Let eta_{1} be the efficiency of a Carnot eng | vedclass

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(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_{L}}{T_{H}}$,where temperatures must be in Kelvin $(K = ^{\circ}C + 273)$.For $\

Vedclass · Accepted Answer

$0.56$

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(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_{L}}{T_{H}}$,where temperatures must be in Kelvin $(K = ^{\circ}C + 273)$.For $\eta_{1}$:$T_{H} = 447 + 273 = 720 \ K$$T_{L} = 147 + 273 = 420 \ K$$\eta_{1} = 1 - \frac{420}{720} = 1 - \frac{42}{72} = 1 - \frac{7}{12} = \frac{5}{12}$.For $\eta_{2}$:$T_{H} = 947 + 273 = 1220 \ K$$T_{L} = 47 + 273 = 320 \ K$$\eta_{2} = 1 - \frac{320}{1220} = 1 - \frac{32}{122} = 1 - \frac{16}{61} = \frac{45}{61}$.Now,the ratio $\frac{\eta_{1}}{\eta_{2}}$ is:$\frac{\eta_{1}}{\eta_{2}} = \frac{5/12}{45/61} = \frac{5}{12} 	imes \frac{61}{45} = \frac{61}{12 	imes 9} = \frac{61}{108} \approx 0.5648$.Rounding to two decimal places,the ratio is $0.56$.

Let $\eta_{1}$ be the efficiency of a Carnot engine at $T_{H}=447^{\circ}C$ and $T_{L}=147^{\circ}C$,while $\eta_{2}$ is the efficiency at $T_{H}=947^{\circ}C$ and $T_{L}=47^{\circ}C$. The ratio $\frac{\eta_{1}}{\eta_{2}}$ will be:

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