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11.Thermodynamics
normal
A carnot engine having an efficiency of as heat engine, is used as a refrigerator. 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
$100 $
B
$99 $
C
$90$
D
$1$
Solution
The relation between coefficient of performance and efficiency of camot engine is given as
$\beta=\frac{1-\eta}{\eta}$
Given $\eta=\frac{1}{10}, W=10 \mathrm{\,J}$
$\beta=\frac{1-\frac{1}{10}}{\frac{1}{10}}=\frac{9}{10} \cdot 10=9$
Since, $\beta=\frac{Q_{2}}{W},$ where $Q_{2}$ is the amount of energy absorbed from the reservoir
$\therefore Q_{2}=\beta W=9 \times 10=90 \mathrm{\,J}$
Standard 11
Physics
