Question

A refrigerator is to maintain eatables kept inside at 9°C. If the room temperature is 36°C, calculate the coefficient of performance.

Answer 1

Hint: Effectiveness of a refrigerator or a heat pump is determined by the coefficient of performance. Here we have to determine the coefficient of performance i.e. the effectiveness of the Air conditioner. We are given the inside temperature which is 9°C and the outside temperature of the room which is 36°C Apply the formula for the coefficient of performance which relates the outside and inside temperature.

Formula used:
The formula for finding out the coefficient of performance is given below.
($COP$)$_R$= $\dfrac{{{T_1}}}{{({T_2} - {T_1})}}$)
Here, (COP)$_R$= Coefficient of performance of the refrigerator.
${T_1}$= Temperature inside the refrigerator (9°C)
${T_2}$= Temperature of the room (36°C)

Complete step by step answer:
Given, ${T_1}$= 9+273 = 282K and
${T_2}$= 36+273 = 309K
We need to find the coefficient of performance.
($COP$)$_R$= $\dfrac{{{T_1}}}{{({T_2} - {T_1})}}$)
Put the values in the above equation
($COP$)$_R$= $\dfrac{{282}}{{(309 - 282)}}$
Do the necessary calculation
($COP$)$_R$= $\dfrac{{282}}{{27}}$
The coefficient of performance comes out to be:
($COP$)$_R$= 10.44

The coefficient of performance is 10.44

Additional Information:
The coefficient of performance or COP is a measurement of the efficiency of refrigerators and many other heating devices. The higher the value of COP the higher would be the value of its efficiency. For example, a refrigerator that requires less work to remove a given amount of heat will possess a larger value of COP.

Note:
To measure the effectiveness of the refrigerator. We have to find a formula that relates the coefficient of performance, inside temperature, and outside temperature. Make sure to make no calculation error.