A system is provided 50 J of heat and work done on the system is 10 J. The change in internal energy during the process is

Options

A. 40 J
B. 60 J
C. 80 J
D. 50 J.

Explanation

To find the change in internal energy of the system, 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 added to the system,
- \(W\) is the work done on the system.

### Step-by-Step Solution:

1. **Identify the heat added to the system (Q)**:
- The problem states that the system is provided with 50 J of heat.
- According to the sign convention, heat added to the system is positive.
- Therefore, \(Q = +50 \, \text{J}\).

2. **Identify the work done on the system (W)**:
- The problem states that the work done on the system is 10 J.
- According to the sign convention, work done on the system is also positive.
- Therefore, \(W = +10 \, \text{J}\).

3. **Apply the First Law of Thermodynamics**:
- Substitute the values of \(Q\) and \(W\) into the equation:
\[ \Delta U = Q + W = 50 \, \text{J} + 10 \, \text{J} \]

4. **Calculate the change in internal energy (\(\Delta U\))**:
- Perform the addition:
\[ \Delta U = 60 \, \text{J} \]

5. **Conclusion**:
- The change in internal energy during the process is \(60 \, \text{J}\).

### Final Answer:
The change in internal energy of the system is \(60 \, \text{J}\).

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