A carnot engine, having an efficiency of eta = 1/10 as heat engine, is used as a refrigetator.

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11.Thermodynamics

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A carnot engine, having an efficiency of $\eta = 1/10$ as heat engine, is used as a refrigetator. If the work done on the system is $10\,J$ , the amount of energy absorbed from the reservoir at lower temperature is .......... $\mathrm{J}$

A

$99$

B

$90$

C

$1$

D

$100$

Solution

$\eta=1-\frac{T_{2}}{T_{1}}=\frac{1}{10}$

$\mathrm{COP}=\frac{Q_{\text {cooling }}}{\mathrm{W}}=\frac{1}{\frac{\mathrm{T}_{1}}{\mathrm{T}_{2}}-1}$

${Q_{{\rm{cooling }}}} = \frac{{\frac{{\frac{W}{{\frac{{{T_1}}}{{{T_2}}} – 1}}}}{{10}}}}{{\frac{{10}}{9} – 1}}$

$ = 90{\rm{\,J}}$

Standard 11

Physics

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