Change in internal energy when `4 kJ` of work is done on the system and `1 kJ` of heat is given out of the system is

Options

A. `+ 1 kJ`
B. `- 5 kJ`
C. `+ 5 kJ`
D. `+ 3 kJ`

Explanation

To find the change in internal energy (ΔU) of a system when work is done on it and heat is given out, we can use the first law of thermodynamics, which is expressed as:

\[ \Delta U = Q + W \]

Where:
- ΔU is the change in internal energy,
- Q is the heat exchanged,
- W is the work done on the system.

### Step-by-Step Solution:

1. **Identify the values given in the problem:**
- Work done on the system (W) = 4 kJ (since work is done on the system, it is considered positive).
- Heat given out by the system (Q) = -1 kJ (since the system is losing heat, it is considered negative).

2. **Substitute the values into the first law of thermodynamics equation:**
\[ \Delta U = Q + W \]
Substituting the known values:
\[ \Delta U = (-1 \, \text{kJ}) + (4 \, \text{kJ}) \]

3. **Perform the calculation:**
\[ \Delta U = -1 + 4 = 3 \, \text{kJ} \]

4. **Conclusion:**
The change in internal energy (ΔU) is +3 kJ.

### Final Answer:
The change in internal energy is **+3 kJ**.

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