91Ó°ÊÓ

Here we have to calculate two things

1. Heat transfer to the engine which can be find out with the help of relation $\mathbf{W}\mathbf{=}{\mathbf{Q}}_{\mathbf{1}}\mathbf{-}{\mathbf{Q}}_{\mathbf{2}}$where W is the amount of work done by the Heat engine , Q₁is the amount of heat drawn by the engine and Q₂ is the amount of heat rejected by the engine.
2. Engine efficiency can be calculated with the help of formula,$\mathbf{η}\mathbf{=}\frac{\mathbf{W}}{{\mathbf{Q}}_{\mathbf{1}}}\mathbf{.}$

Heat Engine is a system which converts heat into work by taking heat from hot body(reservoir) to carry out some work and there by discharging some of it at a very low temperature to cold body (sink). First and second law of thermodynamics helps in the operation of Heat engine.

Work done by the heat Engine $W=10.0\mathrm{kJ}$

Heat Transfer $Q_2=8.50\mathrm{kJ}$

Relation to be used $W=Q_1-Q_2$

Here

$Q_1$ is the amount of heat rejected by the heat engine

To find Efficiency (ɳ)

Relation to be used $h=\frac{W}{Q_1}$

$$\begin{gathered} h-\mathrm{Engine}\mathrm{Efficiency}. \\ W-\mathrm{Work}\mathrm{done}\mathrm{by}\mathrm{the}\mathrm{Engine}. \\ {Q}_1-\mathrm{Amount}\mathrm{of}\mathrm{Heat}\mathrm{Extracted}\mathrm{By}\mathrm{the}\mathrm{engine}. \end{gathered}$$

$$\begin{gathered} W=Q_1-Q_2 \\ {Q}_1=W+Q_2 \\ {Q}_1=10kJ+8.50\mathrm{kJ} \\ {Q}_1=18.50\mathrm{kJ} \end{gathered}$$

$$\begin{gathered} h=\frac{W}{Q_1} \\ h=\frac{10}{18.50} \\ h=0.5405 \\ h\%=54.1\% \end{gathered}$$

Therefore, Amount of heat Drawn by the heat engine is $18.50\mathrm{kJ}$ and engine efficiency is$54.1\%$.