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