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

Solution (By Examveda Team)

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:
$$\eqalign{ & \mathop \eta \limits^\iota = \frac{W}{{{Q_1}}} = \frac{{{T_1} - {T_2}}}{{{T_1}}} \cr & \Rightarrow W = 1000\left( {\frac{{300 - 100}}{{573}}} \right) \cr & \Rightarrow W = 349J \cr} $$
Here $${{T_1}}$$ and $${{T_2}}$$ are temperatures in thermodynamic scale or kelvin temperatures.