In a closed insulated container, a liquid is stirred with a paddle to increase the temperature. Which of the following is true?

Options

A. `DeltaE = W ne 0 ,q=0`
B. `DeltaE = W =0 ,qne 0`
C. `DeltaE =0, W = qne0`
D. `W=0,DeltaE=q ne 0`

Explanation

To solve the problem, we will analyze the situation step by step using the principles of thermodynamics, particularly the first law of thermodynamics.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a closed insulated container filled with a liquid. The container is insulated, which means there is no heat exchange with the surroundings.
- A paddle is used to stir the liquid, which increases its temperature.

2. **Applying the First Law of Thermodynamics**:
- The first law of thermodynamics states:
\[ \Delta E = Q + W \]
where:
- \(\Delta E\) = change in internal energy of the system
- \(Q\) = heat added to the system
- \(W\) = work done on the system

3. **Analyzing Heat Transfer (\(Q\))**:
- Since the container is insulated, there is no heat transfer into or out of the system. Therefore, we can conclude:
\[ Q = 0 \]

4. **Analyzing Work Done (\(W\))**:
- When the paddle stirs the liquid, work is done on the liquid. This stirring action increases the kinetic energy of the liquid molecules, which contributes to an increase in temperature.
- Thus, we can say that:
\[ W \neq 0 \]
(Work is done on the system by stirring the liquid.)

5. **Analyzing Change in Internal Energy (\(\Delta E\))**:
- As the liquid is stirred, its temperature increases, which means the internal energy of the liquid also increases. Therefore:
\[ \Delta E \neq 0 \]
(The internal energy of the system increases due to the work done on it.)

6. **Conclusion**:
- From the analysis, we find:
- \(Q = 0\)
- \(W \neq 0\)
- \(\Delta E \neq 0\)
- Therefore, the correct statement regarding the system is that both the change in internal energy and the work done on the system are not equal to zero.

### Final Answer:
The correct option is **A**: \(\Delta E \neq 0\) and \(W \neq 0\).