A Carnot engine works as a refrigerator in between `250K` and `300K`. If it acquires `750` calories form heat source at low temperature, then what is the heat generated at higher temperature (in calories)?

Explanation

To solve the problem, we will use the concept of the Carnot engine and the relationship between the heat absorbed and released by the refrigerator.

### Step-by-Step Solution:

1. **Identify the temperatures**:
- The high temperature (T1) is given as 300 K.
- The low temperature (T2) is given as 250 K.

2. **Understand the Carnot efficiency**:
- The efficiency (η) of a Carnot engine can be expressed as:
\[ \eta = \frac{T2}{T1 - T2} \]
- For a refrigerator, we use the coefficient of performance (COP), which is given by:
\[ COP = \frac{Q_{L}}{Q_{H} - Q_{L}} \]
- Here, \(Q_{L}\) is the heat absorbed from the low temperature (750 calories), and \(Q_{H}\) is the heat rejected at the high temperature.

3. **Set up the equation**:
- We know that:
\[ COP = \frac{T2}{T1 - T2} \]
- Substituting the values:
\[ COP = \frac{250}{300 - 250} = \frac{250}{50} = 5 \]

4. **Relate COP to heat quantities**:
- We can express COP in terms of heat:
\[ 5 = \frac{750}{Q_{H} - 750} \]

5. **Cross-multiply to solve for \(Q_{H}\)**:
- Rearranging gives:
\[ 5(Q_{H} - 750) = 750 \]
- Expanding this:
\[ 5Q_{H} - 3750 = 750 \]
- Adding 3750 to both sides:
\[ 5Q_{H} = 750 + 3750 = 4500 \]

6. **Solve for \(Q_{H}\)**:
- Dividing both sides by 5:
\[ Q_{H} = \frac{4500}{5} = 900 \text{ calories} \]

### Final Answer:
The heat generated at the higher temperature (Q_H) is **900 calories**.

---