Question

If the coefficient of performance of a refrigerator is 5 and operates at the room temperature ${27^o}C$, the temperature inside the refrigerator is:
$\left( A \right)$ 240 K
$\left( B \right)$ 250 K
$\left( C \right)$ 230 K
$\left( D \right)$ 260 K

Answer 1

Hint: In this question use the concept that coefficient of performance of the refrigerator is the ratio of inside the refrigerator temperature to the temperature difference at the outside of the refrigerator both in kelvin so use this property to reach the solution of the question.

Complete Step-by-Step solution:
Given data:
Coefficient of performance of a refrigerator (C.O.P) = 5
Room temperature = ${27^o}C$
Now as we know that ${1^o}C = 273K$
$ \Rightarrow {27^o}C = 273 + 27 = 300K$
Now as we know that the coefficient of performance of the refrigerator is the ratio of inside the refrigerator temperature to the temperature difference at the outside of the refrigerator.
Let inside the refrigerator temperature = ${T_{cold}}$
And outside the refrigerator temperature = ${T_{hot}}$
$ \Rightarrow {T_{hot}} = 300K$
So the temperature difference at the outside of the refrigerator = ${T_{hot}} - {T_{cold}}$
$ \Rightarrow C.O.P = \dfrac{{{T_{cold}}}}{{{T_{hot}} - {T_{cold}}}}$
Now substitute the value in this equation we have,
\[ \Rightarrow 5 = \dfrac{{{T_{cold}}}}{{300 - {T_{cold}}}}\]
Now simplify it we have,
\[ \Rightarrow 1500 - 5{T_{cold}} = {T_{cold}}\]
$ \Rightarrow 6{T_{cold}} = 1500$
$ \Rightarrow {T_{cold}} = \dfrac{{1500}}{6} = 250K$
So this is the required temperature inside the refrigerator.
Hence option (B) is the correct answer.

Note – Whenever we face such types of questions always recall the relation between coefficient of performance of the refrigerator and the temperature between inside and outside the refrigerator which is stated above so first convert the degree Celsius into kelvin as above then substitute all the values in the formula and simplify, we will get the required answer.