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. `372K` and `310K`
B. `372` and `300K`
C. `330K` and `268K`
D. `310K` and `248K`

Explanation

To solve the problem step by step, we will use the efficiency formula for a Carnot engine and the information given in the question.

### Step 1: Write down the efficiency formula for the Carnot engine
The efficiency (η) of a Carnot engine operating between two temperatures \( T_1 \) (hot reservoir) and \( T_2 \) (cold reservoir) is given by:
\[ \eta = 1 - \frac{T_2}{T_1} \]

### Step 2: Set up the first equation using the given efficiency
From the problem, we know that the efficiency is \( \frac{1}{6} \):
\[ \frac{1}{6} = 1 - \frac{T_2}{T_1} \]
Rearranging this gives:
\[ \frac{T_2}{T_1} = 1 - \frac{1}{6} = \frac{5}{6} \]
Thus, we can express \( T_2 \) in terms of \( T_1 \):
\[ T_2 = \frac{5}{6} T_1 \quad \text{(Equation 1)} \]

### Step 3: Set up the second equation with the new efficiency
When \( T_2 \) is lowered by 62 K, the new temperature \( T_2' \) becomes:
\[ T_2' = T_2 - 62 \]
The new efficiency is given as \( \frac{1}{3} \):
\[ \frac{1}{3} = 1 - \frac{T_2 - 62}{T_1} \]
Rearranging this gives:
\[ \frac{T_2 - 62}{T_1} = 1 - \frac{1}{3} = \frac{2}{3} \]
Thus, we can express \( T_2 - 62 \) in terms of \( T_1 \):
\[ T_2 - 62 = \frac{2}{3} T_1 \quad \text{(Equation 2)} \]

### Step 4: Substitute Equation 1 into Equation 2
We substitute \( T_2 \) from Equation 1 into Equation 2:
\[ \frac{5}{6} T_1 - 62 = \frac{2}{3} T_1 \]

### Step 5: Solve for \( T_1 \)
To solve for \( T_1 \), we first eliminate the fractions by multiplying through by 6 (the least common multiple of 6 and 3):
\[ 5 T_1 - 372 = 4 T_1 \]
Now, rearranging gives:
\[ 5 T_1 - 4 T_1 = 372 \]
\[ T_1 = 372 \text{ K} \]

### Step 6: Find \( T_2 \) using \( T_1 \)
Now that we have \( T_1 \), 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:
\[ T_1 = 372 \text{ K}, \quad T_2 = 310 \text{ K} \]