Question

An engine is designed to operate between $480\text{K}$ and $300\text{K}$. Assuming that the engine actually produces $1.2\text{kJ}$ of mechanical energy per $\text{kcal}$ of heat absorbed. Find the ratio of actual efficiency of engine to the theoretical maximum efficiency.
(Take $1\text{cal}=4.2\text{J}$)

• A. $\frac{4}{3}$
• B. $\frac{8}{28}$
• C. $\frac{16}{21}$
  
• D. None of the these

Answer 1

The correct option is C $\frac{16}{21}$


Temperature of source, $T_1=480\text{K}$
Temperature of sink, $T_2=300\text{K}$
For $1\text{kcal}$ energy input, work output is $1.2\text{kJ}$
$\therefore$ Maximum efficiency of the engine will correspond for Carnot's engine at given conditions.
${\eta }_{\text{max}}=1-\frac{T_2}{T_1}=1-\frac{300}{480}$
${\eta }_{\text{max}}=\frac{180}{480}=\frac{3}{8}$
Actual efficiency of the engine is given as,
$\left({\eta }^{\prime }\right)=\frac{\text{Work output}}{\text{Energy input}}=\frac{1.2\text{kJ}}{1\text{kcal}}$
${\eta }^{\prime }=\frac{1.2\text{kJ}}{4.2\text{kJ}}=\frac{2}{7}$
Taking the ratio of efficiencies,
$\frac{{\eta }^{\prime }}{\eta }=\frac{\left(\frac{2}{7}\right)}{\left(\frac{3}{8}\right)}$
$\therefore \frac{{\eta }^{\prime }}{\eta }=\frac{16}{21}$