Example Problem with Complete Solution

4C-1 : Application of the 1st Law to a Cannonball Falling Into Water 5 pts
A circus performer drops a cannonball with a mass of 50 kg from a platform 12 m above a drum containing 25 kg of water. Initially, the cannonball and the water are at the same temperature, state1.
Calculate ΔU, ΔEkin, ΔEpot, Q and W for each of the following changes of state and for the entire process.
a.) From state 1 until the cannonball is about to enter the water, state 2.
b.) From state 2 until the instant the cannonball comes to rest on the bottom of the drum, state 3.
c.) From state 3 until heat has been transferred to the surroundings in such an amount that the cannonball and water in the drum have returned to their initial temperature, state 4, T4 = T1.
Assume
g = 9.8066 m/s2.

Read : Choose the combination of the cannonball and the water as the system.
In step 1-2, if no friction or heat transfer exist, potential energy is converted into kinetic energy.
In step 2-3, if the water has negligible depth, kinetic energy is converted into internal energy by friction between the cannonball and the water.
In step 3-4, heat transfer from the system to the surroundings reduces the internal energy of the system back to its initial value.
Given: T1 = T4 Find: DU ??? kJ
h1 12 m DEkin ??? kJ
mw 25 kg DEpot ??? kJ
mCB 50 kg Q ??? kJ
Diagram:
Assumptions: 1 - Friction between the air and the cannonball is negligible.
2 - The air and the cannonball are at the same temperature.
3 - The depth of the water is very small, compared to h1.
4 - g 9.8066 m/s2
Equations / Data / Solve:
The starting point for this problem is the integral form of the1st Law :
Eqn 1
Step 1-2 As the cannonball falls through the air, it experiences some air friction, but we can assume that this is negligible.  Consequently, there is no change in the temperature or internal energy of the cannonball.  If we further assume that the air and cannonball are at the same temperature, then no heat transfer occurs during step 1-2.  Finally, if we consider the cannonball and the water to be our system, then no work has crosses the system boundary either.
Eqn 2 Eqn 3
Eqn 4
This allows us to simplify the 1st Law to : Eqn 5
Next, we can evaluate DEpot from its definition. Eqn 6
When we apply this equation to our problem, Dz = h1 = -12 m.  So we can now plug values into Eqn 3.
gc 1 kg-m/N-s2 DEpot -5884.0 J
Now, we can use Eqn 2 to evaluate DEkin : DEkin 5884.0 J
Step 2-3 Apply the 1st Law, Eqn 1,  to a process from State 2 to State 3, again using the cannonball and the water as our system.
Assume that the depth of the water is negligible so that: Eqn 7
Because the water and cannonball are at the same temperature, no heat transfer occurs, therefore :
Eqn 8
Just as in Step 1-2, no work crosses the boundary of the system (the cannonball and the water) :
Eqn 9
Now, use Eqns 7 - 9 to simplify the 1st Law, Eqn 1 to : Eqn 10
Eqn 10 tells us that in Step 2-3, all of the kinetic energy of the cannonball is converted into internal energy in both the cannonball and the water.
Since the kinetic energy of the cannonball in state 1 is zero: Ekin,1 0 J
We conclude from part (a) that: Ekin,2 5884.0 J
After the cannonball hits the bottom of the tank,
it has zero kinetic energy:
Ekin,3 0 J
Therefore, for Step 2-3: DEkin -5884.0 J
Plug this value of DEkin into Eqn 10 to get: DU 5884.0 J
Step 3-4 Apply the 1st Law, Eqn 1,  to a process from State 3 to State 4, again using the cannonball and the water as our system.
In Step 3-4, there is no change in either the kinetic or the potential energy of the system.  No work crosses the boundary of the system.  Therefore :
Eqn 11 Eqn 12
Eqn 13
This allows us to simplify the 1st Law to : Eqn 14
Because in Step 3-4 the system returns to its original temperature:
Eqn 15
DU -5884.0 J
Finally, we can plug this value for DU back into Eqn 14 to evaluate Q34: Q34 -5884.0 J
Note that the negative value for Q34 means that heat is transferred from the system to the surroundings.
Step 1-4 We can determine the values Q, W, DU, DEkin and DEpot for the process from state 1 to state 4 by adding the results from parts (a) through (c).
Q14 -5883.96 J DEkin 0.0 J
W14 0.0 J DEkin -5883.96 J
DU14 0.0 J
Verify: The assumptions made in this problem solution cannot be verified.
Answers :   Q (J) W (J) DU (J) DEkin (J) DEpot (J)
a.) 1-2 0 0 0 5884 -5884
b.) 2-3 0 0 5884 -5884 0
c.) 3-4 -5884 0 -5884 0 0
d.) 1-4 -5884 0 0 0 -5884