(A) The efficiency of a Carnot cycle is given by the formula:$\eta = 1 - \frac{T_{cold}}{T_{hot}}$Here,the cold reservoir temperature $T_{cold} = 4^{\

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(A) The efficiency of a Carnot cycle is given by the formula:$\eta = 1 - \frac{T_{cold}}{T_{hot}}$Here,the cold reservoir temperature $T_{cold} = 4^{\circ} C = 273 + 4 = 277 \ K$.The hot reservoir temperature $T_{hot} = 15^{\circ} C = 273 + 15 = 288 \ K$.Substituting these values into the formula:$\eta = 1 - \frac{277}{288}$$\eta = 1 - 0.9618$$\eta \approx 0.0382$Rounding to three decimal places,we get $\eta = 0.038$.

Class 11 Physics Thermodynamics Heat Engine and Carnot Cycle

Medium

The temperature of food material in a refrigerator is $4^{\circ} C$ and the temperature of the environment is $15^{\circ} C$. If a Carnot cycle is used in its working gas,find its Carnot efficiency.

AIIMS 2019 All AIIMS

A
$0.038$
B
$0.028$
C
$0.053$
D
$0.072$

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The temperature of food material in a refrigerator is $4^{\circ} C$ and the temperature of the environment is $15^{\circ} C$. If a Carnot cycle is used in its working gas,find its Carnot efficiency.

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If a Carnot engine functions at a source temperature of $127^{\circ}C$ and a sink temperature of $87^{\circ}C$,what is its efficiency in $\%$?

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