A 3.00 gram lead bullet travels at a speed of 240 m/s and hits a wooden post. If half the heat energy generated remain with the bullet, what is the increase in temperature of the embedded bullet? (clead = 128 J/kg C)
KE= 1/2 mv^2 = 1/2 (3x10^-3)(240)^2 = 86.4J
Q=mct
t= Q/mc = 86.4J/ (3x10^-3)(128J) = 225celsius
Divide by half, 112.5, round it off to 113 degrees celsius.
Well, it sounds like that bullet really made an impact! Now, let's calculate how much its temperature increased.
To do this, we need to use the formula:
ΔQ = mcΔT
Where:
ΔQ is the heat energy generated
m is the mass of the bullet
c is the specific heat capacity of lead
ΔT is the change in temperature we're trying to find
Now, we're given that the mass of the bullet is 3.00 grams, which is equivalent to 0.003 kg, and the specific heat capacity of lead is 128 J/kg C.
We also know that half of the heat energy generated remains with the bullet. So, we can set up the equation like this:
(1/2)ΔQ = mcΔT
Now, let's calculate the heat energy generated first:
ΔQ = (1/2)mv²
ΔQ = (1/2) * 0.003 kg * (240 m/s)²
Using a calculator, this gives us ΔQ = 8.64 J.
Now, let's substitute this value back into the previous equation:
(1/2)(8.64 J) = (0.003 kg)(128 J/kg C)ΔT
Simplifying this, we find:
4.32 J = 0.384 ΔT
Now, let's solve for ΔT:
ΔT = 4.32 J / 0.384
ΔT ≈ 11.25°C
So, the increase in temperature of the embedded bullet is approximately 11.25 degrees Celsius. Making it quite a hot shot!
To find the increase in temperature of the embedded bullet, we can use the equation:
Q = mcΔT
Where:
Q is the heat energy absorbed by the bullet
m is the mass of the bullet
c is the specific heat capacity of lead
ΔT is the change in temperature
First, let's find the heat energy absorbed by the bullet. Since half of the heat energy generated remains with the bullet, we need to take this into account.
Q = (1/2) * (mass of the bullet) * (heat energy generated)
The heat energy generated can be calculated using the equation:
heat energy generated = (mass of the bullet) * (bullet's initial velocity)^2
Given:
Mass of the bullet (m) = 3.00 grams = 0.003 kg
Bullet's initial velocity = 240 m/s
Using these values, we can calculate the heat energy generated:
heat energy generated = (0.003 kg) * (240 m/s)^2
Now let's substitute this value into the equation for Q:
Q = (1/2) * (0.003 kg * (240 m/s)^2)
Next, we can calculate the change in temperature using the formula:
ΔT = Q / (m * c)
The specific heat capacity of lead (c) is given as 128 J/kg°C.
Substituting the known values into the equation:
ΔT = [(1/2) * (0.003 kg * (240 m/s)^2)] / (0.003 kg * 128 J/kg°C)
Now, let's simplify and solve the equation:
ΔT = [(1/2) * 0.003 * 240^2] / (0.003 * 128)
= (0.5 * 0.003 * 240^2) / (0.003 * 128)
= (0.5 * 240^2) / 128
Calculating the final answer:
ΔT = (0.5 * 57600) / 128
= 28800 / 128
= 225°C
Therefore, the increase in temperature of the embedded bullet is 225°C.
To calculate the increase in temperature of the embedded bullet, we can use the formula:
Q = m * c * ΔT
Where:
Q is the heat energy absorbed by the bullet (since half the heat energy generated remains with the bullet)
m is the mass of the bullet
c is the specific heat capacity of lead
ΔT is the change in temperature
First, we need to calculate the heat energy absorbed by the bullet. We can use the formula:
Q = 0.5 * KE
Where:
Q is the heat energy absorbed by the bullet
KE is the kinetic energy of the bullet
The kinetic energy of the bullet can be calculated using the formula:
KE = 0.5 * m * v^2
Where:
m is the mass of the bullet
v is the velocity of the bullet
Now let's calculate step by step:
1. Calculate the kinetic energy (KE) of the bullet:
KE = 0.5 * m * v^2
= 0.5 * 0.003 kg * (240 m/s)^2
2. Calculate the heat energy (Q) absorbed by the bullet:
Q = 0.5 * KE
3. Calculate the change in temperature (ΔT):
Q = m * c * ΔT
ΔT = Q / (m * c)
4. Substitute the values into the formula to find the change in temperature:
ΔT = (0.5 * KE) / (m * c)
Now you can plug in the values and calculate the increase in temperature of the embedded bullet.