A Carno'ts heat engine working between the temperature limits 400 K and 800 K has a work input of 1000 J per cycle. The heat energy supplied by the source to the engine per cycle is

Options

A. 1500J
B. 100J
C. 2460J
D. 200J

Explanation

To solve the problem of finding the heat energy supplied by the source to a Carnot heat engine working between temperature limits of 400 K and 800 K with a work input of 1000 J per cycle, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Temperatures**:
- The higher temperature (source) \( T_1 = 800 \, \text{K} \)
- The lower temperature (sink) \( T_2 = 400 \, \text{K} \)

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:
\[ \eta = 1 - \frac{400}{800} = 1 - 0.5 = 0.5 \]

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

4. **Substitute the Known Values**:
Given that the work input \( W = 1000 \, \text{J} \):
\[ Q_1 = \frac{1000 \, \text{J}}{0.5} = 2000 \, \text{J} \]

5. **Conclusion**:
The heat energy supplied by the source to the engine per cycle is:
\[ Q_1 = 2000 \, \text{J} \]

### Final Answer:
The heat energy supplied by the source to the engine per cycle is **2000 J**.

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