91Ó°ÊÓ

Heat Engine is a device by which a system is made to undergo a cyclic process. It results conversion of heat into work. First and second law helps in the operation of heat engine.

Efficiency of heat engine $\mathrm{η}=42\%$

Heat transfer to the engine $Q_1=1.25×{10}^{14}\mathrm{J}$

(a) Relation to be used-

$$\begin{gathered} h=\frac{W}{Q_1} \\ W=Q_1-Q_2 \end{gathered}$$

Here

role="math" localid="1668683443178" $$\begin{gathered} h-\mathrm{Efficiency}\mathrm{of}\mathrm{the}\mathrm{engine} \\ {Q}_1-\mathrm{Amount}\mathrm{of}\mathrm{Heat}\mathrm{transfered}\mathrm{to}\mathrm{the}\mathrm{engine}. \\ {Q}_2-\mathrm{Amount}\mathrm{of}\mathrm{Heat}\mathrm{transfered}\mathrm{to}\mathrm{the}\mathrm{enviroment} \\ W-\mathrm{Amount}\mathrm{of}\mathrm{work}\mathrm{done}\mathrm{by}\mathrm{the}\mathrm{heat}\mathrm{engine}. \end{gathered}$$

$$\begin{gathered} h=\frac{W}{Q_1} \\ h=\frac{Q_1-Q_2}{Q_1} \\ \frac{42}{100}=1-\frac{Q_2}{Q_1} \\ .58=\frac{Q_2}{1.25×{10}^{14}} \\ {Q}_2=0.725×{10}^{14}\mathrm{J} \end{gathered}$$

(b)

Ratio $\left(R\right)$of Heat transfer to the environment to the work output

$$\begin{gathered} R=\frac{Q_2}{W} \\ =\frac{Q_2}{Q_1-Q_2} \\ =\frac{.725×{10}^{14}}{1.25×{10}^{14}-1.25×{10}^{14}} \\ =1.38\mathbf{:}1 \end{gathered}$$

(c) Amount of work done by the heat engine

$$\begin{gathered} W=Q_1-Q_2 \\ =1.25×{10}^{14}-.725×{10}^{14} \\ =.525×{10}^{14}\mathrm{J} \end{gathered}$$

Therefore (a) Amount of heat transfer from heat engine to the environment is $.725×{10}^{14}$(b) Ratio of heat transfer to the environment to work output is $1.38\mathbf{:}1$(c) Amount of work output is$.525×{10}^{14}\mathrm{J}$