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Q. A Carnot engine operates between $227^{\circ} C$ and $27^{\circ} C$. Efficiency of the engine will be :

BHUBHU 2004

Solution:

Efficiency of heat engine is the fraction of the total heat supplied to the engine
which is converted into work that is
$\eta=1-\frac{Q_{2}}{Q_{1}} $
$=1-\frac{T_{2}}{T_{1}} $
$\Rightarrow \eta=\frac{T_{1}-T_{2}}{T_{1}} $
Given, $T_{1}=227^{\circ} C$
$=227+273=500\, K$
$T_{2}=27^{\circ} C$
$=273+27=300\, K $
$\therefore \eta=\frac{500-300}{500} $
$=\frac{200}{500}=\frac{2}{5}$