Work done by a Carnot engine operating between temperatures 127$$^\circ$$C and 27$$^\circ$$C is 2 kJ. The amount of heat transferred to the engine by the reservoir is:

We have a Carnot engine operating between a hot reservoir at $$T_H = 127°C = 400$$ K and a cold reservoir at $$T_C = 27°C = 300$$ K, producing work $$W = 2$$ kJ.

The efficiency of the Carnot engine is:

$$\eta = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{400} = 1 - 0.75 = 0.25$$

Now, since $$\eta = \frac{W}{Q_H}$$, the heat absorbed from the hot reservoir is:

$$Q_H = \frac{W}{\eta} = \frac{2}{0.25} = 8 \text{ kJ}$$

Hence, the correct answer is Option 1.