Find the efficiency of carnot engine whose source and sink are at `927^(@)C` and `27^(@)C`.

Explanation

To find the efficiency of a Carnot engine, we can use the formula:

\[ \eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}} \]

where:
- \( \eta \) is the efficiency,
- \( T_{\text{sink}} \) is the temperature of the sink in Kelvin,
- \( T_{\text{source}} \) is the temperature of the source in Kelvin.

### Step 1: Convert the temperatures from Celsius to Kelvin
The temperatures given are:
- Source temperature \( T_{\text{source}} = 927^\circ C \)
- Sink temperature \( T_{\text{sink}} = 27^\circ C \)

To convert Celsius to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]

Calculating the temperatures:
- For the source:
\[ T_{\text{source}} = 927 + 273 = 1200 \, K \]
- For the sink:
\[ T_{\text{sink}} = 27 + 273 = 300 \, K \]

### Step 2: Substitute the temperatures into the efficiency formula
Now that we have both temperatures in Kelvin, we can substitute them into the efficiency formula:
\[ \eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}} = 1 - \frac{300}{1200} \]

### Step 3: Simplify the fraction
Calculating the fraction:
\[ \frac{300}{1200} = \frac{1}{4} = 0.25 \]

### Step 4: Calculate the efficiency
Now substitute back into the efficiency equation:
\[ \eta = 1 - 0.25 = 0.75 \]

### Step 5: Convert efficiency to percentage
To express efficiency as a percentage, multiply by 100:
\[ \eta = 0.75 \times 100 = 75\% \]

### Final Answer
The efficiency of the Carnot engine is \( 75\% \).

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