A Carnot's engine working between 400 K and 800 K has a work output of 1200 J per cycle. The amount of heat energy supplied to the engine from the source in each cycle is:

The efficiency of a Carnot engine operating between a cold reservoir at temperature $$T_C = 400$$ K and a hot reservoir at $$T_H = 800$$ K is $$\eta = 1 - \frac{T_C}{T_H} = 1 - \frac{400}{800} = \frac{1}{2}$$.

Since efficiency is also defined as $$\eta = \frac{W}{Q_H}$$, where $$W = 1200$$ J is the work output and $$Q_H$$ is the heat absorbed from the source, we have $$\frac{1}{2} = \frac{1200}{Q_H}$$, giving $$Q_H = 2400$$ J.

The correct answer is option 4: 2400 J.