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Standard XII

Physics

Carnot Cycle

An ideal Carn...

Question

An ideal Carnot engine having efficiency $η$ works between temperatures $127 ° C$ and $- 73 ° C$ . $η_{1}$ denotes the efficiency when the temperature of the source increases by $50 ° C$ and $η_{2}$ denotes efficiency when temperature of the sink decreases by $50 ° C$ . Then the correct order is

η > η_{1} > η_{2}

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η_{1} > η > η_{2}

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η_{2} > η_{1} > η

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η = η_{1} = η_{2}

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Solution

The correct option is C $η_{2} > η_{1} > η$

$T_{1} = 127^{o} = 400 K T_{2} = - 73^{o} = 200 K$
effeciency ( $η$ ) $= 1 - \frac{T_{2}}{T_{1}}$
$η = 1 - \frac{200}{400} = 1 - \frac{1}{2}$
= $50$ %

Case 1: $T_{1} = 450 K T_{2} = 200 K$
$η_{1} = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{200}{450}$
$= 55$ %

Case 2: $T_{1} = 400 K T_{2} = 150 K$
$η_{2} = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{150}{400}$
$= 62.5$ %

From above observation, $η_{2} > η_{1} > η$

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Similar questions

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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 $\frac{η_{1}}{η_{2}}$

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An ideal Carnot engine having efficiency η works between temperatures 127°C and −73°C. η1 denotes the efficiency when the temperature of the source increases by 50°C and η2 denotes efficiency when temperature of the sink decreases by 50°C. Then the correct order is

The correct option is C η2>η1>η T1=127o=400 KT2=−73o=200 K effeciency (η)=1−T2T1 η=1−200400 =1−12 =50% Case 1: T1=450 KT2=200 K η1=1−T2T1 =1−200450 =55% Case 2: T1=400 KT2=150 K η2=1−T2T1 =1−150400 =62.5% From above observation, η2>η1>η