A Carnot engine,having an efficiency of eta = | vedclass

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

Question

(C) The efficiency $(\eta)$ of a Carnot engine and the coefficient of performance $(\beta)$ of a refrigerator are related as:$\beta = \frac{1 - \eta}{

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(C) The efficiency $(\eta)$ of a Carnot engine and the coefficient of performance $(\beta)$ of a refrigerator are related as:$\beta = \frac{1 - \eta}{\eta}$Given $\eta = 1/10$,we calculate the coefficient of performance:$\beta = \frac{1 - 1/10}{1/10} = \frac{9/10}{1/10} = 9$The coefficient of performance $(\beta)$ is also defined as the ratio of heat absorbed from the cold reservoir $(Q_2)$ to the work done $(W)$ on the system:$\beta = \frac{Q_2}{W}$Given $W = 10 \ J$ and $\beta = 9$,we have:$9 = \frac{Q_2}{10 \ J}$$Q_2 = 9 	imes 10 \ J = 90 \ J$Therefore,the energy absorbed from the reservoir at a lower temperature is $90 \ J$.

$A$ 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 ....... $J$.

Similar Questions

If the door of a refrigerator is kept open in a room,will it make the room warm or cool?

Why does the coefficient of performance of a refrigerator become infinity?

The temperatures inside and outside of a refrigerator are $260 \ K$ and $315 \ K$ respectively. Assuming that the refrigerator cycle is reversible,calculate the heat delivered to the surroundings for every joule of work done. (in $.73$)

For a refrigerator,heat absorbed from the cold reservoir (sink) is $500\, J$ and heat rejected to the hot reservoir (source) is $800\, J$. The coefficient of performance is:

If the door of a refrigerator is kept open in an insulated room,the temperature of the air in the room will . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module