Question

A heat engine operating between a hot reservoir at $500°\mathrm{C}$and a cold reservoir at $0°\mathrm{C}$is $60\%$as efficient as a Carnot engine. If this heat engine and the Carnot engine do the same amount of work, what is the ratio${\mathrm{Q}}_{\mathrm{H}}/{\left({\mathrm{Q}}_{\mathrm{H}}\right)}_{\text{Carnot}}$$?$

Short answer

The ratio of heat engine and the carnot engine with same work$\frac{{\mathrm{Q}}_{\mathrm{h}}}{{\mathrm{Q}}_{\mathrm{h},\mathrm{C}}}$is$\frac{5}{3}$.

Step-by-step solution

Formula for efficiency

Efficiencies be $\mathrm{\eta }$for the real engine,

And${\mathrm{\eta }}_{\mathrm{C}}$for the Carnot one.

Heat efficiency is,

$\mathrm{\eta }=\frac{\mathrm{W}}{{\mathrm{Q}}_{\mathrm{h}}}\Rightarrow {\mathrm{Q}}_{\mathrm{h}}=\frac{W}{\eta }$

Carnot efficiency is,

${\mathrm{\eta }}_{\mathrm{C}}=\frac{\mathrm{W}}{{\mathrm{Q}}_{\mathrm{h},\mathrm{C}}}\Rightarrow {\mathrm{Q}}_{\mathrm{h},\mathrm{C}}=\frac{\mathrm{W}}{{\mathrm{\eta }}_{\mathrm{C}}}$

Calculation for ratio of heat energy and carnot energyQhQh,C

Taking the ratio we get,

$\frac{{\mathrm{Q}}_{\mathrm{h}}}{{\mathrm{Q}}_{\mathrm{h},\mathrm{C}}}=\frac{\mathrm{W}/\mathrm{\eta }}{\mathrm{W}/{\mathrm{\eta }}_{\mathrm{C}}}=\frac{{\mathrm{\eta }}_{\mathrm{C}}}{\mathrm{\eta }}$

$\mathrm{\eta }=0.6·{\mathrm{\eta }}_{\mathrm{C}}$.

Substituting this we get,

$\frac{{\mathrm{Q}}_{\mathrm{h}}}{{\mathrm{Q}}_{\mathrm{h},\mathrm{C}}}=\frac{{\mathrm{\eta }}_{\mathrm{C}}}{0.6{\mathrm{\eta }}_{\mathrm{C}}}=\frac{5}{3}$