A Carnot engine operates with a source at 500 K and sink at 375 K. Engine consumes 600 kcal of heat per cycle. The heat rejected to sink per cycle is

Options

A. 250 kcal
B. 350 kcal
C. 450 kcal
D. 550 kcal

Explanation

To find the heat rejected to the sink per cycle by a Carnot engine, we can use the relationship derived from the Carnot theorem. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Source temperature (TH) = 500 K
- Sink temperature (TL) = 375 K
- Heat consumed by the engine per cycle (QH) = 600 kcal

### Step 2: Use the Carnot theorem
According to the Carnot theorem, the ratio of the heat rejected (QL) to the heat absorbed (QH) is equal to the ratio of the temperatures of the sink and source:
\[ \frac{QL}{QH} = \frac{TL}{TH} \]

### Step 3: Rearrange the formula to find QL
We can rearrange the formula to solve for QL:
\[ QL = QH \times \frac{TL}{TH} \]

### Step 4: Substitute the known values
Now, substitute the known values into the equation:
\[ QL = 600 \, \text{kcal} \times \frac{375 \, \text{K}}{500 \, \text{K}} \]

### Step 5: Simplify the equation
First, simplify the fraction:
\[ \frac{375}{500} = 0.75 \]
Now substitute this back into the equation:
\[ QL = 600 \, \text{kcal} \times 0.75 \]

### Step 6: Calculate QL
Now perform the multiplication:
\[ QL = 600 \times 0.75 = 450 \, \text{kcal} \]

### Conclusion
The heat rejected to the sink per cycle (QL) is **450 kcal**.