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

Explanation

To find the temperature of the source for a Carnot engine with an efficiency of 80% and a sink temperature of 127°C, we can follow these steps:

### Step-by-Step Solution:

1. **Convert Sink Temperature to Kelvin:**
The temperature of the sink is given as 127°C. To convert this to Kelvin, we use the formula:
\[ T_{\text{sink}} = 127 + 273.15 = 400.15 \text{ K} \approx 400 \text{ K} \]

2. **Use the Efficiency Formula:**
The efficiency (η) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}} \]
Given that the efficiency is 80%, we can express this as:
\[ 0.8 = 1 - \frac{400}{T_{\text{source}}} \]

3. **Rearrange the Equation:**
Rearranging the equation to isolate the term involving the source temperature:
\[ 0.8 = 1 - \frac{400}{T_{\text{source}}} \]
\[ \frac{400}{T_{\text{source}}} = 1 - 0.8 = 0.2 \]

4. **Solve for the Source Temperature:**
Now we can solve for \(T_{\text{source}}\):
\[ T_{\text{source}} = \frac{400}{0.2} = 2000 \text{ K} \]

5. **Convert Source Temperature to Celsius:**
Finally, to convert the source temperature back to degrees Celsius:
\[ T_{\text{source}} = 2000 - 273.15 \approx 1726.85 \text{ °C} \approx 1727 \text{ °C} \]

### Final Answer:
The temperature of the source is approximately **1727°C**.