If 2500 cal. of heat is added to a system while the system does work equivalent to 3500 cal by expanding against the surrounding atmosphere, the value of `Delta U` for the system is

Options

A. 4000 cal
B. `- 4000 cal `
C. 1000 cal
D. `- 1000 cal `

Explanation

To find the change in internal energy (ΔU) of the system, we can use the first law of thermodynamics, which is expressed as:

\[ \Delta U = Q + W \]

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 heat added to the system (Q)**:
- Given that 2500 cal of heat is added to the system, we have:
\[ Q = +2500 \text{ cal} \]

2. **Identify the work done by the system (W)**:
- The system does work equivalent to 3500 cal while expanding against the atmosphere. Since work is done by the system, it is considered negative in this context:
\[ W = -3500 \text{ cal} \]

3. **Apply the first law of thermodynamics**:
- Substitute the values of \( Q \) and \( W \) into the equation:
\[ \Delta U = Q + W \]
\[ \Delta U = 2500 \text{ cal} + (-3500 \text{ cal}) \]

4. **Calculate ΔU**:
- Perform the addition:
\[ \Delta U = 2500 \text{ cal} - 3500 \text{ cal} = -1000 \text{ cal} \]

5. **Conclusion**:
- The value of ΔU for the system is:
\[ \Delta U = -1000 \text{ cal} \]

### Final Answer:
The change in internal energy (ΔU) for the system is \(-1000 \text{ cal}\).

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