During a process work equivalent to 400J is done on a system, which gives out of 125J of energy. The change in internal energy is

Options

A. 525J
B. 375J
C. 275J
D. 200J

Answer

To solve the problem, we will use the first law of thermodynamics, which states:

\[ \Delta U = q + w \]

where:
- \(\Delta U\) is the change in internal energy,
- \(q\) is the heat exchanged,
- \(w\) is the work done on or by the system.

### Step-by-Step Solution:

1. **Identify the Work Done on the System**:
- The problem states that work equivalent to 400 J is done on the system. According to the convention used in thermodynamics, when work is done on the system, it is considered positive.
- Therefore, \(w = +400 \, \text{J}\).

2. **Identify the Heat Exchanged**:
- The system gives out 125 J of energy, which means it releases heat. When heat is released by the system, it is considered negative.
- Thus, \(q = -125 \, \text{J}\).

3. **Apply the First Law of Thermodynamics**:
- Substitute the values of \(q\) and \(w\) into the equation:
\[ \Delta U = q + w = -125 \, \text{J} + 400 \, \text{J} \]

4. **Calculate the Change in Internal Energy**:
- Perform the calculation:
\[ \Delta U = -125 + 400 = 275 \, \text{J} \]

5. **Final Result**:
- The change in internal energy \(\Delta U\) is \(275 \, \text{J}\).

### Summary:
The change in internal energy of the system is \(275 \, \text{J}\).

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