A Carnot engine having an efficiency of 1/5 a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The efficiency of a Carnot engine is given by $\eta = 1/5$. When this engine is used as a refrigerator,its coefficient of performance $\beta$ is r

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(C) The efficiency of a Carnot engine is given by $\eta = 1/5$. When this engine is used as a refrigerator,its coefficient of performance $\beta$ is related to the efficiency $\eta$ by the formula: $\beta = \frac{1 - \eta}{\eta}$. Substituting the value of $\eta$: $\beta = \frac{1 - 1/5}{1/5} = \frac{4/5}{1/5} = 4$. The coefficient of performance is also defined as the ratio of heat absorbed from the cold reservoir $(Q)$ to the work done on the system $(W)$: $\beta = \frac{Q}{W}$. Given $W = 50 \ J$,we have $4 = \frac{Q}{50}$. Therefore,$Q = 4 	imes 50 = 200 \ J$.

$A$ Carnot engine having an efficiency of $1/5$ as a heat engine is used as a refrigerator. If the work done on the system is $50 \ J$,then the amount of energy absorbed from the reservoir at the lower temperature is (in $J$)

Similar Questions

$A$ Carnot engine,having an efficiency of $\eta = 1/10$ as a 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 a lower temperature is ....... $J$.

The door of a working refrigerator is left open in a well-insulated room. The temperature of the air in the room will

Determine the coefficient of performance for the given temperature limits.
$T_{1} = 27^{\circ}C$ [outside fridge]
$T_{2} = -23^{\circ}C$ [inside fridge]

If the door of a refrigerator is kept open in a closed room,will the room become cooler or warmer?

If a refrigerator of coefficient of performance of $5$ has a freezer at a temperature of $-13^{\circ} C$,then the room temperature is (in $^{\circ} C$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module