Question

What are (a) the thermal efficiency and (b) the heat extracted from the hot reservoir for the heat engine shown in $\mathrm{FIGURE}\mathrm{EX}21.15?$

Short answer

a.Thermal efficiency heat engine is$13\%$.

b.The heat extracted from the hot reservoir for the heat engine is$320\mathrm{J}$.

Step-by-step solution

Calculation for efficiency of heat engine (part a)

a.

To calculate the efficiency we can use the following formula,

$\mathrm{\eta }=\frac{{\mathrm{W}}_{\mathrm{out}}}{\mathrm{Q}}=\frac{{\mathrm{W}}_{\mathrm{out}}}{{\mathrm{W}}_{\mathrm{out}}+{\mathrm{Q}}_{\mathrm{c}}}$

where $Q_c$is the heat that was given to the cold reservoir.

On the other hand, the work done by the gas in one cycle will be equal to the region encircled by the graph.

Work is ,

localid="1648643811450" role="math" ${\mathrm{W}}_{\text{out}}=\frac{1}{2}·(300-100)·{10}^3·(600-200)·{10}^{-6}=40\mathrm{J}$

On the other hand, the total amount of heat lost is equal to the sum of the two losses.

localid="1648643864648" role="math" ${\mathrm{Q}}_{\mathrm{c}}=100+180=280\mathrm{J}$

The efficiency ,

localid="1648643885687" $\mathrm{\eta }=\frac{40}{40+280}\approx 13\%$

Calculation for the heat extracted from the hot reservoir for the heat engine (part b)

b.

The heat extracted from the hot reservoir will be as follows:

From solution a,

${\mathrm{W}}_{\mathrm{out}}=40\mathrm{J}$

${\mathrm{Q}}_{\mathrm{c}=}280$

Substitute above value,

we get,

${\mathrm{Q}}_{\mathrm{c}}={\mathrm{W}}_{\text{out}}+{\mathrm{Q}}_{\mathrm{c}}=40+280=320\mathrm{J}$