A lead ball, with an initial temperature of 25.0oC, is released from a height of 96.0 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.
This seems easy enough, but I'm lost as to which equation to use.
N/m I figured it out!
To find the final temperature of the lead ball after it hits the ground, we can use the equation:
Q = mcΔT
Where:
Q is the heat transferred to the lead ball
m is the mass of the lead ball
c is the specific heat capacity of lead
ΔT is the change in temperature
Given that the ball falls from a height of 96.0 m and all the energy goes into heating the lead, we can assume that the potential energy lost by the ball is converted into heat energy.
The potential energy lost by the ball as it falls can be calculated using the formula:
PE = mgh
Where:
PE is the potential energy
m is the mass of the lead ball
g is the acceleration due to gravity
h is the height
Since the ball does not bounce, we can assume all the potential energy is converted into heat energy. Therefore, we have:
Q = PE
Now, we can calculate the potential energy lost by the ball:
PE = mgh
Substituting the given values: m = mass of lead ball (unknown), g = 9.8 m/s^2, and h = 96.0 m, we obtain:
PE = mg(96.0)
Next, we need to calculate the heat transferred to the lead ball. We know the heat transferred is equal to the potential energy:
Q = PE = mg(96.0)
Now, substituting the specific heat capacity of lead, c = 128 J/(kg⋅°C), we can write:
mg(96.0) = mcΔT
Since the mass of the lead ball cancels out, we can solve for ΔT:
(96.0) = cΔT
Finally, we can solve for ΔT, which will give us the change in temperature:
ΔT = (96.0) / c
Substituting the given value of c = 128 J/(kg⋅°C), we can calculate ΔT:
ΔT = (96.0) / 128
Thus, the change in temperature, ΔT, is equal to 0.75 °C.
To find the final temperature, we need to add ΔT to the initial temperature of the lead ball:
Final temperature = Initial temperature + ΔT
Substituting the given initial temperature of 25.0 °C, we can calculate the final temperature:
Final temperature = 25.0 + 0.75
Therefore, the temperature of the lead ball after it hits is 25.75 °C.