Calculate `DeltaU` if `2kJ` heat is released and `10kJ` work is done on the system.

Options

A. `12kJ`
B. `8kJ`
C. `-8kJ`
D. `-12kJ`

Answer

To calculate the change in internal energy (ΔU) of the system, we can use the first law of thermodynamics, which is expressed as:

\[ \Delta U = Q + W \]

where:
- \( \Delta U \) is the change in internal energy,
- \( Q \) is the heat added to the system, and
- \( W \) is the work done on the system.

### Step-by-Step Solution:

1. **Identify the heat (Q) released:**
- The problem states that 2 kJ of heat is released. Since heat is released, it is considered negative in the context of the system.
- Therefore, \( Q = -2 \, \text{kJ} \).

2. **Identify the work (W) done on the system:**
- The problem states that 10 kJ of work is done on the system. Work done on the system is considered positive.
- Therefore, \( W = +10 \, \text{kJ} \).

3. **Apply the first law of thermodynamics:**
- Substitute the values of \( Q \) and \( W \) into the equation:
\[ \Delta U = Q + W = (-2 \, \text{kJ}) + (10 \, \text{kJ}) \]

4. **Perform the calculation:**
- Calculate \( \Delta U \):
\[ \Delta U = -2 + 10 = 8 \, \text{kJ} \]

5. **Final Result:**
- The change in internal energy \( \Delta U \) is \( 8 \, \text{kJ} \).

### Summary:
The change in internal energy of the system is \( \Delta U = 8 \, \text{kJ} \).