A carnot engine has efficiency of 80%. If its sink is at `127^(@)C`, the find the temperature of source.

Explanation

To solve the problem, we need to find the temperature of the source (T_high) of a Carnot engine given its efficiency and the temperature of the sink (T_low).

### Step-by-Step Solution:

1. **Understand the Efficiency of a Carnot Engine**:
The efficiency (η) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T_{low}}{T_{high}} \]
where:
- \( T_{low} \) is the temperature of the sink (in Kelvin),
- \( T_{high} \) is the temperature of the source (in Kelvin).

2. **Convert Sink Temperature to Kelvin**:
The temperature of the sink is given as \( 127^\circ C \). To convert this to Kelvin, we use the formula:
\[ T_{low} = 127 + 273 = 400 \, K \]

3. **Substitute the Given Efficiency**:
The efficiency of the Carnot engine is given as 80%, which can be expressed as:
\[ \eta = 0.8 \]

4. **Set Up the Equation**:
Using the efficiency formula, we can set up the equation:
\[ 0.8 = 1 - \frac{400}{T_{high}} \]

5. **Rearrange the Equation**:
Rearranging the equation to isolate \( T_{high} \):
\[ 0.8 = 1 - \frac{400}{T_{high}} \implies \frac{400}{T_{high}} = 1 - 0.8 = 0.2 \]

6. **Solve for \( T_{high} \)**:
Now, we can solve for \( T_{high} \):
\[ T_{high} = \frac{400}{0.2} = 2000 \, K \]

7. **Convert Temperature Back to Celsius**:
Finally, we convert the temperature from Kelvin back to Celsius:
\[ T_{high} = 2000 - 273 = 1727^\circ C \]

### Final Answer:
The temperature of the source is \( 1727^\circ C \).