What is `DeltaE` for system that does 500 cal of work on surrounding and 300 cal of heat is absorbed by the system ?

Options

A. `-200 cal`
B. `-300` cal
C. `+200` cal
D. `+300` cal

Explanation

To solve for the change in internal energy (ΔE) of the system, we will use the first law of thermodynamics, which is represented by the equation:

\[ \Delta E = q + w \]

Where:
- \( \Delta E \) = change in internal energy
- \( q \) = heat absorbed by the system
- \( w \) = work done on the system

### Step 1: Identify the values of \( q \) and \( w \)

From the question:
- The system absorbs 300 cal of heat, so \( q = +300 \, \text{cal} \).
- The system does 500 cal of work on the surroundings, so \( w = -500 \, \text{cal} \) (since work done by the system is considered negative).

### Step 2: Substitute the values into the equation

Now we can substitute the values of \( q \) and \( w \) into the equation:

\[ \Delta E = q + w \]
\[ \Delta E = 300 \, \text{cal} + (-500 \, \text{cal}) \]

### Step 3: Perform the calculation

Now, we perform the addition:

\[ \Delta E = 300 \, \text{cal} - 500 \, \text{cal} \]
\[ \Delta E = -200 \, \text{cal} \]

### Final Answer

Thus, the change in internal energy (ΔE) of the system is:

\[ \Delta E = -200 \, \text{cal} \]

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