The coefficient of performance of a Carnot re | vedclass

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

Question

(C) The coefficient of performance $(COP)$ of a refrigerator is given by $\alpha = \frac{Q_2}{Q_1 - Q_2}$.For a Carnot refrigerator,the ratio of heat

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(C) The coefficient of performance $(COP)$ of a refrigerator is given by $\alpha = \frac{Q_2}{Q_1 - Q_2}$.For a Carnot refrigerator,the ratio of heat exchanged is proportional to the absolute temperatures: $\frac{Q_1}{Q_2} = \frac{T_1}{T_2}$.This implies $\frac{Q_1 - Q_2}{Q_2} = \frac{T_1 - T_2}{T_2}$.Therefore,the $COP$ is $\alpha = \frac{T_2}{T_1 - T_2}$.Given temperatures are $T_1 = 30 + 273 = 303 \, 	ext{K}$ and $T_2 = 0 + 273 = 273 \, 	ext{K}$.Substituting these values: $\alpha = \frac{273}{303 - 273} = \frac{273}{30} = 9.1$.Rounding to the nearest integer provided in the options,we get $9$.

The coefficient of performance of a Carnot refrigerator working between temperatures $30 \, ^\circ\text{C}$ and $0 \, ^\circ\text{C}$ is:

Similar Questions

In a mechanical refrigerator, the low temperature coils are at a temperature of $-23^{\circ}C$ and the compressed gas in the condenser has a temperature of $27^{\circ}C$. The theoretical coefficient of performance is

An ideal heat engine has an efficiency $\eta$. The coefficient of performance of the engine when driven backward will be

What is the coefficient of performance of a refrigerator operating between $30^{\circ}C$ and $0^{\circ}C$?

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

$A$ Carnot's engine works as a refrigerator between $250\, K$ and $300\, K$. It receives $500\, cal$ heat from the reservoir at the lower temperature. The amount of work done in each cycle to operate the refrigerator is ..... $J$

Vedclass Test Series

Exam Paper Generator

Online Exam Module