A reversible engine has an efficiency of $$\frac{1}{4}$$. If the temperature of the sink is reduced by 58°C, its efficiency becomes double. Calculate the temperature of the sink:

For a completely reversible (Carnot) heat engine, the efficiency $$\eta$$ is related to the source temperature $$T_1$$ and the sink (cold reservoir) temperature $$T_2$$ by the well-known Carnot relation:

$$\eta \;=\;1-\dfrac{T_2}{T_1}$$

We are first told that the efficiency is $$\dfrac14$$. Substituting this value, we have

$$\dfrac14 \;=\;1-\dfrac{T_2}{T_1}$$

Now we isolate the fraction $$\dfrac{T_2}{T_1}$$. Moving terms carefully,

$$\dfrac{T_2}{T_1}=1-\dfrac14=\dfrac34$$

Multiplying both sides by $$T_1$$ gives

$$T_2=\dfrac34\,T_1$$

Next, the problem says that the sink temperature is lowered by 58 °C, and the efficiency becomes double. Because temperature differences are identical in the Celsius and Kelvin scales, we can use the same numerical value in kelvin. Let the new sink temperature be $$T_2'$$ and the (unchanged) source temperature remain $$T_1$$. We therefore have

$$T_2'=T_2-58$$

The efficiency is now doubled, becoming $$2\times\dfrac14=\dfrac12$$. Using the Carnot formula again,

$$\dfrac12 \;=\;1-\dfrac{T_2'}{T_1}$$

Solving for $$T_2'$$,

$$\dfrac{T_2'}{T_1}=1-\dfrac12=\dfrac12\quad\Longrightarrow\quad T_2'=\dfrac12\,T_1$$

But we also have $$T_2' = T_2 - 58$$. Substituting $$T_2=\dfrac34\,T_1$$ obtained earlier,

$$\dfrac12\,T_1 \;=\;\dfrac34\,T_1 - 58$$

Bringing the terms containing $$T_1$$ to one side,

$$\dfrac34\,T_1 - \dfrac12\,T_1 = 58$$

The left-hand side simplifies since $$\dfrac34-\dfrac12=\dfrac14$$, giving

$$\dfrac14\,T_1 = 58$$

Multiplying both sides by 4, we find the source temperature:

$$T_1 = 4 \times 58 = 232\text{ K}$$

Finally, we determine the original sink temperature from $$T_2=\dfrac34\,T_1$$:

$$T_2=\dfrac34 \times 232 = \dfrac{696}{4}=174\text{ K}$$

Hence, the correct answer is Option C.