A Carnot engine whose heat sinks at $$27°$$C, has an efficiency of $$25\%$$. By how many degrees should the temperature of the source be changed to increase the efficiency by $$100\%$$ of the original efficiency?

We are given the heat sink temperature $$T_{sink} = 27°C = 300$$ K, the initial efficiency $$\eta_1 = 25\% = 0.25$$, and we wish to increase the efficiency by 100% of the original, doubling it to $$50\%$$. The Carnot efficiency is given by $$\eta = 1 - \frac{T_{sink}}{T_{source}}$$.

Substituting $$\eta_1 = 0.25$$ leads to $$0.25 = 1 - \frac{300}{T_1}$$, which gives $$\frac{300}{T_1} = 0.75$$ and hence $$T_1 = \frac{300}{0.75} = 400 \text{ K}$$.

For the doubled efficiency $$\eta_2 = 0.50$$ we have $$0.50 = 1 - \frac{300}{T_2}$$, so $$\frac{300}{T_2} = 0.50$$, giving $$T_2 = \frac{300}{0.50} = 600 \text{ K}$$.

Therefore, the required increase in source temperature is $$\Delta T = T_2 - T_1 = 600 - 400 = 200 \text{ K} = 200°C$$, meaning the source temperature must be raised by $$200°C$$.

Final Answer: Option B.