A carnot engine works between temperatures `327^(@)C` and `27^(@)C`. If the engine takes 1600 J of heat from the higher temperature reservoir, the work done by the engine per cycle (in Joule) is equal to __________.

Explanation

To solve the problem, we will use the concept of a Carnot engine and the formula for its efficiency. Here are the steps:

### Step-by-Step Solution:

1. **Identify the temperatures**:
- The higher temperature reservoir \( T_1 = 327^\circ C \)
- The lower temperature reservoir \( T_2 = 27^\circ C \)

2. **Convert temperatures to Kelvin**:
- To convert Celsius to Kelvin, use the formula:
\[ T(K) = T(°C) + 273 \]
- Therefore,
\[ T_1 = 327 + 273 = 600 \, K \]
- And,
\[ T_2 = 27 + 273 = 300 \, K \]

3. **Use the efficiency formula of a Carnot engine**:
- The efficiency \( \eta \) of a Carnot engine is given by:
\[ \eta = 1 - \frac{T_2}{T_1} \]
- Substituting the values:
\[ \eta = 1 - \frac{300}{600} = 1 - 0.5 = 0.5 \]

4. **Calculate the work done**:
- The work done \( W \) by the engine can be calculated using the relationship:
\[ W = \eta \times Q_1 \]
- Where \( Q_1 \) is the heat absorbed from the higher temperature reservoir, which is given as \( 1600 \, J \).
- Therefore,
\[ W = 0.5 \times 1600 \, J = 800 \, J \]

5. **Final Answer**:
- The work done by the engine per cycle is \( 800 \, J \).