The temperature of sink of Carnot engine is `27^(@)C`. Efficiency of engine is `25%`. Then temeperature of source is

Options

A. `227^(@)C`
B. `327^(@)C`
C. `127^(@)C`
D. `27^(@)C`

Explanation

To find the temperature of the source (T1) in a Carnot engine given the temperature of the sink (T2) and the efficiency (η), we can use the formula for the efficiency of a Carnot engine:

\[ \eta = 1 - \frac{T2}{T1} \]

### Step-by-Step Solution:

1. **Convert the Sink Temperature to Kelvin**:
The temperature of the sink (T2) is given as 27°C. To convert this to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]
Therefore,
\[ T2 = 27 + 273 = 300 \text{ K} \]

2. **Express the Efficiency in Fraction**:
The efficiency of the engine is given as 25%. We can express this as a fraction:
\[ \eta = \frac{25}{100} = \frac{1}{4} \]

3. **Substitute Values into the Efficiency Formula**:
Now we can substitute the values of η and T2 into the efficiency formula:
\[ \frac{1}{4} = 1 - \frac{300}{T1} \]

4. **Rearranging the Equation**:
Rearranging the equation to isolate T1:
\[ \frac{300}{T1} = 1 - \frac{1}{4} \]
Simplifying the right side:
\[ 1 - \frac{1}{4} = \frac{3}{4} \]
Thus, we have:
\[ \frac{300}{T1} = \frac{3}{4} \]

5. **Cross-Multiply to Solve for T1**:
Cross-multiplying gives:
\[ 300 \cdot 4 = 3 \cdot T1 \]
Simplifying this:
\[ 1200 = 3T1 \]

6. **Divide to Find T1**:
Now, divide both sides by 3:
\[ T1 = \frac{1200}{3} = 400 \text{ K} \]

7. **Convert T1 Back to Celsius**:
Finally, convert T1 back to degrees Celsius:
\[ T1(°C) = T1(K) - 273 = 400 - 273 = 127°C \]

### Final Answer:
The temperature of the source (T1) is **127°C**.