calculate the heat absorbed by the bullet:
Q = 80% of the heat produced
Since we know the speed of the bullet, we can use the kinetic energy equation to calculate the heat produced:
KE = (1/2)mv^2
where:
KE = kinetic energy (in J)
m = mass of the bullet (in kg)
v = velocity of the bullet (in m/s)
We can rearrange this equation to solve for the mass:
m = KE / ((1/2)v^2)
Substituting the given values:
m = (0.5 * 3.00x10^-3 kg * (2.00x10^2 m/s)^2) / ((1/2)*(2.00x10^2 m/s)^2)
m = 3.00x10^-3 kg
Now we can calculate the heat absorbed by the bullet:
Q = 80% * KE
Q = 0.80 * (0.5 * 3.00x10^-3 kg * (2.00x10^2 m/s)^2)
Q = 0.80 * 0.6 J
Q = 0.48 J
Finally, let's calculate the change in temperature:
Q = mcΔT
ΔT = Q / (mc)
Substituting the values:
ΔT = (0.48 J) / (3.00x10^-3 kg * 1.30x10^2 J/kg K)
ΔT = 0.48 J / 3.90x10^-1 J/K
ΔT ≈ 1.23 K
Therefore, the rise in temperature of the bullet is approximately 1.23 K.
To calculate the rise in temperature of the bullet, we need to use the formula for heat transfer:
Q = mcΔT
Where:
Q = heat transferred (in joules)
m = mass of the bullet (in kg)
c = specific heat capacity of lead (in J/kg K)
ΔT = change in temperature (in Kelvin)
First, let's calculate the heat transferred. The heat transferred is equal to the energy lost by the bullet:
Q = KE_lost
= (1/2)mv^2
Where:
m = mass of the bullet (in kg)
v = velocity of the bullet (in m/s)
Plugging in the values:
Q = (1/2)(3.00x10^-3 kg)(2.00x10^2 m/s)^2
= 0.60 J
Now, we can calculate the rise in temperature using the formula:
Q = mcΔT
Rearranging the formula, we get:
ΔT = Q / (mc)
Plugging in the values:
ΔT = (0.60 J) / (0.80)(3.00x10^-3 kg)(1.30x10^2 J/kg K)
= 1.25 K
Therefore, the rise in temperature of the bullet is 1.25 Kelvin.