A heat engine is operating between 500 K to 300 K and it absorbs 10 kcal of heat from 500 K reservoir reversibly per cycle. Calculate the work done (in kcal) per cycle.

Answer

To solve the problem of calculating the work done by the heat engine operating between two temperature reservoirs, we can follow these steps:

### Step 1: Identify the temperatures and heat absorbed
The heat engine operates between two temperatures:
- High-temperature reservoir (T1) = 500 K
- Low-temperature reservoir (T2) = 300 K

The heat absorbed from the high-temperature reservoir (Q1) is given as:
- Q1 = 10 kcal

### Step 2: Calculate the efficiency of the heat engine
The efficiency (η) of a reversible heat engine operating between two temperatures is given by the formula:
\[ \eta = 1 - \frac{T2}{T1} \]
Substituting the values:
\[ \eta = 1 - \frac{300}{500} = 1 - 0.6 = 0.4 \]
Thus, the efficiency of the engine is 0.4 or 40%.

### Step 3: Relate work done to heat absorbed
The work done (W) by the heat engine can be calculated using the relationship between work, heat absorbed, and efficiency:
\[ \eta = \frac{W}{Q1} \]
Rearranging this gives:
\[ W = \eta \times Q1 \]
Substituting the values we have:
\[ W = 0.4 \times 10 \text{ kcal} = 4 \text{ kcal} \]

### Conclusion
The work done per cycle by the heat engine is:
\[ \boxed{4 \text{ kcal}} \]