A Carnot engine has an efficiency of $$\frac{1}{6}$$. When the temperature of the sink is reduced by 62°C, its efficiency is doubled. The temperatures of the source and the sink are, respectively,

We are given that a Carnot engine has an efficiency of $$\frac{1}{6}$$, and when the temperature of the sink is reduced by $$62^\circ C$$, its efficiency is doubled.

Step 1: Set up equations using Carnot efficiency

The efficiency of a Carnot engine is given by:

$$\eta = 1 - \frac{T_2}{T_1}$$

where $$T_1$$ is the source temperature and $$T_2$$ is the sink temperature (both in Kelvin).

Step 2: First condition

$$\frac{1}{6} = 1 - \frac{T_2}{T_1}$$

$$\frac{T_2}{T_1} = 1 - \frac{1}{6} = \frac{5}{6}$$

$$T_2 = \frac{5}{6}T_1$$ $$-(1)$$

Step 3: Second condition

When the sink temperature is reduced by $$62^\circ C$$ (i.e., by 62 K), the new sink temperature is $$T_2 - 62$$ and the efficiency doubles to $$\frac{1}{3}$$:

$$\frac{1}{3} = 1 - \frac{T_2 - 62}{T_1}$$

$$\frac{T_2 - 62}{T_1} = \frac{2}{3}$$

$$T_2 - 62 = \frac{2}{3}T_1$$ $$-(2)$$

Step 4: Solve the equations

Substituting $$(1)$$ into $$(2)$$:

$$\frac{5}{6}T_1 - 62 = \frac{2}{3}T_1$$

$$\frac{5}{6}T_1 - \frac{2}{3}T_1 = 62$$

$$\frac{5T_1 - 4T_1}{6} = 62$$

$$\frac{T_1}{6} = 62$$

$$T_1 = 372 \text{ K}$$

From $$(1)$$:

$$T_2 = \frac{5}{6} \times 372 = 310 \text{ K}$$

Step 5: Convert to Celsius

$$T_1 = 372 - 273 = 99^\circ C$$ (source temperature)

$$T_2 = 310 - 273 = 37^\circ C$$ (sink temperature)

The temperatures of the source and sink are $$99^\circ C$$ and $$37^\circ C$$, respectively.

The correct answer is Option C.