
The efficiency of a Carnot engine is mentioned to be as eighty percent. When the direction of the cycle be reversed, what will be the value of C.O.P. of reversed Carnot Cycle?
$\begin{align}
& A.0.25 \\
& B.0.5 \\
& C.0.8 \\
& D.1.25 \\
\end{align}$
Answer
485.4k+ views
Hint: The C.O.P of this condition will be equivalent to the ratio of the difference of one and the efficiency and the Carnot cycle to the efficiency of the Carnot cycle. Substitute the value of efficiency in this equation. This will help you in answering this question.
Complete answer:
It has been mentioned in the question that the efficiency of the Carnot cycle be,
$\eta =0.8$
If the direction of the cycle has been reversed, then the relation between the C.O.P and the efficiency can be found. The C.O.P of this condition will be equivalent to the ratio of the difference of one and the efficiency and the Carnot cycle to the efficiency of the Carnot cycle. This can be expressed in the form of an equation which can be written as,
$COP=\dfrac{1-\eta }{\eta }$
Substituting the values in this equation can be written as,
$COP=\dfrac{1-0.8}{0.8}=0.25$
Therefore the COP of the Carnot cycle has been calculated.
This answer has been mentioned as option A.
Note:
The Carnot engine is defined as a theoretically proven thermodynamic cycle. This cycle has been proposed by Leonard Carnot. It is providing the estimate of the maximum efficiency possible for a heat engine during the conversion process of heat into work and in the opposite manner also. The efficiency of this kind of engine will be independent of the nature of the working substance in the system and will be only dependent on the temperature of the hot and cold reservoirs. A Carnot cycle can be taken as an ideal reversible closed thermodynamic cycle.
Complete answer:
It has been mentioned in the question that the efficiency of the Carnot cycle be,
$\eta =0.8$
If the direction of the cycle has been reversed, then the relation between the C.O.P and the efficiency can be found. The C.O.P of this condition will be equivalent to the ratio of the difference of one and the efficiency and the Carnot cycle to the efficiency of the Carnot cycle. This can be expressed in the form of an equation which can be written as,
$COP=\dfrac{1-\eta }{\eta }$
Substituting the values in this equation can be written as,
$COP=\dfrac{1-0.8}{0.8}=0.25$
Therefore the COP of the Carnot cycle has been calculated.
This answer has been mentioned as option A.
Note:
The Carnot engine is defined as a theoretically proven thermodynamic cycle. This cycle has been proposed by Leonard Carnot. It is providing the estimate of the maximum efficiency possible for a heat engine during the conversion process of heat into work and in the opposite manner also. The efficiency of this kind of engine will be independent of the nature of the working substance in the system and will be only dependent on the temperature of the hot and cold reservoirs. A Carnot cycle can be taken as an ideal reversible closed thermodynamic cycle.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Trending doubts
What is the difference between superposition and e class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

If the HCF of 657 and 963 is expressible in the fo-class-11-maths-CBSE

How do I convert ms to kmh Give an example class 11 physics CBSE

Convert the following into basic units a 287pm b 1515pm class 11 chemistry CBSE

Number of oneone functions from A to B where nA 4 and class 11 maths CBSE
