Question

An ideal Carnot refrigerator with a performance coefficient (COP) of 2.1 cools items inside of it to 5.0° C. What is the high temperature needed to operate this refrigerator?
A) 7° C
B) 589° C
C) 137° C
D) 16° C

Answer 1

Answer:

the high temperature needed to operate this refrigerator is C) 137.4° C

Explanation:

Hello!

The carnot refrigeration cycle is one in which a machine absorbs heat from an enclosure and expels it to the surroundings, the equation that defines the COP performance coefficient for this cycle is:

$COP=\frac{T1}{T2-T1}$

COP=performance coefficient =2.1

T1=  Low temperature

T2=high temperature

Now use algebra to find the high temperature

$COP=\frac{T1}{T2-T1}\\(T2-T1)=\frac{T1}{COP}\\T2=\frac{T1}{COP}+T1\\T2=T1(\frac{1}{COP} +1)$

If we replace the values:

note = remember that the temperature must be in absolute units, for which we must add 273.15 to the low temperature to find the temperature in Kelvin

T1 = 5 + 273.15 = 278.15K

$T2=278.15(\frac{1}{2.1} +1)=410.60k$

In celsius

T2=410.60-273.15=137.4° C

the high temperature needed to operate this refrigerator is C) 137.4° C