If work done by the system is 300 joule when 100 cal. Heat is supplied to it. The change in internal energy during the process is :-

Options

A. `-200` Joul
B. 400 Joul
C. 720 Joul
D. 120 Joul

Explanation

To solve the problem, we will use the first law of thermodynamics, which states:

\[ \Delta U = Q + W \]

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

### Step-by-Step Solution:

1. **Identify the given values:**
- Work done by the system, \(W = -300 \, \text{J}\) (since work done by the system is considered negative in thermodynamics).
- Heat supplied to the system, \(Q = 100 \, \text{cal}\).

2. **Convert heat from calories to joules:**
- We know that \(1 \, \text{cal} = 4.2 \, \text{J}\).
- Therefore, \(Q = 100 \, \text{cal} \times 4.2 \, \text{J/cal} = 420 \, \text{J}\).

3. **Substitute the values into the first law of thermodynamics equation:**
\[ \Delta U = Q + W \]
\[ \Delta U = 420 \, \text{J} + (-300 \, \text{J}) \]

4. **Calculate the change in internal energy:**
\[ \Delta U = 420 \, \text{J} - 300 \, \text{J} = 120 \, \text{J} \]

5. **Final Result:**
The change in internal energy during the process is \(\Delta U = 120 \, \text{J}\).