A system absorbs 300 cal of heat. The work done by the system is 200 cal. `Delta U` for the above change is

Options

A. 100 cal
B. 500 cal
C. `-500` cal
D. `-100` cal

Answer

To find the change in internal energy (ΔU) for the given system, we will use the first law of thermodynamics, which states:

\[ \Delta U = Q + W \]

Where:
- \( \Delta U \) = Change in internal energy
- \( Q \) = Heat absorbed by the system
- \( W \) = Work done by the system

### Step-by-Step Solution:

1. **Identify the heat absorbed (Q)**:
- The system absorbs 300 cal of heat.
- Since the heat is absorbed by the system, it is considered positive:
\[ Q = +300 \, \text{cal} \]

2. **Identify the work done (W)**:
- The work done by the system is 200 cal.
- According to the convention, work done by the system is taken as negative:
\[ W = -200 \, \text{cal} \]

3. **Apply the first law of thermodynamics**:
- Substitute the values of \( Q \) and \( W \) into the equation:
\[ \Delta U = Q + W \]
\[ \Delta U = 300 \, \text{cal} + (-200 \, \text{cal}) \]

4. **Calculate ΔU**:
- Perform the addition:
\[ \Delta U = 300 \, \text{cal} - 200 \, \text{cal} = 100 \, \text{cal} \]

5. **Final Result**:
- Therefore, the change in internal energy \( \Delta U \) is:
\[ \Delta U = +100 \, \text{cal} \]

### Summary:
The change in internal energy \( \Delta U \) for the system is \( +100 \, \text{cal} \).

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