An engine takes heat from a reservior and converts its `1//6` part into work. By decreasing temperature of sink by `62^(@)C`, its efficiency becomes double. The temperatures of source and sink must be

Options

A. `90^(@)C, 37^(@)C`
B. `99^(@)C, 37^(@)C`
C. `372^(@)C, 37^(@)C`
D. `206^(@), 37^(@)C`

Answer

To solve the problem, we need to find the temperatures of the source (T1) and sink (T2) of a heat engine based on the given conditions. Let's break it down step by step.

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

### Step 2: Given efficiency
According to the problem, the engine converts \(\frac{1}{6}\) of the heat taken into work. Therefore, the efficiency can also be expressed as:
\[ η = \frac{W}{Q} = \frac{1}{6} \]
Thus, we can set up the equation:
\[ 1 - \frac{T_2}{T_1} = \frac{1}{6} \]

### Step 3: Rearranging the equation
Rearranging the equation gives us:
\[ \frac{T_2}{T_1} = 1 - \frac{1}{6} = \frac{5}{6} \]
From this, we can express \(T_2\) in terms of \(T_1\):
\[ T_2 = \frac{5}{6} T_1 \]

### Step 4: New efficiency after decreasing sink temperature
When the temperature of the sink is decreased by \(62^\circ C\), the new temperature of the sink becomes:
\[ T_2' = T_2 - 62 \]
The new efficiency becomes double the original efficiency:
\[ η' = 2 \times \frac{1}{6} = \frac{1}{3} \]
Using the efficiency formula again:
\[ η' = 1 - \frac{T_2'}{T_1} \]
Substituting for \(T_2'\):
\[ \frac{1}{3} = 1 - \frac{T_2 - 62}{T_1} \]

### Step 5: Substitute \(T_2\) in the new efficiency equation
Substituting \(T_2 = \frac{5}{6} T_1\) into the equation:
\[ \frac{1}{3} = 1 - \frac{\frac{5}{6} T_1 - 62}{T_1} \]
This simplifies to:
\[ \frac{1}{3} = 1 - \left(\frac{5}{6} - \frac{62}{T_1}\right) \]
\[ \frac{1}{3} = \frac{1}{6} + \frac{62}{T_1} \]

### Step 6: Solve for \(T_1\)
Rearranging gives:
\[ \frac{1}{3} - \frac{1}{6} = \frac{62}{T_1} \]
Finding a common denominator:
\[ \frac{2}{6} - \frac{1}{6} = \frac{62}{T_1} \]
\[ \frac{1}{6} = \frac{62}{T_1} \]
Cross-multiplying gives:
\[ T_1 = 62 \times 6 = 372 \text{ K} \]

### Step 7: Convert \(T_1\) to Celsius
To convert Kelvin to Celsius:
\[ T_1 = 372 - 273 = 99^\circ C \]

### Step 8: Find \(T_2\)
Using \(T_2 = \frac{5}{6} T_1\):
\[ T_2 = \frac{5}{6} \times 372 = 310 \text{ K} \]
Convert \(T_2\) to Celsius:
\[ T_2 = 310 - 273 = 37^\circ C \]

### Final Answer
The temperatures of the source and sink are:
- \(T_1 = 99^\circ C\)
- \(T_2 = 37^\circ C\)