A carnot engine has efficiency of 50% when its sink is at `227^(@)C`. What is the temperature of source

Options

A. `727^(@)C`
B. `454^(@)C`
C. `113.5^(@)C`
D. `500^(@)C`.

Answer

To solve the problem, we will use the formula for the efficiency of a Carnot engine, which is given by:

\[ \text{Efficiency} = 1 - \frac{T_2}{T_1} \]

where:
- \( T_1 \) is the temperature of the source (in Kelvin),
- \( T_2 \) is the temperature of the sink (in Kelvin).

### Step-by-Step Solution:

1. **Convert the Sink Temperature to Kelvin**:
The sink temperature \( T_2 \) is given as \( 227^\circ C \). To convert this to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273.15 \]
Therefore,
\[ T_2 = 227 + 273.15 = 500.15 \approx 500 \text{ K} \quad (\text{for simplicity, we can round to 500 K}) \]

2. **Set Up the Efficiency Equation**:
We know the efficiency of the Carnot engine is \( 50\% \) or \( 0.5 \). Plugging this into the efficiency formula gives us:
\[ 0.5 = 1 - \frac{500}{T_1} \]

3. **Rearrange the Equation**:
Rearranging the equation to isolate \( T_1 \):
\[ \frac{500}{T_1} = 1 - 0.5 \]
\[ \frac{500}{T_1} = 0.5 \]

4. **Solve for \( T_1 \)**:
Now, we can solve for \( T_1 \):
\[ T_1 = \frac{500}{0.5} = 1000 \text{ K} \]

5. **Convert \( T_1 \) to Celsius**:
To convert \( T_1 \) back to Celsius:
\[ T_1(°C) = T_1(K) - 273.15 = 1000 - 273.15 \approx 726.85 \approx 727^\circ C \]

### Final Answer:
The temperature of the source is approximately \( 727^\circ C \).

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