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(D) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the source temperature and $T_2$ is the sink temperature

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$40 \ K, 20 \ K$

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(D) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the source temperature and $T_2$ is the sink temperature.To maximize $\eta$,the ratio $\frac{T_2}{T_1}$ must be minimized.Calculating the ratio for each option:$A: \frac{60}{80} = 0.75$$B: \frac{80}{100} = 0.80$$C: \frac{40}{60} \approx 0.67$$D: \frac{20}{40} = 0.50$Since the ratio is minimum for option $D$,the efficiency is highest for $T_1 = 40 \ K$ and $T_2 = 20 \ K$.

For which of the following options will the efficiency of a Carnot engine be the highest?

Similar Questions

In a Carnot engine,the temperature of the reservoir is $927\ ^oC$ and that of the sink is $27\ ^oC.$ If the work done by the engine is $12.6 \times 10^6 \ J,$ the quantity of heat absorbed by the engine from the reservoir is:

In practice,all heat engines have an efficiency less than that of a Carnot engine because

$Assertion :$ The Carnot cycle is useful in understanding the performance of heat engines.
$Reason :$ The Carnot cycle provides a way of determining the maximum possible efficiency achievable with reservoirs of given temperatures.

Explain heat engines and their working.

An ideal Carnot engine whose efficiency is $50 \%$ receives heat at $500 \ K$. If the efficiency is to be $60 \%$, then the intake temperature for the same exhaust temperature is: (in $K$)

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