Question

A Carnot engine receiving heat at 400 K has an efficiency of 50 %. What is the COP of a Carnot refrigerator working between the same temperature limits?

1. 4
2. 1
3. 2
4. 3

Answer 1

Correct Answer - Option 2 : 1

Concept:

\(η_{carnot}=\frac{T_H-T_L}{T_H}=1-\frac{T_L}{T_H}\)

\((COP)_{carnot}=\frac{T_L}{T_H-T_L}\)

Calculation:

Given:

η_Carnot = 50 % = 0.5, T_H = 400 K

\(η_{carnot}=\frac{T_H-T_L}{T_H}=1-\frac{T_L}{T_H}\)

\(\Rightarrow\frac{T_L}{T_H}=0.5=\frac{1}{2}\)

Now,

\((COP)_{carnot}=\frac{T_L}{T_H-T_L}=\frac{1}{\frac{T_H}{T_L}-1}=\frac{1}{2-1}=1\)