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 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 added to the system,
- \(W\) is the work done on the system.

### Step-by-Step Solution:

1. **Identify the values of Q and W**:
- Given that 10 J of heat is given to the system, we have:
\[ Q = +10 \, \text{J} \quad (\text{since heat is added to the system}) \]
- Given that 20 J of work is done by the system, we have:
\[ W = -20 \, \text{J} \quad (\text{since work is done by the system, it is negative}) \]

2. **Substitute the values into the first law of thermodynamics**:
\[ \Delta U = Q + W \]
Substituting the values we identified:
\[ \Delta U = 10 \, \text{J} + (-20 \, \text{J}) \]

3. **Perform the calculation**:
\[ \Delta U = 10 \, \text{J} - 20 \, \text{J} = -10 \, \text{J} \]

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

### Final Answer:
The change in internal energy is \(-10 \, \text{J}\).

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