A Carnot engine, whose temperature of the source is `400K` receives `200 calories` of heat at this temperature and rejects `150 calaroies` of heat to the sink. What is the temperature of the sink? Also, calculate the efficiency of the engine

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Temperature of the source (T1) = 400 K
- Heat received from the source (Q1) = 200 calories
- Heat rejected to the sink (Q2) = 150 calories

### Step 2: Use the Carnot engine formula
For a Carnot engine, the relationship between the heat absorbed (Q1), the heat rejected (Q2), and the temperatures of the source (T1) and sink (T2) is given by:
\[ \frac{Q1}{Q2} = \frac{T1}{T2} \]

### Step 3: Rearrange the formula to find T2
We can rearrange the formula to solve for T2:
\[ T2 = \frac{T1 \cdot Q2}{Q1} \]

### Step 4: Substitute the known values into the equation
Substituting the known values into the equation:
\[ T2 = \frac{400 \, \text{K} \cdot 150 \, \text{cal}}{200 \, \text{cal}} \]

### Step 5: Calculate T2
Calculating T2:
\[ T2 = \frac{400 \cdot 150}{200} = \frac{60000}{200} = 300 \, \text{K} \]

### Step 6: Calculate the efficiency of the engine
The efficiency (η) of a Carnot engine is given by:
\[ \eta = 1 - \frac{T2}{T1} \]

### Step 7: Substitute the values to find η
Substituting the values we found:
\[ \eta = 1 - \frac{300 \, \text{K}}{400 \, \text{K}} \]

### Step 8: Calculate η
Calculating η:
\[ \eta = 1 - 0.75 = 0.25 \]

### Step 9: Convert efficiency to percentage
To express efficiency as a percentage:
\[ \eta = 0.25 \times 100 = 25\% \]

### Final Answer
- The temperature of the sink (T2) is **300 K**.
- The efficiency of the engine is **25%**.

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