A system absorbs reversibly 600 J of heat and performs 250 J of work. The increase in the internal energy of system is :-

Options

A. 850 J
B. 250 J
C. 600 J
D. 350 J

Answer

To find the increase in the internal energy of the system, we can use the first law of thermodynamics, which is given by the equation:

\[ \Delta U = Q + W \]

Where:
- \(\Delta U\) = change in internal energy
- \(Q\) = heat absorbed by the system
- \(W\) = work done by the system

### Step-by-step Solution:

1. **Identify the Heat Absorbed (Q)**:
The system absorbs 600 J of heat. Since the system absorbs heat, we take this value as positive.
\[ Q = +600 \, \text{J} \]

2. **Identify the Work Done (W)**:
The system performs 250 J of work. When the system does work, we consider this as energy leaving the system, so we take this value as negative.
\[ W = -250 \, \text{J} \]

3. **Apply the First Law of Thermodynamics**:
Substitute the values of \(Q\) and \(W\) into the first law equation.
\[ \Delta U = Q + W = 600 \, \text{J} + (-250 \, \text{J}) \]

4. **Calculate the Change in Internal Energy (\(\Delta U\))**:
Simplifying the equation gives:
\[ \Delta U = 600 \, \text{J} - 250 \, \text{J} = 350 \, \text{J} \]

5. **Conclusion**:
The increase in the internal energy of the system is:
\[ \Delta U = 350 \, \text{J} \]

### Final Answer:
The increase in the internal energy of the system is **350 J**.

---