A reversible heat engine converts one-fourth of the heat input into work. When the temperature of the sink is reduced by 52 K, its efficiency is doubled. The temperature in Kelvin of the source will be ______.

For a reversible (Carnot) heat engine, the efficiency is $$\eta = 1 - \frac{T_2}{T_1}$$, where $$T_1$$ is the source temperature and $$T_2$$ is the sink temperature.

Given that the engine converts one-fourth of heat input into work, the initial efficiency is $$\eta_1 = \frac{1}{4}$$. So $$1 - \frac{T_2}{T_1} = \frac{1}{4}$$, which gives $$\frac{T_2}{T_1} = \frac{3}{4}$$, hence $$T_2 = \frac{3}{4}T_1$$.

When the sink temperature is reduced by 52 K, the new sink temperature is $$T_2' = T_2 - 52$$ and the new efficiency is doubled: $$\eta_2 = \frac{1}{2}$$. So $$1 - \frac{T_2 - 52}{T_1} = \frac{1}{2}$$, which gives $$\frac{T_2 - 52}{T_1} = \frac{1}{2}$$, hence $$T_2 - 52 = \frac{T_1}{2}$$.

Substituting $$T_2 = \frac{3}{4}T_1$$: $$\frac{3}{4}T_1 - 52 = \frac{1}{2}T_1$$, so $$\frac{3}{4}T_1 - \frac{1}{2}T_1 = 52$$, giving $$\frac{1}{4}T_1 = 52$$.

Therefore $$T_1 = 208$$ K.