A Carnot engine takes $$5000$$ kcal of heat from a reservoir at $$727°$$C and gives heat to a sink at $$127°$$C. The work done by the engine is

Heat absorbed $$Q_H = 5000$$ kcal, source temperature $$T_H = 727°$$C, sink temperature $$T_C = 127°$$C. Converting the temperatures to Kelvin gives $$T_H = 727 + 273 = 1000 \text{ K}$$ and $$T_C = 127 + 273 = 400 \text{ K}$$.

The efficiency of a Carnot engine is $$\eta = 1 - \frac{T_C}{T_H} = 1 - \frac{400}{1000} = 1 - 0.4 = 0.6$$.

The heat input in joules is $$Q_H = 5000 \text{ kcal} = 5000 \times 4200 \text{ J} = 21 \times 10^6 \text{ J}$$.

The work done by the engine is $$W = \eta \times Q_H = 0.6 \times 21 \times 10^6$$ and therefore $$= 12.6 \times 10^6 \text{ J}$$. The work done by the Carnot engine is $$12.6 \times 10^6$$ J.

The correct answer is Option D.