The temperature of sink of Carnot engine is `127^@C`. Efficiency of engine is `50%`. Then, temperature of source is

Options

A. `227^@C`
B. `327^@C`
C. `527^@C`
D. `427^@C`

Answer

To find the temperature of the source (TH) in a Carnot engine given the temperature of the sink (TL) and the efficiency (η), we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Temperature of the sink (TL) = 127°C
- Efficiency (η) = 50% = 0.5

2. **Convert Sink Temperature to Kelvin**:
- The temperature in Kelvin (TK) is given by the formula:
\[ TK = T°C + 273 \]
- Therefore,
\[ TL = 127 + 273 = 400 \text{ K} \]

3. **Use the Efficiency Formula**:
- The efficiency of a Carnot engine is given by:
\[ η = 1 - \frac{TL}{TH} \]
- Rearranging this formula to solve for TH gives:
\[ TH = \frac{TL}{1 - η} \]

4. **Substitute the Known Values**:
- Substitute TL = 400 K and η = 0.5 into the equation:
\[ TH = \frac{400}{1 - 0.5} = \frac{400}{0.5} = 800 \text{ K} \]

5. **Convert the Temperature of the Source Back to Celsius**:
- To convert from Kelvin back to Celsius:
\[ T°C = TK - 273 \]
- Therefore,
\[ TH = 800 - 273 = 527°C \]

6. **Final Answer**:
- The temperature of the source (TH) is **527°C**.