In a certain chemical process, a lab technician supplies 254 J of heat to a system. At the same time, 73 J of work are done on the system by its surroundings. What is the increase in the internal energy of the system?

Explanation

To find the increase in the internal energy of the system, we can 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 supplied (Q)**:
The technician supplies \(Q = 254 \, \text{J}\) of heat to the system.

2. **Identify the work done (W)**:
The work done on the system by its surroundings is \(W = +73 \, \text{J}\) (since work is done on the system, we consider it positive).

3. **Apply the first law of thermodynamics**:
According to the first law, we can express the change in internal energy as:
\[ \Delta U = Q + W \]

4. **Substitute the values into the equation**:
\[ \Delta U = 254 \, \text{J} + 73 \, \text{J} \]

5. **Calculate the increase in internal energy**:
\[ \Delta U = 254 \, \text{J} + 73 \, \text{J} = 327 \, \text{J} \]

Thus, the increase in the internal energy of the system is \(327 \, \text{J}\).