Q23PE - College Physics (Urone)

Question

(a)-What is the efficiency of a cyclical heat engine in which 75.0 kJ of heat transfer occurs to the environment for every 95.0 kJ of Heat transfer into the engine (b) How much work does it produce for 100.0 kJ of heat transfer into the engine?

Step-by-Step Solution

Answer

Efficiency (ɳ) of a cyclical heat engine is $0.211$ and amount of work it produce for $100.0 \mathrm{kJ}$ of heat transfer into the engine is $21.1 \mathrm{kJ}$.

Step: 1 Introduction-

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.

Step: 2 Given parameters and formula to be used.

Heat transfer into the engine from source $\left(Q_1\right)=9.5.0 \mathrm{kJ}$

Heat transfer to the environment from engine $\left(Q_2\right)=75.0 \mathrm{kJ}$

Relation to be used      $h=\frac{Q_1-Q_2}{Q_1}$

Work done by the engine, when heat transferred to the engine is $100 \mathrm{kJ}$ is-

Relation to be used        $W=\eta \times Q_1$

Here

$$\begin{gathered} h-\mathrm{Efficinency} \mathrm{of} \mathrm{the} \mathrm{Heat} \mathrm{Engine}. \\ W-\mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{the} \mathrm{Heat} \mathrm{Engine}. \\ {Q}_1-\mathrm{Amount} \mathrm{of} \mathrm{Heat} \mathrm{Extracted} \mathrm{from} \mathrm{Heat} \mathrm{Engine}. \\ {Q}_2-\mathrm{Amount} \mathrm{of} \mathrm{Heat} \mathrm{transfered} \mathrm{to} \mathrm{Envoirnment}. \end{gathered}$$

Step: 3 Calculations

(a)

$$\begin{gathered} h=\frac{Q_1-Q_2}{Q_1} \\ h=\frac{95-75}{95} \\ h=.211 \end{gathered}$$

(b)  

$$\begin{gathered} h=\frac{W}{Q_1} \\ W=h\times Q_1 \\ W=.211\times 100 \\ W=21.1 \mathrm{kJ} \end{gathered}$$

Therefore, Efficiency of cyclic heat engine is $.211$.  and amount of work done is $21.1\mathrm{kJ}$.