Do the energy-transfer diagrams in$\mathrm{FIGURE}\mathrm{Q}21.8$represent possible heat engines? If not, what is wrong?

Consider the first part a in $\mathrm{FIGURE}\mathrm{Q}21.8$

The output value of a heat engine should always be greater than the input value.

From the above figure the input temperature value is $600K$,

And the output temperature value is $300K$.

The input range is greater than the output value,

So the part a in $\mathrm{FIGURE}\mathrm{Q}21.8$is not represent possible of heat engine.

Consider the part b of $\mathrm{FIGURE}\mathrm{Q}21.8$

First to calculate efficiency of normal heat engine,

Efficiency $\mathrm{\eta }$$=\mathrm{output}/\mathrm{input}$output value is

$\mathrm{\eta }=4/10=0.4$

Calculate efficiency for carnot heat engine,

${\mathrm{\eta }}_{\mathrm{c}}=1-\frac{{\mathrm{T}}_{\mathrm{c}}}{{\mathrm{T}}_{\mathrm{h}}}$

${\mathrm{\eta }}_{\mathrm{c}}=1-300/600$

${\mathrm{\eta }}_{\mathrm{c}}=0.5$

So,$\mathrm{\eta }<{\mathrm{\eta }}_{\mathrm{c}}$.

Consider the part c in figure,

Calculate the efficiency of heat engine,

$\mathrm{\eta }=\mathrm{output}/\mathrm{input}$

$\mathrm{\eta }=6/10$

$\mathrm{\eta }=0.6$

Calculate the efficiency for carnot heat engine,

${\mathrm{\eta }}_{\mathrm{c}}=1-\frac{{\mathrm{T}}_{\mathrm{c}}}{{\mathrm{T}}_{\mathrm{h}}}$

${\mathrm{\eta }}_{\mathrm{c}}=1-\frac{300}{600}=0.5$

So,$\mathrm{\eta }>{\mathrm{\eta }}_{\mathrm{c}}$