An engine has an efficiency of `1/6`. When the temperature of sink is reduced by `62^(@)C`, its efficiency is doubled. Temperature of the source is

Options

A. `127^(@)C`
B. `37^(@)C`
C. `62^(@)C`
D. `99^(@)C`

Explanation

To solve the problem, we need to determine the temperature of the source (T1) given the efficiency of the engine and the change in the temperature of the sink (T2).

### Step-by-Step Solution:

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

2. **Initial Efficiency**:
We are given that the initial efficiency is \( \frac{1}{6} \):
\[ \frac{1}{6} = 1 - \frac{T_2}{T_1} \]
Rearranging this gives:
\[ \frac{T_2}{T_1} = 1 - \frac{1}{6} = \frac{5}{6} \]
Thus, we can express \(T_2\) in terms of \(T_1\):
\[ T_2 = \frac{5}{6} T_1 \quad \text{(Equation 1)} \]

3. **Change in Efficiency**:
When the temperature of the sink is reduced by \(62^\circ C\), the new efficiency becomes:
\[ \eta' = 2 \cdot \frac{1}{6} = \frac{1}{3} \]
The new temperature of the sink is:
\[ T_2' = T_2 - 62 \]

4. **New Efficiency Equation**:
Using the new efficiency:
\[ \frac{1}{3} = 1 - \frac{T_2 - 62}{T_1} \]
Rearranging gives:
\[ \frac{T_2 - 62}{T_1} = 1 - \frac{1}{3} = \frac{2}{3} \]
Therefore:
\[ T_2 - 62 = \frac{2}{3} T_1 \quad \text{(Equation 2)} \]

5. **Substituting Equation 1 into Equation 2**:
Substitute \(T_2\) from Equation 1 into Equation 2:
\[ \frac{5}{6} T_1 - 62 = \frac{2}{3} T_1 \]

6. **Solving for T1**:
To eliminate the fractions, we can multiply the entire equation by 6 (the least common multiple of the denominators):
\[ 5T_1 - 372 = 4T_1 \]
Rearranging gives:
\[ 5T_1 - 4T_1 = 372 \]
Thus:
\[ T_1 = 372 \text{ K} \]

7. **Converting Kelvin to Celsius**:
To convert \(T_1\) from Kelvin to Celsius:
\[ T_1 (^\circ C) = T_1 (K) - 273 = 372 - 273 = 99^\circ C \]

### Final Answer:
The temperature of the source is \(99^\circ C\).