Let η1 is the efficiency of an engine at T1=447∘C and T2=147∘C while η2 is the efficiency at T1=947∘C and T2=47∘C. The ratio η1η2 will be :

Efficiency η=1−TCTHη1=1−147+273447+273=1−420720η1=300720η2=1−47+273947+273=1−3201220η2=9001220η1η2=300720×1220900=12272×3η1η2=0.56

Let η1 is the efficiency of an engine at T1=447∘C and T2=147∘C while η2 is the efficiency at T1=947∘C and T2=47∘C. The ratio η1η2 will be :

Efficiency η=1−TCTHη1=1−147+273447+273=1−420720η1=300720η2=1−47+273947+273=1−3201220η2=9001220η1η2=300720×1220900=12272×3η1η2=0.56

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Let $η_{1}$ is the efficiency of an engine at $T_{1} = 447^{\circ} C and T_{2} = 147^{\circ} C$ while $η_{2}$ is the efficiency at $T_{1} = 947^{\circ} C and T_{2} = 47^{\circ} C$ . The ratio $\frac{η_{1}}{η_{2}}$ will be :

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0.56

0.70

0.41

0.73

answer is B.

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Detailed Solution

$\begin{array}{l} Efficiency η = 1 - \frac{T_{C}}{T_{H}} \\ η_{1} = 1 - \frac{147 + 273}{447 + 273} = 1 - \frac{420}{720} \\ η_{1} = \frac{300}{720} \\ η_{2} = 1 - \frac{47 + 273}{947 + 273} = 1 - \frac{320}{1220} \\ η_{2} = \frac{900}{1220} \\ \frac{η_{1}}{η_{2}} = \frac{300}{720} \times \frac{1220}{900} = \frac{122}{72 \times 3} \\ \frac{η_{1}}{η_{2}} = 0 .56 \end{array}$