In a Carnot engine, the absolute temperature of the source is $25 \%$ more than the absolute temperature of…
In a Carnot engine, the absolute temperature of the source is $25 \%$ more than the absolute temperature of the sink. The efficiency of the engine is
- $10 \%$
- $50 \%$
- $25 \%$
- $20 \%$
Solution
$\mathrm{T}_2=\mathrm{T}, \mathrm{~T}_1=1.25 \mathrm{~T}$
$\therefore$ Efficiency of cannot engine is
$\eta=1-\frac{T_2}{T_1}=1-\frac{T}{1.25 T}=\frac{1}{5}$
$\therefore \quad \% \eta=\frac{1}{5} \times 100=20 \%$
Asked in: AP EAMCET 2024 (18 May Shift 1)
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