In a Carnot engine- when T2-0-C and T1-200-C- its efficiency is - 1 and when T1-0-C and T2-200-C- Its efficiency is - 2- then what is - 1 - - 2

The correct option is A 0.577Using the efficiency of carnot cycle η =1 T_2T_1⇒η _1=(473 273)473=200473 and η _2=273 73273=200273 #160;so required ...

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Standard XII

Physics

Coefficient of Viscosity

In a Carnot e...

Question

In a Carnot engine, when $T_{2} = 0^{\circ} C$ and $T_{1} = 200^{\circ} C$ , its efficiency is $η_{1}$ and when $T_{1} = 0^{\circ} C$ and $T_{2} = 200^{\circ} C$ , Its efficiency is $η_{2}$ , then what is $η_{1}$ / $η_{2}$

0.577

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0.733

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0.638

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Can not be calculated

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Solution

The correct option is A 0.577
Using the efficiency of carnot cycle
$η = 1 - \frac{T_{2}}{T_{1}} \Rightarrow η_{1} = \frac{(473 - 273)}{473} = \frac{200}{473}$ and $η_{2} = \frac{273 - 73}{273} = \frac{200}{273}$
so required ratio $\frac{η_{1}}{η_{2}} = \frac{273}{473} = 0.577$

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In a Carnot engine, when T2=0∘C and T1=200∘C, its efficiency is η1 and when T1=0∘C and T2=200∘C, Its efficiency is η2, then what is η1 /η2

The correct option is A 0.577Using the efficiency of carnot cycleη=1−T2T1⇒η1=(473−273)473=200473 and η2=273−73273=200273 so required ratio η1η2=273473=0.577

In a Carnot engine, when $T_{2} = 0^{\circ} C$ and $T_{1} = 200^{\circ} C$ , its efficiency is $η_{1}$ and when $T_{1} = 0^{\circ} C$ and $T_{2} = 200^{\circ} C$ , Its efficiency is $η_{2}$ , then what is $η_{1}$ / $η_{2}$

The correct option is A 0.577
Using the efficiency of carnot cycle
$η = 1 - \frac{T_{2}}{T_{1}} \Rightarrow η_{1} = \frac{(473 - 273)}{473} = \frac{200}{473}$ and $η_{2} = \frac{273 - 73}{273} = \frac{200}{273}$
so required ratio $\frac{η_{1}}{η_{2}} = \frac{273}{473} = 0.577$