A Carnot engine has an efficiency of `1//6`. When the temperature of the sink is reduced by `62^(@)C`, its efficiency is doubled. The temperature of the source and the sink are, respectively.

Options

A. `62^(@)C,124^(@)C`
B. `99^(@)C,37^(@)C`
C. `124^(@)C,62^(@)C`
D. `37^(@)C,99^(@)C`

Answer

To solve the problem, we need to find the temperatures of the source (T2) and the sink (T1) for a Carnot engine with the given efficiency and conditions.

### Step-by-Step Solution:

1. **Understanding the Efficiency of the Carnot Engine**:
The efficiency (η) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T_1}{T_2} \]
where \(T_1\) is the temperature of the sink and \(T_2\) is the temperature of the source.

2. **Setting Up the Initial Efficiency**:
We know from the problem that the initial efficiency is:
\[ \eta = \frac{1}{6} \]
Therefore, we can write:
\[ 1 - \frac{T_1}{T_2} = \frac{1}{6} \]
Rearranging gives:
\[ \frac{T_1}{T_2} = 1 - \frac{1}{6} = \frac{5}{6} \]
This leads to:
\[ T_1 = \frac{5}{6} T_2 \quad \text{(Equation 1)} \]

3. **Considering the Change in Sink Temperature**:
When the sink temperature is reduced by \(62^\circ C\), the new efficiency becomes:
\[ \eta' = 2 \times \frac{1}{6} = \frac{1}{3} \]
The new efficiency can also be expressed as:
\[ \eta' = 1 - \frac{T_1 - 62}{T_2} \]
Setting this equal to \(\frac{1}{3}\):
\[ 1 - \frac{T_1 - 62}{T_2} = \frac{1}{3} \]
Rearranging gives:
\[ \frac{T_1 - 62}{T_2} = 1 - \frac{1}{3} = \frac{2}{3} \]
Thus:
\[ T_1 - 62 = \frac{2}{3} T_2 \quad \text{(Equation 2)} \]

4. **Substituting Equation 1 into Equation 2**:
Now we substitute \(T_1\) from Equation 1 into Equation 2:
\[ \frac{5}{6} T_2 - 62 = \frac{2}{3} T_2 \]
To eliminate the fractions, multiply the entire equation by 6:
\[ 5T_2 - 372 = 4T_2 \]
Rearranging gives:
\[ 5T_2 - 4T_2 = 372 \]
Thus:
\[ T_2 = 372 \text{ K} \]

5. **Finding T1**:
Now we can find \(T_1\) using Equation 1:
\[ T_1 = \frac{5}{6} T_2 = \frac{5}{6} \times 372 = 310 \text{ K} \]

6. **Converting Kelvin to Celsius**:
Finally, we convert both temperatures from Kelvin to Celsius:
\[ T_1 = 310 \text{ K} - 273 = 37^\circ C \]
\[ T_2 = 372 \text{ K} - 273 = 99^\circ C \]

### Final Answer:
The temperatures of the sink and source are:
- Sink Temperature (T1) = \(37^\circ C\)
- Source Temperature (T2) = \(99^\circ C\)