If the temperature of the sink of a Carnot engine having an efficiency `(1)/( 6)` is reduced by `62^(@)C`, then its efficiency is doubled. Find the temperature of the sink and source respectively.

Options

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

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the efficiency of a Carnot engine
The efficiency (η) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}} \]
where \(T_{\text{sink}}\) is the temperature of the sink and \(T_{\text{source}}\) is the temperature of the source.

### Step 2: Set up the initial equation
We are given that the initial efficiency of the Carnot engine is \(\frac{1}{6}\). Therefore, we can write:
\[ \frac{1}{6} = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}} \]
Rearranging gives:
\[ \frac{T_{\text{sink}}}{T_{\text{source}}} = 1 - \frac{1}{6} = \frac{5}{6} \]

### Step 3: Express \(T_{\text{sink}}\) in terms of \(T_{\text{source}}\)
From the equation above, we can express the temperature of the sink as:
\[ T_{\text{sink}} = \frac{5}{6} T_{\text{source}} \]

### Step 4: Set up the equation after reducing the sink temperature
When the temperature of the sink is reduced by \(62^\circ C\), the new efficiency becomes double the initial efficiency:
\[ \text{New Efficiency} = 2 \times \frac{1}{6} = \frac{1}{3} \]
Thus, we can write:
\[ \frac{1}{3} = 1 - \frac{T_{\text{sink}} - 62}{T_{\text{source}}} \]
Rearranging gives:
\[ \frac{T_{\text{sink}} - 62}{T_{\text{source}}} = 1 - \frac{1}{3} = \frac{2}{3} \]

### Step 5: Express the new sink temperature equation
From the equation above, we can express it as:
\[ T_{\text{sink}} - 62 = \frac{2}{3} T_{\text{source}} \]
Thus,
\[ T_{\text{sink}} = \frac{2}{3} T_{\text{source}} + 62 \]

### Step 6: Substitute \(T_{\text{sink}}\) from Step 3 into Step 5
Now we substitute \(T_{\text{sink}} = \frac{5}{6} T_{\text{source}}\) into the equation:
\[ \frac{5}{6} T_{\text{source}} = \frac{2}{3} T_{\text{source}} + 62 \]

### Step 7: Solve for \(T_{\text{source}}\)
To solve for \(T_{\text{source}}\), we first find a common denominator:
\[ \frac{5}{6} T_{\text{source}} - \frac{2}{3} T_{\text{source}} = 62 \]
Converting \(\frac{2}{3}\) to sixths:
\[ \frac{5}{6} T_{\text{source}} - \frac{4}{6} T_{\text{source}} = 62 \]
This simplifies to:
\[ \frac{1}{6} T_{\text{source}} = 62 \]
Thus,
\[ T_{\text{source}} = 62 \times 6 = 372 \text{ K} \]

### Step 8: Find \(T_{\text{sink}}\)
Now we can find \(T_{\text{sink}}\) using \(T_{\text{sink}} = \frac{5}{6} T_{\text{source}}\):
\[ T_{\text{sink}} = \frac{5}{6} \times 372 = 310 \text{ K} \]

### Step 9: Convert temperatures to Celsius
To convert Kelvin to Celsius:
\[ T_{\text{source}} = 372 \text{ K} - 273 = 99^\circ C \]
\[ T_{\text{sink}} = 310 \text{ K} - 273 = 37^\circ C \]

### Final Answer
The temperatures of the sink and source are:
- \(T_{\text{sink}} = 37^\circ C\)
- \(T_{\text{source}} = 99^\circ C\)