An electrical refrigerator with beta=5 extrac | vedclass

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

Question

(A) The coefficient of performance $(\beta)$ of a refrigerator is defined as the ratio of the heat extracted from the cold reservoir $(Q)$ to the work

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) The coefficient of performance $(\beta)$ of a refrigerator is defined as the ratio of the heat extracted from the cold reservoir $(Q)$ to the work done on the system $(W)$: $\beta = \frac{Q}{W}$ Given: $\beta = 5$ $Q = 5000 \,J$ We need to find the electrical energy utilised by the motor, which is equal to the work done $(W)$: $W = \frac{Q}{\beta}$ $W = \frac{5000 \,J}{5} = 1000 \,J$ Since $1000 \,J = 1 \,kJ$, the electrical energy utilised is $1 \,kJ$.

An electrical refrigerator with $\beta=5$ extracts $5000 \,J$ from the contents of the refrigerator. During this process, find the electrical energy utilised by its motor. (in $\,kJ$)

Similar Questions

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

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

The coefficient of performance of a Carnot refrigerator working between $30^oC$ and $0^oC$ is

Explain the working of refrigerators/heat pumps and their coefficient of performance.

The efficiency of a Carnot engine acting as a heat engine is $\eta = 1/10$. If it is used as a refrigerator and the work done on the system is $10 \ J$,then the heat absorbed from the low-temperature reservoir is ...... $J$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module