A system absorbs `300cal` of heat , its volume doubles and temperature rises from `273` to `298 k`, the work done on the surrounding is `200 cal`. `Delta E` for the above reaction is `:`

Options

A. 100 cal
B. 500 cal
C. `-5.00` cal
D. `-100` cal

Explanation

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

\[ \Delta E = Q + W \]

where:
- \( Q \) is the heat absorbed by the system,
- \( W \) is the work done on the surroundings.

### Step-by-Step Solution:

1. **Identify the values given in the problem:**
- Heat absorbed by the system, \( Q = 300 \, \text{cal} \) (since the system absorbs heat, this value is positive).
- Work done on the surroundings, \( W = -200 \, \text{cal} \) (since work is done on the surroundings, it is negative).

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

3. **Perform the calculation:**
\[ \Delta E = 300 \, \text{cal} - 200 \, \text{cal} \]
\[ \Delta E = 100 \, \text{cal} \]

4. **Conclusion:**
The change in internal energy \( \Delta E \) for the reaction is \( 100 \, \text{cal} \).