A carnot engine having an efficiency of (1/10) 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 :

Question

A carnot engine having an efficiency of (1/10) 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 : (A) 90 J (B) 99 J (C) 100 J (D) 1 J

Tardigrade Admin · Accepted Answer

β=(1-η/η) =(1-(1/10)/(1)10)=((9/10)/(1)10) β=9 β=(Q2/W) Q2=9 × 10=90 J

Q. A carnot engine having an efficiency of $\frac{1}{10}$ 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 :

90 J

99 J

100 J

1 J

$β = \frac{1 - η}{η}$
$= \frac{1 - \frac{1}{10}}{\frac{1}{10}} = \frac{\frac{9}{10}}{\frac{1}{10}}$
$β = 9$
$β = \frac{Q _{2}}{W}$
$Q_{2} = 9 \times 10 = 90 J$

Q. A carnot engine having an efficiency of 101​ as heat engine, is used as a refrigerator. If the work done on the system is 10J, the amount of energy absorbed from the reservoir at lower temperature is :

90 J

99 J

100 J

1 J

β=η1−η​ =101​1−101​​=101​109​​ β=9 β=WQ2​​ Q2​=9×10=90J

Q. A carnot engine having an efficiency of $\frac{1}{10}$ 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 :

$β = \frac{1 - η}{η}$
$= \frac{1 - \frac{1}{10}}{\frac{1}{10}} = \frac{\frac{9}{10}}{\frac{1}{10}}$
$β = 9$
$β = \frac{Q _{2}}{W}$
$Q_{2} = 9 \times 10 = 90 J$