91Ó°ÊÓ

$\left(a\right)$The area under the graph will be quantitatively equal to the work done by the gas. As a result, this work is calculated as follows.

$$\begin{gathered} W_{out}=300â‹\dots {10}^3â‹\dots 100â‹\dots {10}^{−6} \\ {W}_{out}=30\mathrm{J}. \end{gathered}$$

$Q_H$is heat energy that the machine takes in . It will be equal to work done by gas and total heat evolved by the system . It can be expressed as follows.

localid="1649318298079" $Q_H=W_{net}+Q_c$

Total amount of heat evolved or lost $Q_c$is sum of 90 J and 25 J . They are negative heat , so must have been given out by gas.

$Q_c=90+25=115\mathrm{J}$

The heat that is added to - or taken away from - the heat reservoir is

localid="1649318324387" $Q_H=30+115=145\mathrm{J}.$ $$\begin{gathered} Q_h=30+115 \\ {Q}_h=145\mathrm{J}. \end{gathered}$$

$\left(b\right)$The efficiency can be calculated as

localid="1649318348602" $\begin{array}{c}\mathrm{η}=\frac{W_{out}}{Q_H}\\ \mathrm{η}=\frac{30}{145}=21\mathrm{\%}.\end{array}$