The heat reservoir of an ideal carnot engine is at 800K and its sink is at 400K. The amount of heat taken in it in one second to produce useful mechanical work at the rate of 750J is

Options

A. 2250J
B. 1125J
C. 1500J
D. 750J

Explanation

To solve the problem of how much heat \( Q_1 \) is taken in by a Carnot engine operating between a hot reservoir at temperature \( T_1 = 800 \, K \) and a cold sink at temperature \( T_2 = 400 \, K \), while producing useful work at the rate of \( W = 750 \, J \), we can use the efficiency formula for a Carnot engine.

### Step-by-Step Solution:

1. **Understand the Efficiency of a Carnot Engine**:
The efficiency \( \eta \) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T_2}{T_1} \]
where \( T_1 \) is the temperature of the hot reservoir and \( T_2 \) is the temperature of the cold sink.

2. **Substitute the Given Values**:
Substitute \( T_1 = 800 \, K \) and \( T_2 = 400 \, K \) into the efficiency formula:
\[ \eta = 1 - \frac{400}{800} = 1 - 0.5 = 0.5 \]

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

4. **Substitute the Work Done and Efficiency**:
We know \( W = 750 \, J \) and \( \eta = 0.5 \):
\[ Q_1 = \frac{750 \, J}{0.5} = 1500 \, J \]

5. **Conclusion**:
The amount of heat taken in by the Carnot engine in one second is \( Q_1 = 1500 \, J \).

### Final Answer:
The amount of heat taken in by the Carnot engine is \( 1500 \, J \).

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