A system absorbs 'xJ' heat and does "yJ" work. Its `Delta E` is +Ve when

Options

A. y gt x
B. x gt y
C. y = 2x
D. x = y

Answer

To determine when the change in energy (ΔE) of a system is positive, we can use the first law of thermodynamics, which states:

\[ \Delta E = q + w \]

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

### Step-by-Step Solution:

1. **Identify the values**:
- The system absorbs \( x \) joules of heat, so \( q = +x \).
- The system does \( y \) joules of work. Since work done by the system is considered negative in this context, we have \( w = -y \).

2. **Substitute into the first law equation**:
\[ \Delta E = q + w = x - y \]

3. **Set the condition for positive ΔE**:
For ΔE to be positive:
\[ x - y > 0 \]

4. **Rearranging the inequality**:
\[ x > y \]

Thus, the change in energy (ΔE) is positive when the heat absorbed by the system (\( x \)) is greater than the work done by the system (\( y \)).

### Conclusion:
The correct condition for ΔE to be positive is \( x > y \).