A carnot engine having an efficiency of frac{1}{10} is being used as a refrigerator. If work done on

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11.Thermodynamics

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A carnot engine having an efficiency of $\frac{1}{10}$ is being used as a refrigerator. If the work done on the refrigerator is $10 \;\mathrm{J},$ the amount of heat absorbed from the reservoir at lower temperature is .............. $\mathrm{J}$

A

$99$

B

$100$

C

$90$

D

$1$

Solution

Refrigerator cycle is

$\eta=\frac{W}{Q_{+}}=\frac{W}{W+Q_{-}}$

$\frac{1}{10}=\frac{10}{10+\mathrm{Q}_{-}}$

$Q_{-}=90 \mathrm{J}$

Heat absorbed from the reservoir at lower temperature is $90 \mathrm{J}$

Standard 11

Physics

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