A Carnot reversible engine converts `1//6` of heat input into work. When the temperature of the sink is redused by 62 K, the efficiency of Carnot’s cycle becomes `1//3`. The sum of temperature (in kelvin) of the source and sink will be

Answer

To solve the problem step by step, we will use the efficiency formula for a Carnot engine and the information provided in the question.

### Step 1: Define Variables
Let:
- \( T_1 \) = Temperature of the source (in Kelvin)
- \( T_2 \) = Temperature of the sink (in Kelvin)

### Step 2: Write the Efficiency Equation for the First Case
The efficiency \( \eta \) of a Carnot engine is given by:
\[ \eta = \frac{T_1 - T_2}{T_1} \]
From the problem, we know that the efficiency is \( \frac{1}{6} \):
\[ \frac{1}{6} = \frac{T_1 - T_2}{T_1} \]
Cross-multiplying gives:
\[ T_1 - T_2 = \frac{1}{6} T_1 \]
Rearranging this, we get:
\[ T_1 - \frac{1}{6} T_1 = T_2 \]
\[ \frac{5}{6} T_1 = T_2 \]
Thus, we can express \( T_2 \) in terms of \( T_1 \):
\[ T_2 = \frac{5}{6} T_1 \quad \text{(Equation 1)} \]

### Step 3: Write the Efficiency Equation for the Second Case
When the sink temperature is reduced by 62 K, the new sink temperature becomes \( T_2 - 62 \). The new efficiency is \( \frac{1}{3} \):
\[ \frac{1}{3} = \frac{T_1 - (T_2 - 62)}{T_1} \]
Cross-multiplying gives:
\[ T_1 - (T_2 - 62) = \frac{1}{3} T_1 \]
Rearranging this, we find:
\[ T_1 - T_2 + 62 = \frac{1}{3} T_1 \]
\[ T_1 - \frac{1}{3} T_1 = T_2 - 62 \]
\[ \frac{2}{3} T_1 = T_2 - 62 \]
Thus, we can express \( T_2 \) in terms of \( T_1 \):
\[ T_2 = \frac{2}{3} T_1 + 62 \quad \text{(Equation 2)} \]

### Step 4: Equate the Two Expressions for \( T_2 \)
Now we have two expressions for \( T_2 \):
1. \( T_2 = \frac{5}{6} T_1 \)
2. \( T_2 = \frac{2}{3} T_1 + 62 \)

Setting them equal to each other:
\[ \frac{5}{6} T_1 = \frac{2}{3} T_1 + 62 \]

### Step 5: Solve for \( T_1 \)
To solve for \( T_1 \), we first eliminate the fractions by finding a common denominator, which is 6:
\[ 5T_1 = 4T_1 + 372 \]
Subtracting \( 4T_1 \) from both sides gives:
\[ T_1 = 372 \text{ K} \]

### Step 6: Substitute \( T_1 \) Back to Find \( T_2 \)
Using Equation 1 to find \( T_2 \):
\[ T_2 = \frac{5}{6} T_1 = \frac{5}{6} \times 372 = 310 \text{ K} \]

### Step 7: Calculate the Sum of \( T_1 \) and \( T_2 \)
Now, we can find the sum of the temperatures:
\[ T_1 + T_2 = 372 + 310 = 682 \text{ K} \]

### Final Answer
The sum of the temperature of the source and sink is:
\[ \boxed{682 \text{ K}} \]