A Carnot engine operating between temperature `T_1 and T_2` has efficiency 1/6. When `T_2` is lowered by 62K its efficiency increase to 1/3. Then `T_1 and T_2` are, respectively:

Options

A. 372 K and 310 K
B. 372 K and 330 K
C. 330 K and 268 K
D. 310 K and 248 K

Explanation

To solve the problem, we will use the efficiency formula of a Carnot engine and set up equations based on the given conditions.

### Step 1: Write the efficiency equation for the first case
The efficiency (η) of a Carnot engine is given by:
\[ \eta = 1 - \frac{T_2}{T_1} \]
For the first case, we know the efficiency is \( \frac{1}{6} \):
\[ \frac{1}{6} = 1 - \frac{T_2}{T_1} \]
Rearranging this, we get:
\[ \frac{T_2}{T_1} = 1 - \frac{1}{6} = \frac{5}{6} \]
This gives us our first equation:
\[ T_2 = \frac{5}{6} T_1 \quad \text{(Equation 1)} \]

### Step 2: Write the efficiency equation for the second case
In the second case, \( T_2 \) is lowered by 62 K, and the new efficiency becomes \( \frac{1}{3} \):
\[ \frac{1}{3} = 1 - \frac{T_2 - 62}{T_1} \]
Rearranging this, we have:
\[ \frac{T_2 - 62}{T_1} = 1 - \frac{1}{3} = \frac{2}{3} \]
This gives us our second equation:
\[ T_2 - 62 = \frac{2}{3} T_1 \quad \text{(Equation 2)} \]

### Step 3: Substitute Equation 1 into Equation 2
Now we can substitute \( T_2 \) from Equation 1 into Equation 2:
\[ \frac{5}{6} T_1 - 62 = \frac{2}{3} T_1 \]
To eliminate the fractions, we can multiply the entire equation by 6:
\[ 5T_1 - 372 = 4T_1 \]

### Step 4: Solve for \( T_1 \)
Rearranging gives:
\[ 5T_1 - 4T_1 = 372 \]
\[ T_1 = 372 \, \text{K} \]

### Step 5: Find \( T_2 \)
Now we can find \( T_2 \) using Equation 1:
\[ T_2 = \frac{5}{6} T_1 = \frac{5}{6} \times 372 = 310 \, \text{K} \]

### Final Answer
Thus, the temperatures \( T_1 \) and \( T_2 \) are:
\[ T_1 = 372 \, \text{K}, \quad T_2 = 310 \, \text{K} \]