A reversible engine converts one-sixth of the heat input into work. When the temperature of the sink is reduced by ` 62^ circ C`, the efficiency of the engine is doubled the temperature of the source is

Options

A. 372 K
B. 310 K
C. 400 K
D. 645 K

Answer

To solve the problem step by step, we need to use the concepts of thermodynamics, specifically the efficiency of a reversible engine.

### Step 1: Define the efficiency of the engine
The efficiency (η) of a reversible engine is given by the formula:
\[ \eta = 1 - \frac{T_2}{T_1} \]
where \( T_1 \) is the temperature of the heat source and \( T_2 \) is the temperature of the heat sink.

### Step 2: Set up the initial efficiency equation
Given that the engine converts one-sixth of the heat input into work, we can express this as:
\[ \eta = \frac{1}{6} \]
Substituting this into the efficiency formula, we have:
\[ 1 - \frac{T_2}{T_1} = \frac{1}{6} \]
Rearranging gives:
\[ \frac{T_2}{T_1} = 1 - \frac{1}{6} = \frac{5}{6} \]
Thus, we can write:
\[ T_2 = \frac{5}{6} T_1 \quad \text{(Equation 1)} \]

### Step 3: Consider the change in sink temperature
When the temperature of the sink is reduced by \( 62^\circ C \), the new sink temperature becomes:
\[ T_2' = T_2 - 62 \]
The new efficiency (η') is double the original efficiency:
\[ \eta' = 2 \times \frac{1}{6} = \frac{1}{3} \]
Using the efficiency formula again:
\[ \eta' = 1 - \frac{T_2'}{T_1} \]
Substituting for \( T_2' \):
\[ 1 - \frac{T_2 - 62}{T_1} = \frac{1}{3} \]
Rearranging gives:
\[ \frac{T_2 - 62}{T_1} = 1 - \frac{1}{3} = \frac{2}{3} \]
Thus, we can write:
\[ T_2 - 62 = \frac{2}{3} T_1 \quad \text{(Equation 2)} \]

### Step 4: Solve the equations
Now we have two equations:
1. \( T_2 = \frac{5}{6} T_1 \)
2. \( T_2 - 62 = \frac{2}{3} T_1 \)

Substituting Equation 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:
\[ 5T_1 - 4T_1 = 372 \]
Thus:
\[ T_1 = 372 \text{ K} \]

### Step 5: Convert to Celsius
To find the temperature in Celsius:
\[ T_1 = 372 \text{ K} - 273 = 99^\circ C \]

### Final Answer
The temperature of the source is \( 372 \text{ K} \) or \( 99^\circ C \).