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

Explanation

To solve the problem, we need to use the efficiency formula for a Carnot engine, which is given by:

\[ \eta = 1 - \frac{T_2}{T_1} \]

where:
- \(\eta\) is the efficiency of the engine,
- \(T_1\) is the temperature of the hot reservoir,
- \(T_2\) is the temperature of the cold reservoir.

### Step 1: Set up the equations based on the given efficiencies

1. From the first condition, we know that the efficiency \(\eta_1 = \frac{1}{6}\). Therefore, we can write:
\[ \frac{1}{6} = 1 - \frac{T_2}{T_1} \]
Rearranging 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 2: Set up the second equation with the new efficiency

2. When \(T_2\) is lowered by 62 K, the new efficiency \(\eta_2 = \frac{1}{3}\). The new temperature \(T_2'\) is:
\[ T_2' = T_2 - 62 \]
The efficiency can be expressed as:
\[ \frac{1}{3} = 1 - \frac{T_2 - 62}{T_1} \]
Rearranging 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 3: Substitute Equation 1 into Equation 2

3. Substitute \(T_2\) from Equation 1 into Equation 2:
\[ \frac{5}{6} T_1 - 62 = \frac{2}{3} T_1 \]
To eliminate the fractions, multiply through by 6:
\[ 5T_1 - 372 = 4T_1 \]
Rearranging gives:
\[ 5T_1 - 4T_1 = 372 \]
Thus:
\[ T_1 = 372 \text{ K} \]

### Step 4: Calculate \(T_2\)

4. Now substitute \(T_1\) back into Equation 1 to find \(T_2\):
\[ T_2 = \frac{5}{6} \times 372 = 310 \text{ K} \]

### Final Answer

Thus, the values of \(T_1\) and \(T_2\) are:
\[ T_1 = 372 \text{ K}, \quad T_2 = 310 \text{ K} \]