A Carnot engine operating between two reservoirs has efficiency `1/3` . When the temperature of cold reservoir raised by x, its efficiency decreases to `1/6` . The value of `x,` if the temperature of hot reservoir is `99°C,` will be

Options

A. `62 K`
B. `16.5 K`
C. ` 33 K`
D. `66 K`

Answer

To solve the problem of the Carnot engine's efficiency, we will follow these steps:

### 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 absolute temperature of the hot reservoir and \( T_2 \) is the absolute temperature of the cold reservoir.

### Step 2: Convert the temperature of the hot reservoir to Kelvin
Given that the temperature of the hot reservoir is \( 99^\circ C \), we convert this to Kelvin:
\[ T_1 = 99 + 273 = 372 \, K \]

### Step 3: Set up the equation for the initial efficiency
We know the initial efficiency is \( \frac{1}{3} \). Plugging this into the efficiency formula:
\[ \frac{1}{3} = 1 - \frac{T_2}{372} \]
Rearranging gives:
\[ \frac{T_2}{372} = 1 - \frac{1}{3} = \frac{2}{3} \]
Thus,
\[ T_2 = 372 \times \frac{2}{3} = 248 \, K \]

### Step 4: Set up the equation for the new efficiency after raising the cold reservoir temperature
When the temperature of the cold reservoir is raised by \( x \), the new temperature becomes \( T_2 + x \). The new efficiency is given as \( \frac{1}{6} \):
\[ \frac{1}{6} = 1 - \frac{T_2 + x}{T_1} \]
Rearranging gives:
\[ \frac{T_2 + x}{372} = 1 - \frac{1}{6} = \frac{5}{6} \]
Thus,
\[ T_2 + x = 372 \times \frac{5}{6} \]

### Step 5: Calculate \( T_2 + x \)
Calculating \( T_2 + x \):
\[ T_2 + x = 372 \times \frac{5}{6} = 310 \, K \]

### Step 6: Solve for \( x \)
Now we substitute \( T_2 \) into the equation:
\[ 248 + x = 310 \]
Solving for \( x \):
\[ x = 310 - 248 = 62 \, K \]

### Final Answer
The value of \( x \) is \( 62 \, K \).

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