An ideal Carnot heat engine with an efficiency of `30%`. It absorbs heat from a hot reservoir at `727^(@)C`. The temperature of the cold reservoir is

Options

A. `509^(@)C`
B. `427^(@)C`
C. `273^(@)C`
D. `218^(@)C`

Explanation

To solve the problem, we need to find the temperature of the cold reservoir (T2) given the efficiency of a Carnot heat engine and the temperature of the hot reservoir (T1).

### Step-by-Step Solution:

1. **Identify the given values:**
- Efficiency (η) = 30% = 0.3 (as a decimal)
- Temperature of the hot reservoir (T1) = 727°C

2. **Convert the temperature from Celsius to Kelvin:**
- The formula to convert Celsius to Kelvin is:
\[ T(K) = T(°C) + 273 \]
- Therefore, for T1:
\[ T1 = 727 + 273 = 1000 \, K \]

3. **Use the formula for the efficiency of a Carnot engine:**
- The efficiency of a Carnot engine is given by:
\[ η = 1 - \frac{T2}{T1} \]
- Rearranging this formula to find T2 gives:
\[ \frac{T2}{T1} = 1 - η \]
- Thus:
\[ T2 = T1 \times (1 - η) \]

4. **Substituting the known values into the equation:**
- Substitute T1 = 1000 K and η = 0.3:
\[ T2 = 1000 \times (1 - 0.3) = 1000 \times 0.7 = 700 \, K \]

5. **Convert the temperature of the cold reservoir back to Celsius:**
- Use the conversion formula:
\[ T(°C) = T(K) - 273 \]
- Therefore, for T2:
\[ T2 = 700 - 273 = 427 \, °C \]

### Final Answer:
The temperature of the cold reservoir (T2) is **427°C**.

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