A Carnot's cycle operating betiween `T_(1)=600K and T_(2)=300 K` Producing 1.5 KJ of mechanical work per cycle. The transferred to the engine by the reservoirs is

Options

A. 2.5 KJ
B. 3 KJ
C. 3.5 KJ
D. 4 KJ

Explanation

To solve the problem, we need to determine the heat transferred to the Carnot engine from the hot reservoir, denoted as \( Q_1 \). We will use the efficiency of the Carnot cycle and the relationship between work done, heat absorbed, and heat rejected.

### Step-by-Step Solution:

1. **Identify the given values:**
- Temperature of the hot reservoir, \( T_1 = 600 \, K \)
- Temperature of the cold reservoir, \( T_2 = 300 \, K \)
- Work done per cycle, \( W = 1.5 \, kJ = 1500 \, J \)

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

3. **Relate efficiency to work done and heat absorbed:**
The efficiency can also be expressed in terms of work done and heat absorbed:
\[ \eta = \frac{W}{Q_1} \]
Rearranging this gives:
\[ Q_1 = \frac{W}{\eta} \]

4. **Substituting the values to find \( Q_1 \):**
Now substituting \( W = 1500 \, J \) and \( \eta = 0.5 \):
\[ Q_1 = \frac{1500}{0.5} = 1500 \times 2 = 3000 \, J \]

5. **Convert Joules to kilojoules:**
Since \( 1 \, kJ = 1000 \, J \):
\[ Q_1 = \frac{3000 \, J}{1000} = 3 \, kJ \]

### Final Answer:
The heat transferred to the engine by the reservoirs is \( Q_1 = 3 \, kJ \).