The (W/Q) fo a Carnot engine is 1/6. Now the temperature of sink is reduced by `62^@C`, then this ratio becomes twice, therefore the initial temperatureof the sink and source are respectively

Options

A. `33^@C, 67^@C`
B. `37^@C, 99^@C`
C. `67^@C, 33^@C`
D. `97K, 37K`

Answer

To solve the problem, we need to find the initial temperatures of the sink (T2) and the source (T1) of a Carnot engine given that the ratio of work done (W) to heat absorbed (Q) is initially 1/6. When the temperature of the sink is reduced by 62°C, this ratio becomes 2/6 or 1/3.

### Step-by-Step Solution:

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

2. **Setting Up the Initial Condition**:
Given that the initial efficiency is \( \frac{W}{Q} = \frac{1}{6} \), we can write:
\[ \frac{1}{6} = 1 - \frac{T_2}{T_1} \]
Rearranging gives:
\[ \frac{T_2}{T_1} = 1 - \frac{1}{6} = \frac{5}{6} \]
Thus, we have:
\[ T_2 = \frac{5}{6} T_1 \quad \text{(Equation 1)} \]

3. **Considering the New Condition**:
When the temperature of the sink is reduced by 62°C, the new efficiency becomes:
\[ \frac{W}{Q} = \frac{1}{3} \]
This leads to:
\[ \frac{1}{3} = 1 - \frac{T_2 - 62}{T_1} \]
Rearranging gives:
\[ \frac{T_2 - 62}{T_1} = 1 - \frac{1}{3} = \frac{2}{3} \]
Thus, we have:
\[ T_2 - 62 = \frac{2}{3} T_1 \quad \text{(Equation 2)} \]

4. **Substituting Equation 1 into Equation 2**:
Substitute \( T_2 = \frac{5}{6} T_1 \) into Equation 2:
\[ \frac{5}{6} T_1 - 62 = \frac{2}{3} T_1 \]
To eliminate the fractions, multiply through by 6:
\[ 5T_1 - 372 = 4T_1 \]
Rearranging gives:
\[ 5T_1 - 4T_1 = 372 \implies T_1 = 372 \text{ K} \]

5. **Finding T2**:
Now, substitute \( T_1 \) back into Equation 1 to find \( T_2 \):
\[ T_2 = \frac{5}{6} \times 372 = 310 \text{ K} \]

6. **Converting to Celsius**:
Convert both temperatures from Kelvin to Celsius:
\[ T_1 = 372 - 273 = 99 \text{ °C} \]
\[ T_2 = 310 - 273 = 37 \text{ °C} \]

### Final Answer:
The initial temperatures of the sink and source are:
- Temperature of the source (T1): **99 °C**
- Temperature of the sink (T2): **37 °C**