Question

The coefficient of performance of a refrigerator is 5. If the temperature inside the freezer is -20°C, the temperature of the surroundings to which it rejects heat is?

Answer 1

Hint: As we know that while using a refrigerating device, it is necessary to determine how much cooling effect that the device generates as compared to the energy it is absorbing to function from the external source.

Complete step by step answer:
We have given that,
The coefficient of performance (COP) of a refrigerator $COP = 5$.
The temperature inside the freezer is ${T_L} = - 20^\circ {\text{C}}$.
We have to determine where the surrounding temperature since the heat and work applied is not given so we can assume this refrigerator as the Carnot refrigerator.
We know that $COP$ of a Carnot refrigerator is given by,
$COP = \dfrac{{{T_L}}}{{{T_H} - {T_L}}}$…… (I)
Here, ${T_L}$ is the freezer temperature, ${T_H}$ is the surrounding temperature.
We can substitute here the value of lower temperature and $COP$ to find the temperature of the surroundings. Therefore, we can substitute ${T_L} = - 20^\circ {\text{C}}$ and $COP = 5$ to find the temperature of the surroundings.
Here we can change the following temperature into the kelvin scale for proper calculation.
We know the relation between the Kelvin scale and Celsius scale is given by,
$T\left( K \right) = T\left( {^\circ {\text{C}}} \right) + 273$…… (II)
Now we can substitute ${T_L} = - 20^\circ {\text{C}}$ in equation in (II) to find the value in kelvin scale.
${T_{L1}} = \left( { - 20 + 273} \right)\,{\text{K}}$
$ \Rightarrow {T_{L1}} = 253\,{\text{K}}$
Now we can substitute ${T_{L1}} = 253\,{\text{K}}$ for ${T_L}$ in equation (I) to find the value of higher body temperature.
$
  5 = \dfrac{{253}}{{{T_H} - 253}} \\
   \Rightarrow 5{T_H} - 1265 = 253 \\
   \Rightarrow 5{T_H} = 1518 \\
   \Rightarrow {T_H} = 303.6\,{\text{K}} \\
 $
Now we can convert the kelvin scale to the Celsius scale using equation (II). So we can substitute ${T_H} = 303.6\,{\text{K}}$ in equation (II) to find the value of ${T_H}$ in Celsius scale.
$
  {T_H}^\iota = \left( {303.6 - 273} \right)^\circ {\text{C}} \\
  {T_H}^\iota = 30.6^\circ {\text{C}} \\
 $
$\therefore$ So we can say that the temperature of the surroundings to which the refrigerator rejects heat is $30.6^\circ {\text{C}}$.

Note:
We can say that here the above problem can be also seen with the help of the absolute thermodynamic scale concept for finding the heat rejected to the warmer surface or heat extracted from the colder surface. We can say that by absolute thermodynamic temperature scale concept,
$\dfrac{{{Q_H}}}{{{Q_L}}} = \dfrac{{{T_H}}}{{{T_L}}}$.
Here, ${Q_H}$ is the heat rejected to the surroundings, and ${Q_L}$ is the heat extracted from the cold surface. ${T_L}$ is the temperature of the cold region in kelvin, and ${T_H}$ is the temperature of the hot region in kelvin.