A Carnot engine efficiency is equal to `1/7`. If the temperature of the sink is reduced by 65 K , the efficiency becomes `1/4`. The temperature of the source and the sink in the first case are respectively

Options

A. 620 K , 520 K
B. 520 K , 606.67 K
C. 606.67 K, 520 K
D. 520 K, 610 K

Answer

To solve the problem, we will use the formula for the efficiency of a Carnot engine, which is given by:

\[ \eta = 1 - \frac{T_C}{T_H} \]

where \( T_H \) is the temperature of the hot reservoir (source) and \( T_C \) is the temperature of the cold reservoir (sink).

### Step 1: Set up the equations based on the given efficiencies.

1. **First Case**: The efficiency is \( \frac{1}{7} \).
\[ \frac{1}{7} = 1 - \frac{T_C}{T_H} \]
Rearranging gives:
\[ \frac{T_C}{T_H} = 1 - \frac{1}{7} = \frac{6}{7} \]
This leads to the equation:
\[ 7T_C = 6T_H \quad \text{(Equation 1)} \]

2. **Second Case**: The efficiency becomes \( \frac{1}{4} \) when the sink temperature is reduced by 65 K.
\[ \frac{1}{4} = 1 - \frac{T_C - 65}{T_H} \]
Rearranging gives:
\[ \frac{T_C - 65}{T_H} = 1 - \frac{1}{4} = \frac{3}{4} \]
This leads to the equation:
\[ 4(T_C - 65) = 3T_H \]
Simplifying gives:
\[ 4T_C - 260 = 3T_H \quad \text{(Equation 2)} \]

### Step 2: Solve the equations simultaneously.

We have two equations:
1. \( 7T_C = 6T_H \) (Equation 1)
2. \( 4T_C - 260 = 3T_H \) (Equation 2)

From Equation 1, we can express \( T_H \) in terms of \( T_C \):
\[ T_H = \frac{7}{6} T_C \]

Now, substitute \( T_H \) into Equation 2:
\[ 4T_C - 260 = 3\left(\frac{7}{6} T_C\right) \]
Multiplying through by 6 to eliminate the fraction:
\[ 24T_C - 1560 = 21T_C \]
Rearranging gives:
\[ 24T_C - 21T_C = 1560 \]
\[ 3T_C = 1560 \]
\[ T_C = 520 \, \text{K} \]

### Step 3: Find \( T_H \).

Now substitute \( T_C \) back into Equation 1 to find \( T_H \):
\[ 7T_C = 6T_H \]
\[ 7(520) = 6T_H \]
\[ 3640 = 6T_H \]
\[ T_H = \frac{3640}{6} = 606.67 \, \text{K} \]

### Final Answer:
The temperatures of the source and sink in the first case are:
- \( T_H = 606.67 \, \text{K} \)
- \( T_C = 520 \, \text{K} \)