A reversible heat engine converts one-fourth of the heat input into work. When the temperature of the sink is reduced by 52 K, its efficiency is doubled. The temperature in Kelvin of the source will be _________.

Answer

To solve the problem step by step, let's define the variables and use the information given in the question.

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

### Step 2: Write the Efficiency Formula
The efficiency (\( \eta \)) of a reversible heat engine is given by:
\[ \eta = 1 - \frac{T_2}{T_1} \]
According to the problem, the efficiency is one-fourth of the heat input, which means:
\[ \eta = \frac{1}{4} \]

### Step 3: Set Up the Equation
From the efficiency formula, we can set up the equation:
\[ 1 - \frac{T_2}{T_1} = \frac{1}{4} \]
Rearranging gives:
\[ \frac{T_2}{T_1} = 1 - \frac{1}{4} = \frac{3}{4} \]
Cross-multiplying gives:
\[ 4T_2 = 3T_1 \quad \text{(Equation 1)} \]

### Step 4: Consider the Change in Efficiency
The problem states that when the temperature of the sink is reduced by 52 K, the efficiency is doubled. Thus, the new efficiency is:
\[ \eta' = 2 \times \frac{1}{4} = \frac{1}{2} \]
Now, the new temperature of the sink is:
\[ T_2' = T_2 - 52 \]
Using the efficiency formula again:
\[ \frac{1}{2} = 1 - \frac{T_2 - 52}{T_1} \]
Rearranging gives:
\[ \frac{T_2 - 52}{T_1} = \frac{1}{2} \]
Cross-multiplying gives:
\[ 2(T_2 - 52) = T_1 \]
Expanding this gives:
\[ 2T_2 - 104 = T_1 \quad \text{(Equation 2)} \]

### Step 5: Solve the Equations
Now we have two equations:
1. \( 4T_2 = 3T_1 \)
2. \( 2T_2 - 104 = T_1 \)

Substituting Equation 2 into Equation 1:
\[ 4T_2 = 3(2T_2 - 104) \]
Expanding this gives:
\[ 4T_2 = 6T_2 - 312 \]
Rearranging gives:
\[ 312 = 6T_2 - 4T_2 \]
Thus:
\[ 2T_2 = 312 \implies T_2 = 156 \, \text{K} \]

### Step 6: Find \( T_1 \)
Now substitute \( T_2 \) back into Equation 1 to find \( T_1 \):
\[ 4(156) = 3T_1 \]
Calculating gives:
\[ 624 = 3T_1 \implies T_1 = \frac{624}{3} = 208 \, \text{K} \]

### Final Answer
The temperature in Kelvin of the source \( T_1 \) is:
\[ \boxed{208 \, \text{K}} \]