In an isulated container water is stirred with a rod to increase the temperature. Which of the following is true?

Options

A. `Delta U = W ne 0, q = 0`
B. `Delta U = W = q ne 0`
C. `Delta U = 0, W = q ne 0`
D. `W = 0, DeltaU = q ne 0`

Answer

To solve the question regarding the behavior of water in an insulated container when stirred, we can analyze the situation using the principles of thermodynamics.

### Step-by-Step Solution:

1. **Understanding the Insulated Container**:
- An insulated container means that there is no heat exchange with the surroundings. Therefore, the heat transfer \( Q = 0 \).

2. **Stirring the Water**:
- When we stir the water, we are doing work on the system. This work done will increase the internal energy of the water.

3. **Applying the First Law of Thermodynamics**:
- The First Law of Thermodynamics states:
\[ \Delta U = Q + W \]
- Where:
- \( \Delta U \) = change in internal energy
- \( Q \) = heat added to the system
- \( W \) = work done on the system

4. **Substituting Values**:
- Since \( Q = 0 \) (no heat exchange in an insulated container), the equation simplifies to:
\[ \Delta U = W \]
- This indicates that any change in internal energy is solely due to the work done on the system.

5. **Conclusion About Internal Energy**:
- As the water is stirred, work is done on it, which increases its internal energy. Therefore, \( \Delta U > 0 \) (the internal energy increases).

6. **Final Statement**:
- Since \( \Delta U > 0 \) and \( Q = 0 \), we conclude that the work done on the water results in an increase in temperature. Thus, the statement that is true in this scenario is that the internal energy of the water increases due to the work done by stirring.

### Answer:
The correct conclusion is that the internal energy of the water increases due to the work done on it by stirring, even though there is no heat exchange with the surroundings.

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