For Carnot engine `W/Q_(in)=1/4`. If sink temperature is decreased by `52^@C` then `W/Q_(in)=1/2`. Find out source temperature in `""^@C`.

Answer

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

### Step 1: Understand the efficiency of the Carnot engine
The efficiency \( \eta \) 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.

### Step 2: Set up the first equation
From the problem, we know that:
\[ \frac{W}{Q_{in}} = \frac{1}{4} \]
This means:
\[ 1 - \frac{T_2}{T_1} = \frac{1}{4} \]
Rearranging gives:
\[ \frac{T_2}{T_1} = 1 - \frac{1}{4} = \frac{3}{4} \]
Thus, we can express \( T_2 \) in terms of \( T_1 \):
\[ T_2 = \frac{3}{4} T_1 \quad \text{(Equation 1)} \]

### Step 3: Set up the second equation
When the sink temperature is decreased by \( 52^\circ C \), the new sink temperature becomes \( T_2 - 52 \). The new efficiency is given as:
\[ \frac{W}{Q_{in}} = \frac{1}{2} \]
This leads to the equation:
\[ 1 - \frac{T_2 - 52}{T_1} = \frac{1}{2} \]
Rearranging gives:
\[ \frac{T_2 - 52}{T_1} = \frac{1}{2} \]
Thus:
\[ T_2 - 52 = \frac{1}{2} T_1 \quad \text{(Equation 2)} \]

### Step 4: Substitute Equation 1 into Equation 2
Now, substitute \( T_2 \) from Equation 1 into Equation 2:
\[ \frac{3}{4} T_1 - 52 = \frac{1}{2} T_1 \]

### Step 5: Solve for \( T_1 \)
Rearranging gives:
\[ \frac{3}{4} T_1 - \frac{1}{2} T_1 = 52 \]
To combine the terms, convert \( \frac{1}{2} T_1 \) to a fraction with a denominator of 4:
\[ \frac{3}{4} T_1 - \frac{2}{4} T_1 = 52 \]
This simplifies to:
\[ \frac{1}{4} T_1 = 52 \]
Multiplying both sides by 4 gives:
\[ T_1 = 208 \quad \text{(in Kelvin)} \]

### Step 6: Convert to Celsius
To convert the temperature from Kelvin to Celsius:
\[ T_1 = 208 - 273 = -65^\circ C \]

### Final Answer
The source temperature \( T_1 \) is:
\[ T_1 = -65^\circ C \]