A liquid is stirred in thermally insulated container , for about 2 hrs. Which of the following si correct ?

Options

A. `w lt 0, q = 0 , DeltaU lt 0 `
B. ` w lt 0 , q = 0 , DeltaU=0`
C. `w lt 0 , q lt 0 , Delta = 0 `
D. `w lt 0, q lt 0 , DeltaU = 0 `

Answer

To solve the question step by step, we will analyze the situation described in the problem and apply the first law of thermodynamics.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a liquid in a thermally insulated container. This means that no heat can be exchanged with the surroundings.

**Hint**: Remember that a thermally insulated container prevents heat transfer.

2. **Analyzing Heat Exchange**:
- Since the container is insulated, the heat exchange (Q) with the surroundings is zero. Therefore, we can state that \( Q = 0 \).

**Hint**: In an insulated system, heat transfer is zero.

3. **Work Done on the System**:
- The liquid is being stirred for about 2 hours. Stirring the liquid involves doing work on the system. Therefore, the work done (W) on the system is positive, i.e., \( W > 0 \).

**Hint**: Work done on the system increases the internal energy.

4. **Applying the First Law of Thermodynamics**:
- The first law of thermodynamics states:
\[ \Delta U = Q + W \]
- Substituting the values we have:
\[ \Delta U = 0 + W \]
- Since \( W > 0 \), it follows that \( \Delta U > 0 \).

**Hint**: The first law of thermodynamics relates internal energy change to heat and work.

5. **Conclusion**:
- Since the change in internal energy (\( \Delta U \)) is greater than zero, we can conclude that the internal energy of the liquid in the container has increased due to the work done on it.

**Hint**: An increase in internal energy indicates that the system has absorbed energy in the form of work.

### Final Answer:
The correct statement is that the change in internal energy (\( \Delta U \)) of the liquid is greater than zero.