An engine has an efficiency of `1/6`. When the temperature of sink is reduced by `62^(@)C`, its efficiency is doubled. Temperature of the source is

Options

A. `124^(@)C`
B. `37^(@)C`
C. `62^(@)C`
D. `99^(@)C`

Answer

To solve the problem, we need to find the temperature of the source (T1) given the efficiency of the engine and the change in the temperature of the sink (T2). Let's go through the solution step by step.

### Step 1: Define the known values
We know that the initial efficiency (n) of the engine is given as:
\[ n = \frac{1}{6} \]

### Step 2: Use the efficiency formula
The efficiency of a heat engine can be defined as:
\[ n = 1 - \frac{T2}{T1} \]
where T1 is the temperature of the source and T2 is the temperature of the sink.

### Step 3: Rearranging the efficiency formula
From the efficiency formula, we can express T2 in terms of T1:
\[ \frac{T2}{T1} = 1 - n \]
Substituting the value of n:
\[ \frac{T2}{T1} = 1 - \frac{1}{6} = \frac{5}{6} \]
This implies:
\[ T2 = \frac{5}{6} T1 \]

### Step 4: Consider the change in efficiency
When the temperature of the sink is reduced by \( 62^{\circ}C \), the new efficiency (n') becomes:
\[ n' = 2n = 2 \times \frac{1}{6} = \frac{1}{3} \]

### Step 5: Write the new efficiency equation
Using the new efficiency:
\[ n' = 1 - \frac{T2 - 62}{T1} \]
Substituting the value of n':
\[ \frac{1}{3} = 1 - \frac{T2 - 62}{T1} \]

### Step 6: Rearranging the new efficiency equation
Rearranging gives:
\[ \frac{T2 - 62}{T1} = 1 - \frac{1}{3} = \frac{2}{3} \]
Thus:
\[ T2 - 62 = \frac{2}{3} T1 \]
So:
\[ T2 = \frac{2}{3} T1 + 62 \]

### Step 7: Substitute T2 from Step 3 into Step 6
Now we substitute \( T2 = \frac{5}{6} T1 \) into the equation:
\[ \frac{5}{6} T1 = \frac{2}{3} T1 + 62 \]

### Step 8: Solve for T1
To solve for T1, we first find a common denominator:
\[ \frac{5}{6} T1 - \frac{2}{3} T1 = 62 \]
Converting \(\frac{2}{3}\) to sixths:
\[ \frac{5}{6} T1 - \frac{4}{6} T1 = 62 \]
This simplifies to:
\[ \frac{1}{6} T1 = 62 \]
Thus:
\[ T1 = 62 \times 6 = 372 \text{ K} \]

### Step 9: Convert T1 to Celsius
To convert the temperature from Kelvin to Celsius:
\[ T1 = 372 \text{ K} - 273 = 99^{\circ}C \]

### Final Answer
The temperature of the source is:
\[ T1 = 99^{\circ}C \]

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