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.
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.
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