4C-1 : | Application of the 1st Law to a Stone Falling Into Water | 5 pts | |||||

Consider a stone having a mass of 10 kg and a bucket containing 100 kg of liquid water. Initially, the stone is 10.2 m above the water and the stone and the water are at the same temperature, state1. Determine ΔU, ΔE_{kin}, ΔE_{pot}, Q and W for the following changes of state, assuming g = 9.8066 m/s^{2}. |
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a.) | From state 1 until the stone is about to enter the water, state 2. | ||||||

b.) | From state 2 until the stone has just come to rest at the bottom of the bucket, state 3. | ||||||

c.) | From state 3 until heat has been transferred to the surroundings in such an amount that the stone and water in state 4 have returned to their initial temperature, T_{4} = T_{1}. |
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Read : | Choose the combination of the stone 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 stone 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: | T_{1} |
100 | ^{} |
Find: | |||

T_{2} |
600 | ^{} |
ΔU | ??? | kJ | ||

h_{1} |
10.2 | m | ΔE_{kin} |
??? | kJ | ||

m_{w} |
100.0 | kg | ΔE_{pot} |
??? | kJ | ||

m_{s} |
10.0 | kg | Q | ??? | kJ | ||

Diagram : | |||||||

Assumptions: | |||||||

1 - | Friction between the air and the stone is negligible. | ||||||

2 - | The air and the stone are at the same temperature. | ||||||

3 - | The depth of the water is very small, compared to h_{1}. |
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Equations / Data / Solve : | |||||||

The starting point for this problem is the integral form of the1st Law : | |||||||

Eqn 1 | |||||||

Part a.) | As the stone 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 stone. If we further assume that the air and stone are at the same temperature, then no heat transfer occurs during this step. Finally, if we consider the stone 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 ΔE_{pot} from its definition. |
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Eqn 6 | |||||||

When we apply this equation to our problem, Δz = h_{1} = -10.2 m. So we can now plug values into Eqn 3. |
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g | 9.8066 | m/s2 | ΔE_{pot} |
-1000.3 | J | ||

g_{c} |
1 | kg-m/N-s^{2} |
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Now, we can use Eqn 2 to evaluate ΔE_{kin} : |
ΔE_{kin} |
1000.3 | J | ||||

Part b.) | Apply the 1st Law, Eqn 1, to a process from State 2 to State 3, again using the stone and the water as our system. | ||||||

Assume that the depth of the water is negligible so that: | |||||||

Eqn 7 | |||||||

Because the water and stone are at the same temperature, no heat transfer occurs, therefore : | |||||||

Eqn 8 | |||||||

Just as in part (a), no work crosses the boundary of the system (the stone 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 stone is converted into internal energy in both the stone and the water. | |||||||

Since the kinetic energy of the stone in state 1 is zero: | |||||||

E_{kin,1} |
0 | J | |||||

We conclude from part (a) that: | E_{kin,2} |
1000.3 | J | ||||

After the stone hits the bottom of the tank, it has zero kinetic energy: |
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E_{kin,3} |
0 | J | |||||

Therefore, for step 2-3: | ΔE_{kin} |
-1000.3 | J | ||||

Plug this value of ΔE_{kin} into Eqn 10 to get: |
ΔU | 1000.3 | J | ||||

Part c.) | Apply the 1st Law, Eqn 1, to a process from State 3 to State 4, again using the stone 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 | |||||||

ΔU | -1000.3 | J | |||||

Finally, we can plug this value for ΔU back into Eqn 14 to evaluate Q_{34}: |
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Q_{34} |
-1000.3 | J | |||||

Note that the negative value for Q_{34} means that heat is transferred from the system to the surroundings. |
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Part d.) | We can determine the values Q, W, ΔU, ΔE_{kin} and ΔE_{pot} for the process from state 1 to state 4 by adding the results from parts (a) through (c). |
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Q_{14} |
-1000.27 | J | |||||

W_{14} |
0.0 | J | ΔU | 0.0 | J | ||

ΔE_{kin} |
0.0 | J | |||||

ΔE_{kin} |
-1000.27 | J | |||||

Verify: | The assumptions made in this problem solution cannot be verified. | ||||||

Answers : | Q (J) | W (J) | ΔU (J) | ΔE_{kin} (J) |
ΔE_{pot} (J) |
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a.) 1-2 | 0 | 0 | 0 | 1000 | -1000 | ||

b.) 2-3 | 0 | 0 | 1000 | -1000 | 0 | ||

c.) 3-4 | -1000 | 0 | -1000 | 0 | 0 | ||

d.) 1-4 | -1000 | 0 | 0 | 0 | -1000.3 | ||