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

Answer

To determine the heat rejected to the sink per cycle by a Carnot engine, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Temperature of the source (T1) = 500 K
- Temperature of the sink (T2) = 375 K
- Heat consumed by the engine (Q1) = 600 kcal

2. **Use the Efficiency Formula for a Carnot Engine:**
The efficiency (η) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T2}{T1} \]
Substituting the values:
\[ \eta = 1 - \frac{375}{500} \]

3. **Calculate the Efficiency:**
\[ \eta = 1 - 0.75 = 0.25 \]
This means the efficiency of the engine is 25%.

4. **Relate Efficiency to Heat Input and Heat Rejected:**
The efficiency can also be expressed in terms of heat absorbed (Q1) and heat rejected (Q2):
\[ \eta = \frac{Q1 - Q2}{Q1} \]
Rearranging this gives:
\[ Q2 = Q1 - \eta \times Q1 \]

5. **Substituting the Values:**
Since we know Q1 = 600 kcal and η = 0.25:
\[ Q2 = 600 - (0.25 \times 600) \]
\[ Q2 = 600 - 150 = 450 \text{ kcal} \]

6. **Conclusion:**
The heat rejected to the sink per cycle (Q2) is 450 kcal.

### Final Answer:
The heat rejected to the sink per cycle is **450 kcal**.

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