A cyclic engine exchanges heat with two reservoirs maintained at 100 and 300°C respectively. The maximum work (in J) that can be obtained from 1000 J of heat extracted from the hot reservoir is

Answer: Option A

Since to obtain the maximum work we have to use an reversible heat engine for an reversible heat engine operating between two reservoirs the efficiency is given by:
$$\begin{array}{rl}& \stackrel{\iota }{\eta }=\frac{W}{Q_1}=\frac{T_1-T_2}{T_1}\\ & \Rightarrow W=1000\left(\frac{300-100}{573}\right)\\ & \Rightarrow W=349J\end{array}$$
Here $$T_1$$ and $$T_2$$ are temperatures in thermodynamic scale or kelvin temperatures.