A carnolt engine works between `27^(@)C` and `127^(@)C` . Heat supplied by the source is 500 J . Then heat ejected to the sink is :

Options

A. 1000 J
B. 667 J
C. 375 J
D. 500 J

Explanation

To solve the problem, we will use the Carnot engine efficiency formula and the relationship between heat supplied and heat ejected. Here’s the step-by-step solution:

### Step 1: Convert temperatures from Celsius to Kelvin
- The temperatures given are:
- \( T_1 = 27^\circ C \)
- \( T_2 = 127^\circ C \)

- To convert these to Kelvin:
\[ T_1 = 27 + 273 = 300 \, K \]
\[ T_2 = 127 + 273 = 400 \, K \]

### Step 2: Use the Carnot theorem
- According to Carnot's theorem, the relationship between the heat absorbed (\( Q_2 \)) and the heat rejected (\( Q_1 \)) is given by:
\[ \frac{Q_1}{Q_2} = \frac{T_1}{T_2} \]

### Step 3: Substitute known values into the equation
- We know:
- \( Q_2 = 500 \, J \) (heat supplied)
- \( T_1 = 300 \, K \)
- \( T_2 = 400 \, K \)

- Substituting these values into the equation:
\[ \frac{Q_1}{500} = \frac{300}{400} \]

### Step 4: Solve for \( Q_1 \)
- Rearranging the equation to find \( Q_1 \):
\[ Q_1 = 500 \times \frac{300}{400} \]
\[ Q_1 = 500 \times \frac{3}{4} \]
\[ Q_1 = 375 \, J \]

### Conclusion
- The heat ejected to the sink is \( Q_1 = 375 \, J \).