The efficiency of a reversible heat engine wo | vedclass

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

Question

(A) The efficiency of a reversible heat engine is given by $\eta = 1 - \frac{T_2}{T_1} = 0.5$. From this,we find $\frac{T_2}{T_1} = 1 - 0.5 = 0.5$. Th

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) The efficiency of a reversible heat engine is given by $\eta = 1 - \frac{T_2}{T_1} = 0.5$. From this,we find $\frac{T_2}{T_1} = 1 - 0.5 = 0.5$. The coefficient of performance $(COP)$ of a refrigerator working between the same temperatures is given by $\beta = \frac{T_2}{T_1 - T_2}$. Dividing the numerator and denominator by $T_1$,we get $\beta = \frac{T_2/T_1}{1 - T_2/T_1}$. Substituting the value $\frac{T_2}{T_1} = 0.5$,we get $\beta = \frac{0.5}{1 - 0.5} = \frac{0.5}{0.5} = 1$.

The efficiency of a reversible heat engine working between two temperatures is $50 \%$. The coefficient of performance of a refrigerator working between the same two temperatures but in reverse direction is

Similar Questions

If a refrigerator's door is kept open,will the room become cool or hot? Explain.

The coefficient of performance of a refrigerator working at a temperature of $27^{\circ} C$ is $5$. Find the temperature inside the refrigerator. (in $K$)

The door of an operating refrigerator is kept open. As a result,the temperature of the room will

The coefficient of performance of a refrigerator is $5$. If the temperature inside the freezer is $-20^{\circ}C$,the temperature of the surroundings to which it rejects heat is ........ $^{\circ}C$.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module