The coefficient performance of a refrigerator is 5. If the temperature inside the freezer is -20℃, calculate the heat rejected to the surrounding

The correct option is 31℃.

Let us break down the step-by-step calculation to find the temperature of the surroundings (which corresponds to the heat rejected to the surroundings in terms of temperature equivalents for a Carnot refrigerator).

First, we identify the given values from the problem:
The Coefficient of Performance (COP) of the refrigerator, $\beta =5$.
The temperature inside the freezer (lower temperature reservoir), $T_2=-20℃$.

To use thermodynamic equations, we must convert the temperature from Celsius to Kelvin:
$T_2=-20+273=253\text{K}$

The relationship between the Coefficient of Performance (COP) and the temperatures of the reservoirs for a refrigerator is given by the formula:

$\beta =\frac{T_2}{T_1-T_2}$

where $T_1$ is the temperature of the surroundings (higher temperature reservoir to which heat is rejected).

Substitute the known values into the equation:

$5=\frac{253}{T_1-253}$

Rearranging the equation to solve for $T_1$:
$5(T_1-253)=253$
$T_1-253=\frac{253}{5}$
$T_1-253=50.6$
$T_1=50.6+253=303.6\text{K}$

Now, convert the temperature of the surroundings back to Celsius:
$T_1\text{(in ℃)}=303.6-273=30.6℃$

Rounding off to the nearest integer, we get:
$T_1\approx 31℃$

Hence, the temperature of the surroundings to which heat is rejected is 31℃.