What is the change in internal energy if 10 J of heat is given to system at constant pressure and 20 J of work is done by the system?

Options

A. 10 J
B. 30 J
C. `-10 J`
D. `-30 J`

Answer

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)**:
- Given that 10 J of heat is given to the system, we consider this as positive:
\[ Q = +10 \, \text{J} \]

2. **Identify the Work Done by the System (W)**:
- The problem states that 20 J of work is done by the system. Since work done by the system is considered negative in this context, we have:
\[ W = -20 \, \text{J} \]

3. **Apply the First Law of Thermodynamics**:
- Substitute the values of \(Q\) and \(W\) into the equation:
\[ \Delta U = Q + W = 10 \, \text{J} + (-20 \, \text{J}) \]

4. **Calculate the Change in Internal Energy**:
- Simplifying the equation:
\[ \Delta U = 10 \, \text{J} - 20 \, \text{J} = -10 \, \text{J} \]

5. **Conclusion**:
- The change in internal energy of the system is:
\[ \Delta U = -10 \, \text{J} \]