A lead ball, with an initial temperature of 25.0oC, is released from a height of 119.0 m. It does not bounce when it hits a hard surface. Assume all the energy of the fall goes into heating the lead. Find the temperature (in degrees C) of the ball after it hits. Data: clead = 128 J/kgoC.
First, we need to determine how much kinetic energy the lead ball acquires as it falls. We can use the following relationship between gravitational potential energy and kinetic energy:
PE = KE
where PE is the gravitational potential energy, KE is the kinetic energy, m is the mass of the ball, g is the gravitational acceleration constant (9.81 m/s^2), and h is the height from which the ball was dropped:
PE = m * g * h
Next, we can reconcile that the acquired kinetic energy must have been converted into heat energy to increase the temperature of the lead ball. We can calculate the heat energy required to raise the temperature of the lead ball using the formula:
Q = m * c * ΔT
where Q is the heat energy required, m is the mass of the ball, c is the specific heat capacity of the lead, and ΔT is the change in temperature.
Since all the potential energy is converted into heat energy, we have:
m * g * h = m * c * ΔT
We are given c = 128 J/kg°C, h = 119 m, and the initial temperature of 25°C. To find the final temperature, we need to solve for ΔT:
ΔT = (m * g * h) / (m * c)
Notice that the mass of the ball (m) cancels, so we are left with:
ΔT = (g * h) / c
ΔT = (9.81 m/s² * 119 m) / (128 J/kg°C)
ΔT ≈ 9.13°C
Now add the initial temperature to find the final temperature:
Final Temperature = Initial Temperature + ΔT
Final Temperature ≈ 25°C + 9.13°C
Final Temperature ≈ 34.13°C
The temperature of the lead ball after it hits the surface is approximately 34.13°C.
To find the temperature of the lead ball after it hits, we need to calculate the change in heat energy and then use the specific heat capacity formula.
First, let's find the change in potential energy of the ball as it falls. We can use the formula:
ΔPE = m * g * h
where ΔPE is the change in potential energy, m is the mass of the ball, g is the acceleration due to gravity (9.8 m/s²), and h is the height (119.0 m).
Since we know the ball is made of lead, we can use its density to find the mass. The density of lead is approximately 11,340 kg/m³.
Next, we need to calculate the mass of the ball. Using the formula:
Density = mass / volume
mass = Density * volume,
where the volume can be calculated using the formula:
Volume = (4/3) * π * r³,
where r is the radius of the lead ball.
Now let's substitute the known values:
Density = 11,340 kg/m³
Radius = ?? (unknown)
Since the radius of the ball is not provided, we cannot proceed with the calculation. Please provide the radius of the lead ball and we can continue solving the problem step-by-step.
To find the temperature of the lead ball after it hits the ground, we need to calculate the change in thermal energy of the ball.
The thermal energy change can be calculated using the formula:
ΔQ = m * c * ΔT
Where:
ΔQ is the change in thermal energy
m is the mass of the ball
c is the specific heat capacity of lead
ΔT is the change in temperature
Since the ball is not bouncing and all the energy of the fall goes into heating the ball, the change in thermal energy is equal to the gravitational potential energy of the ball at the initial height:
ΔQ = m * g * h
Where:
m is the mass of the ball
g is the acceleration due to gravity
h is the initial height
To find the mass of the ball, we can use the formula for gravitational potential energy:
m * g * h = m * c * ΔT
Simplifying further:
g * h = c * ΔT
Rearranging the equation to find ΔT:
ΔT = (g * h) / c
Now we can substitute the given values:
g = 9.8 m/s^2 (acceleration due to gravity)
h = 119.0 m (initial height)
c = 128 J/kg°C (specific heat capacity of lead)
Plugging in these values:
ΔT = (9.8 * 119.0) / 128
Calculating the value:
ΔT ≈ 9.074 °C
Finally, to find the final temperature, we add the change in temperature to the initial temperature:
Final temperature = Initial temperature + ΔT
Final temperature = 25.0 + 9.074
Final temperature ≈ 34.074 °C
Therefore, after the lead ball hits the ground, its temperature will be approximately 34.074 °C.